a) 3,25 < 4;<\/p>\n\n\n\n
b) -5 > -4 1\/4;<\/p>\n\n\n\n
c) -\u221a2 \u2264 3 ?<\/p>\n\n\n\n
L\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n M\u1ec7nh \u0111\u1ec1 \u0111\u00fang l\u00e0 a) 3,25 < 4 v\u00e0 c) -\u221a2 \u2264 3<\/p>\n\n\n\n M\u1ec7nh \u0111\u1ec1 sai l\u00e0 b) -5 > -4 1\/4<\/p>\n\n\n\n Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 1 trang 74<\/strong>: Ch\u1ecdn d\u1ea5u th\u00edch h\u1ee3p (=, <, >) \u0111\u1ec3 khi \u0111i\u1ec1n v\u00e0o ch\u1ed7 tr\u1ed1ng ta \u0111\u01b0\u1ee3c m\u1ed9t m\u1ec7nh \u0111\u1ec1 \u0111\u00fang.<\/p>\n\n\n\n a) 2\u221a2 (…..) 3;<\/ins><\/p>\n\n\n\n b) 4\/3 (…..) 2\/3;<\/p>\n\n\n\n c) 3 + 2\u221a2 (…..) (1 + \u221a2)2<\/sup>;<\/p>\n\n\n\n d) a2<\/sup> + 1 (…..) 0 v\u1edbi a l\u00e0 m\u1ed9t s\u1ed1 \u0111\u00e3 cho.<\/p>\n\n\n\n L\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n a) 2\u221a2 < 3<\/p>\n\n\n\n b) 4\/3 > 2\/3<\/p>\n\n\n\n c) 3 + 2\u221a2 = (1 + \u221a2)2<\/sup><\/p>\n\n\n\n d) a2<\/sup> + 1 > 0 v\u1edbi a l\u00e0 m\u1ed9t s\u1ed1 \u0111\u00e3 cho.<\/p>\n\n\n\n Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 1 trang 75<\/strong>: Ch\u1ee9ng minh r\u1eb1ng a < b \u21d4 a \u2013 b < 0.<\/ins><\/p>\n\n\n\n L\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n a < b \u21d4 a + (-b) < b +(-b) \u21d4 a – b \u2264 0<\/p>\n\n\n\n Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 1 trang 75<\/strong>: N\u00eau v\u00ed d\u1ee5 \u00e1p d\u1ee5ng m\u1ed9t trong c\u00e1c t\u00ednh ch\u1ea5t tr\u00ean.<\/p>\n\n\n\n L\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n x < 3 \u21d4 -2x > -6<\/p>\n\n\n\n Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 1 trang 78<\/strong>: H\u00e3y ch\u1ee9ng minh h\u1ec7 qu\u1ea3 3.<\/ins><\/p>\n\n\n\n L\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n T\u1eeb b\u1ea5t \u0111\u1eb3ng th\u1ee9c C\u00f4- si:<\/p>\n\n\n\n \u221axy \u2264 (x + y)\/2 \u21d4 x + y \u2265 2\u221axy v\u1edbi x,y > 0<\/p>\n\n\n\n D\u1ea5u b\u1eb1ng x\u1ea3y ra khi x = y<\/p>\n\n\n\n Do t\u00edch ab kh\u00f4ng \u0111\u1ed5i n\u00ean 2\u221axy kh\u00f4ng \u0111\u1ed5i \u21d2 T\u1ed5ng x + y nh\u1ecf nh\u1ea5t khi v\u00e0 ch\u1ec9 khi x = y<\/p>\n\n\n\n Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 1 trang 78<\/strong>: <\/p>\n\n\n\n Nh\u1eafc l\u1ea1i \u0111\u1ecbnh ngh\u0129a gi\u00e1 tr\u1ecb tuy\u1ec7t \u0111\u1ed1i v\u00e0 t\u00ednh gi\u00e1 tr\u1ecb tuy\u1ec7t \u0111\u1ed1i c\u1ee7a c\u00e1c s\u1ed1 sau:<\/p>\n\n\n\n a) 0;<\/p>\n\n\n\n b) 1,25;<\/p>\n\n\n\n c) (-3)\/4;<\/p>\n\n\n\n d) -\u03c0.<\/p>\n\n\n\n L\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n Gi\u00e1 tr\u1ecb tuy\u1ec7t \u0111\u1ed1i c\u1ee7a m\u1ed9t s\u1ed1 l\u00e0 kho\u1ea3ng c\u00e1ch c\u1ee7a s\u1ed1 \u0111\u00f3 \u0111\u1ebfn \u0111i\u1ec3m 0 tr\u00ean tr\u1ee5c s\u1ed1 n\u1eb1m ngang.<\/p>\n\n\n\n |0| = 0; |1,25| = 1,25;<\/p>\n\n\n\n |(-3)\/4| = 3\/4;\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 |-\u03c0| = \u03c0<\/p>\n\n\n\n B\u00e0i 1 (trang 79 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Trong c\u00e1c kh\u1eb3ng \u0111\u1ecbnh sau, kh\u1eb3ng \u0111\u1ecbnh n\u00e0o \u0111\u00fang v\u1edbi m\u1ecdi gi\u00e1 tr\u1ecb c\u1ee7a x?<\/p>\n\n\n\n a) 8x > 4x ; b) 4x > 8x<\/p>\n\n\n\n c) 8x2<\/sup> > 4x2<\/sup> ; d) 8 + x > 4 + x<\/p>\n\n\n\n L\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\n a) ch\u1ec9 \u0111\u00fang khi x > 0 (hay n\u00f3i c\u00e1ch kh\u00e1c n\u1ebfu x < 0 th\u00ec a) sai)<\/p>\n\n\n\n b) ch\u1ec9 \u0111\u00fang khi x < 0<\/p>\n\n\n\n c) ch\u1ec9 \u0111\u00fang khi x \u2260 0<\/p>\n\n\n\n d) \u0111\u00fang v\u1edbi m\u1ecdi x.<\/p>\n\n\n\n V\u1eady kh\u1eb3ng \u0111\u1ecbnh d l\u00e0 \u0111\u00fang v\u1edbi m\u1ecdi gi\u00e1 tr\u1ecb c\u1ee7a x.<\/p>\n\n\n\n B\u00e0i 2 (trang 79 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Cho s\u1ed1 x > 5, s\u1ed1 n\u00e0o trong c\u00e1c s\u1ed1 sau \u0111\u00e2y l\u00e0 s\u1ed1 nh\u1ecf nh\u1ea5t?<\/p>\n\n\n\n L\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\n V\u1edbi m\u1ecdi x \u2260 0 ta lu\u00f4n c\u00f3: \n hay C < A < B. <\/p>\n\n\n\n L\u1ea1i c\u00f3 x > 5 \u21d2 x2<\/sup> > 52<\/sup> (B\u00ecnh ph\u01b0\u01a1ng hai v\u1ebf)<\/p>\n\n\n\n \u21d2 \n (Nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a b\u1ea5t \u0111\u1eb3ng th\u1ee9c v\u1edbi \n )<\/p>\n\n\n\n V\u1eady ta c\u00f3 C < A < B v\u00e0 C < A < D n\u00ean trong b\u1ed1n s\u1ed1 tr\u00ean, C l\u00e0 s\u1ed1 nh\u1ecf nh\u1ea5t. <\/p>\n\n\n\n Ki\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n + C\u1ed9ng c\u1ea3 hai v\u1ebf c\u1ee7a B\u0110T v\u1edbi m\u1ed9t s\u1ed1 b\u1ea5t k\u00ec, b\u1ea5t \u0111\u1eb3ng th\u1ee9c kh\u00f4ng \u0111\u1ed5i chi\u1ec1u<\/p>\n\n\n\n a < b \u21d4 a + c < b + c<\/p>\n\n\n\n + N\u00e2ng hai v\u1ebf c\u1ee7a b\u1ea5t \u0111\u1eb3ng th\u1ee9c l\u00ean m\u1ed9t l\u0169y th\u1eeba b\u1eadc ch\u1eb5n: <\/p>\n\n\n\n 0 < a < b \u21d4 a2n<\/sup> < b2n<\/sup> v\u1edbi m\u1ecdi n \u2208 N*.<\/p>\n\n\n\n + Nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a B\u0110T v\u1edbi m\u1ed9t s\u1ed1 d\u01b0\u01a1ng th\u00ec B\u0110T kh\u00f4ng \u0111\u1ed5i chi\u1ec1u: <\/p>\n\n\n\n \u00a0\u00a0\u00a0\u00a0\u00a0a < b \u21d4 a.c < b.c v\u1edbi m\u1ecdi c > 0.<\/p>\n\n\n\n B\u00e0i 3 (trang 79 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Cho a, b, c l\u00e0 \u0111\u1ed9 d\u00e0i ba c\u1ea1nh c\u1ee7a m\u1ed9t tam gi\u00e1c.<\/p>\n\n\n\n a) Ch\u1ee9ng minh (b – c)2<\/sup> < a2<\/sup><\/p>\n\n\n\n b) T\u1eeb \u0111\u00f3 suy ra: a2<\/sup> + b2<\/sup> + c2<\/sup> < 2(ab + bc + ca)<\/p>\n\n\n\n L\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\n a) V\u00ec a, b, c l\u00e0 \u0111\u1ed9 d\u00e0i 3 c\u1ea1nh c\u1ee7a m\u1ed9t tam gi\u00e1c <\/p>\n\n\n\n \u21d2 a + c > b v\u00e0 a + b > c (B\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c) <\/p>\n\n\n\n \u21d2 a + c \u2013 b > 0 v\u00e0 a + b \u2013 c > 0<\/p>\n\n\n\n Ta c\u00f3: (b \u2013 c)2<\/sup> < a2<\/sup> <\/p>\n\n\n\n \u21d4 a2<\/sup> \u2013 (b \u2013 c)2<\/sup> > 0<\/p>\n\n\n\n \u21d4 (a \u2013 (b \u2013 c))(a + (b \u2013 c)) > 0 <\/p>\n\n\n\n \u21d4 (a \u2013 b + c).(a + b \u2013 c) > 0 (Lu\u00f4n \u0111\u00fang v\u00ec a + c \u2013 b > 0 v\u00e0 a + b \u2013 c > 0). <\/p>\n\n\n\n V\u1eady ta c\u00f3 (b \u2013 c)2<\/sup> < a2<\/sup> (1) (\u0111pcm)<\/p>\n\n\n\n b) Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 ph\u1ea7n a) ta c\u00f3 : <\/p>\n\n\n\n ( a \u2013 b)2<\/sup> < c2<\/sup> (2) <\/p>\n\n\n\n (c \u2013 a)2<\/sup> < b2<\/sup> (3)<\/p>\n\n\n\n C\u1ed9ng ba b\u1ea5t \u0111\u1eb3ng th\u1ee9c (1), (2), (3) ta c\u00f3: <\/p>\n\n\n\n (b \u2013 c)2<\/sup> + (c \u2013 a)2<\/sup> + (a \u2013 b)2<\/sup> < a2<\/sup> + b2<\/sup> + c2<\/sup><\/p>\n\n\n\n \u21d2 b2<\/sup> \u2013 2bc + c2<\/sup> + c2<\/sup> \u2013 2ca + a2<\/sup> + a2<\/sup> \u2013 2ab + b2<\/sup> < a2<\/sup> + b2<\/sup> + c2<\/sup> <\/p>\n\n\n\n \u21d2 2(a2<\/sup> + b2<\/sup> + c2<\/sup>) \u2013 2(ab + bc + ca) < a2<\/sup> + b2<\/sup> + c2<\/sup><\/p>\n\n\n\n \u21d2 a2<\/sup> + b2<\/sup> + c2<\/sup> < 2(ab + bc + ca) (\u0111pcm). <\/p>\n\n\n\n B\u00e0i 4 (trang 79 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Ch\u1ee9ng minh r\u1eb1ng:<\/p>\n\n\n\n x3<\/sup> + y3<\/sup> \u2265 x2<\/sup>y + xy2<\/sup>, \u2200x, y \u2265 0<\/p>\n\n\n\n L\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\n Ta c\u00f3: x3 + y3 \u2265 x2<\/sup>y + xy2<\/sup><\/p>\n\n\n\n \u21d4 (x3 + y3) \u2013 (x2<\/sup>y + xy2<\/sup>) \u2265 0 <\/p>\n\n\n\n \u21d4 (x + y)(x2<\/sup> \u2013 xy + y2<\/sup>) \u2013 xy(x + y) \u2265 0<\/p>\n\n\n\n \u21d4 (x + y)(x2<\/sup> \u2013 xy + y2<\/sup> \u2013 xy) \u2265 0 <\/p>\n\n\n\n \u21d4 (x + y)(x2<\/sup> \u2013 2xy + y2<\/sup>) \u2265 0 <\/p>\n\n\n\n \u21d4 (x + y)(x \u2013 y)2<\/sup> \u2265 0 (Lu\u00f4n \u0111\u00fang v\u00ec x + y \u2265 0 ; (x \u2013 y)2<\/sup> \u2265 0)<\/p>\n\n\n\n D\u1ea5u \u00ab = \u00bb x\u1ea3y ra khi (x \u2013 y)2<\/sup> = 0 \u21d4 x = y. <\/p>\n\n\n\n Ki\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n + L\u0169y th\u1eeba b\u1eadc ch\u1eb5n c\u1ee7a m\u1ecdi s\u1ed1 lu\u00f4n \u2265 0. <\/p>\n\n\n\n \u00a0\u00a0\u00a0\u00a0\u00a0A2n<\/sup> \u2265 0 v\u1edbi m\u1ecdi A v\u00e0 n \u2208 N*<\/p>\n\n\n\n B\u00e0i 5 (trang 79 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Ch\u1ee9ng minh r\u1eb1ng:<\/p>\n\n\n\n x4<\/sup> – \u221ax5<\/sup> + x – \u221ax + 1 > 0, \u2200 x \u2265 0<\/p>\n\n\n\n L\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\n \u0110\u1eb7t t = \u221ax (\u0111i\u1ec1u ki\u1ec7n t \u2265 0), khi \u0111\u00f3 <\/p>\n\n\n\n x4<\/sup> \u2013 \u221ax5<\/sup> + x \u2013 \u221ax + 1 = (\u221ax)8<\/sup> \u2013 (\u221ax)5<\/sup> + (\u221ax)2<\/sup> \u2013 (\u221ax) + 1 = t8<\/sup> \u2013 t5<\/sup> + t2<\/sup> \u2013 t + 1<\/p>\n\n\n\n Ta c\u1ea7n ch\u1ee9ng minh : t8<\/sup> \u2013 t5<\/sup> + t2<\/sup> \u2013 t + 1 > 0 <\/p>\n\n\n\n C\u00e1ch 1 (theo h\u01b0\u1edbng d\u1eabn \u1edf \u0111\u1ec1 b\u00e0i).<\/strong> <\/p>\n\n\n\n + X\u00e9t 0 \u2264 t < 1 \u21d2 t3<\/sup> < 1 \u21d2 1 \u2013 t3<\/sup> > 0 ; 1 \u2013 t > 0<\/p>\n\n\n\n t8<\/sup> \u2013 t5<\/sup> + t2<\/sup> \u2013 t + 1 = t8<\/sup> + (t2<\/sup> \u2013 t5<\/sup>) + (1 \u2013 t) <\/p>\n\n\n\n = t8<\/sup> + t2<\/sup>.(1 \u2013 t3<\/sup>) + (1 \u2013 t)<\/p>\n\n\n\n > 0 + 0 + 0 = 0 <\/p>\n\n\n\n + X\u00e9t t \u2265 1 \u21d2 t3<\/sup> \u2265 1 \u21d2 t3<\/sup> \u2013 1 \u2265 0 v\u00e0 t \u2013 1 \u2265 0. <\/p>\n\n\n\n t8<\/sup> \u2013 t5<\/sup> + t2<\/sup> \u2013 t + 1 = t5<\/sup>.(t3<\/sup> \u2013 1) + t.(t \u2013 1) + 1 <\/p>\n\n\n\n \u2265 0 + 0 + 1 > 0<\/p>\n\n\n\n V\u1eady v\u1edbi m\u1ecdi t \u2265 0 th\u00ec t8<\/sup> \u2013 t5<\/sup> + t2<\/sup> \u2013 t + 1 > 0 hay x4<\/sup> \u2013 \u221ax5<\/sup> + x \u2013 \u221ax + 1 > 0, \u2200 x \u2265 0 (\u0111pcm) <\/p>\n\n\n\n C\u00e1ch 2:<\/strong> <\/p>\n\n\n\n 2.(t8<\/sup> \u2013 t5<\/sup> + t2<\/sup> \u2013 t + 1) = t8<\/sup> + t8<\/sup> \u2013 2t5<\/sup> + t2<\/sup> + t2<\/sup> \u2013 2t + 1 + 1<\/p>\n\n\n\n = t8<\/sup> + (t4<\/sup> \u2013 t)2<\/sup> + (t \u2013 1)2<\/sup> + 1. <\/p>\n\n\n\n \u2265 0 + 0 + 0 + 1 = 1. <\/p>\n\n\n\n (V\u00ec t8<\/sup> \u2265 0 ; (t4<\/sup> \u2013 t)2<\/sup> \u2265 0; (t \u2013 1)2<\/sup> \u2265 0) <\/p>\n\n\n\n \u21d2 t8<\/sup> \u2013 t5<\/sup> + t2<\/sup> \u2013 t + 1 \u2265 1\/2 > 0 hay x4<\/sup> \u2013 \u221ax5<\/sup> + x \u2013 \u221ax + 1 > 0, \u2200 x \u2265 0 (\u0111pcm)<\/p>\n\n\n\n B\u00e0i 6 (trang 79 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Trong m\u1eb7t\n ph\u1eb3ng t\u1ecda \u0111\u1ed9 Oxy, tr\u00ean c\u00e1c tia Ox v\u00e0 Oy l\u1ea7n l\u01b0\u1ee3t l\u1ea5y c\u00e1c \u0111i\u1ec3m A v\u00e0 B \nthay \u0111\u1ed5i sao cho \u0111\u01b0\u1eddng th\u1eb3ng AB lu\u00f4n ti\u1ebfp x\u00fac v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m O b\u00e1n \nk\u00ednh 1. X\u00e1c \u0111\u1ecbnh t\u1ecda \u0111\u1ed9 c\u1ee7a A v\u00e0 B \u0111\u1ec3 \u0111o\u1ea1n AB c\u00f3 \u0111\u1ed9 d\u00e0i nh\u1ecf nh\u1ea5t.<\/p>\n\n\n\n L\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\n G\u1ecdi ti\u1ebfp \u0111i\u1ec3m c\u1ee7a AB v\u00e0 \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m O, b\u00e1n k\u00ednh 1 l\u00e0 M, ta c\u00f3: OM \u22a5 AB.<\/p>\n\n\n\n \u0394OAB vu\u00f4ng t\u1ea1i O, c\u00f3 OM l\u00e0 \u0111\u01b0\u1eddng cao n\u00ean MA.MB = MO2<\/sup> = 1 (h\u1eb1ng s\u1ed1)<\/ins><\/p>\n\n\n\n \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c C\u00f4-si ta c\u00f3:<\/p>\n\n\n\n MA + MB \u2265 2\u221aMA.MB = 2. \u221a1 = 2 <\/p>\n\n\n\n D\u1ea5u \u00ab = \u00bb x\u1ea3y ra khi MA = MB = 1. <\/p>\n\n\n\n Khi \u0111\u00f3 OA = \u221a(MA2<\/sup> + MO2<\/sup>) = \u221a2 ; OB = \u221a(OM2<\/sup> + MB2<\/sup>) = \u221a2. <\/p>\n\n\n\n M\u00e0 A, B n\u1eb1m tr\u00ean tia Ox v\u00e0 Oy n\u00ean A(\u221a2; 0); B(0; \u221a2)<\/p>\n\n\n\n V\u1eady t\u1ecda \u0111\u1ed9 l\u00e0 A(\u221a2, 0) v\u00e0 B(0, \u221a2).<\/p>\n","protected":false},"excerpt":{"rendered":" Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 1 trang 74: Trong c\u00e1c m\u1ec7nh \u0111\u1ec1 sau, m\u1ec7nh \u0111\u1ec1 n\u00e0o \u0111\u00fang a) 3,25 < 4; b) -5 > -4 1\/4; c) -\u221a2 \u2264 3 ? L\u1eddi gi\u1ea3i M\u1ec7nh \u0111\u1ec1 \u0111\u00fang l\u00e0 a) 3,25 < 4 v\u00e0 c) -\u221a2 \u2264 3 M\u1ec7nh \u0111\u1ec1 sai l\u00e0 b) […]<\/p>\n","protected":false},"author":12,"featured_media":44333,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-2-trang-79-sgk-dai-so-10.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-2-trang-79-dai-so-10-1.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-2-trang-79-dai-so-10-4.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-2-trang-79-dai-so-10-2.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-2-trang-79-dai-so-10-3.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-6-trang-79-sgk-dai-so-10.PNG\")
<\/figure>\n\n\n\n