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{"id":22867,"date":"2018-03-20T16:37:11","date_gmt":"2018-03-20T09:37:11","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=22867"},"modified":"2018-03-21T16:17:49","modified_gmt":"2018-03-21T09:17:49","slug":"dai-so-chuong-4-bai-2-can-bac-hai-cua-so-phuc-va-phuong-trinh-bac-hai","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/dai-so-chuong-4-bai-2-can-bac-hai-cua-so-phuc-va-phuong-trinh-bac-hai\/","title":{"rendered":"\u0110\u1ea1i s\u1ed1 – Ch\u01b0\u01a1ng 4 – B\u00e0i 2: C\u0103n b\u1eadc hai c\u1ee7a s\u1ed1 ph\u1ee9c v\u00e0 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai"},"content":{"rendered":"

B\u00e0i 17 (trang 195 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao)<\/b><\/span><\/p>\n

T\u00ecm c\u00e1c c\u0103n b\u1eadc hai c\u1ee7a s\u1ed1 ph\u1ee9c sau: -i; 4i; -4;1+4 \u221a3 i<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

G\u1ecdi z=x+yi l\u00e0 c\u0103n b\u1eadc hai c\u1ee7a -I, ta c\u00f3: z2<\/sup>=-i<\/p>\n

<=> (x+yi)2<\/sup>=-i <=> x2<\/sup>-y2<\/sup>+2xyi=-i<\/p>\n

\"\"<\/p>\n

T\u01b0\u01a1ng t\u1ef1, 4i c\u00f3 c\u0103n b\u1eadc hai l\u00e0 z=\u221a2+\u221a2 i v\u00e0 z=-\u221a2-\u221a2 i; -4 c\u00f3 c\u0103n b\u1eadc hai l\u00e0: z = 2i v\u00e0 z = -2i, 1+4 \u221a3 i v\u00e0 z=-2-2 \u221a3 i<\/p>\n

B\u00e0i 18 (trang 196 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao)<\/b><\/span><\/p>\n

Ch\u1ee9ng minh r\u1eb1ng n\u1ebfu z l\u00e0 c\u0103n b\u1eadc hai c\u1ee7a s\u1ed1 ph\u1ee9c w th\u00ec |z| = \u221a(|w| )<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

Gi\u1ea3 s\u1eed w=a+bi c\u00f3 m\u1ed9t c\u0103n b\u1eadc hai l\u00e0 z=x+yi<\/p>\n

V\u00ec z l\u00e0 m\u1ed9t c\u0103n b\u1eadc hai c\u1ee7a w n\u00ean z2<\/sup>=w<\/p>\n

\"\"<\/p>\n

B\u00e0i 19 (trang 196 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b><\/span><\/p>\n

Gi\u1ea3i c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai sau:<\/p>\n

a) z2<\/sup>=z+1 b) z2<\/sup>+2z+5=0 c) z2<\/sup>+(1-3i)z-2(1+i)=0<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

a) z2<\/sup>=z+1 <=> z2<\/sup>-z-1=0, c\u00f3 \u0394=5<\/p>\n

\"Gi\u1ea3i<\/p>\n

V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0<\/p>\n

\"\"<\/p>\n

b) z2<\/sup>+2z+5=0,c\u00f3 \u0394=-16=16i2<\/sup>=(4i)2<\/sup><\/p>\n

\"\"<\/p>\n

V\u1eady Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0: z1<\/sub>=-1-2i;z2<\/sub>=-1+2i<\/p>\n

c) z2<\/sup>+(1-3i)z-2(1+i)=0<\/p>\n

Ta c\u00f3 \u0394=(1-3i)2<\/sup>+8(1+i)=2i<\/p>\n

\u0394 c\u00f3 c\u0103n b\u1eadc hai l\u00e0: a1<\/sub>=1+i,a2<\/sub>=-1-i<\/p>\n

N\u00ean Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 2 nghi\u1ec7m l\u00e0:<\/p>\n

z1<\/sub>=2i;z2<\/sub>=-1+i<\/p>\n

B\u00e0i 20 (trang 196 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b><\/span><\/p>\n

a) H\u1ecfi c\u00f4ng th\u1ee9c Viet v\u1ec1 Ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai v\u1edbi h\u1ec7 s\u1ed1 th\u1ef1c c\u00f3 c\u00f2n \u0111\u00fang cho Ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai v\u1edbi h\u1ec7 s\u1ed1 ph\u1ee9c kh\u00f4ng? V\u00ec sao?<\/p>\n

b) T\u00ecm hai s\u1ed1 ph\u1ee9c, bi\u1ebf t\u1ed5ng c\u1ee7a ch\u00fang b\u1eb1ng 4 \u2013 I v\u00e0 t\u00edch c\u1ee7a ch\u00fang b\u1eb1ng 5(1 \u2013 i).<\/p>\n

c) C\u00f3 ph\u1ea3i m\u1ecdi ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai z2<\/sup>+Bx+C=0 (B, C l\u00e0 hai s\u1ed1 ph\u1ee9c nh\u1eadn hai nghi\u1ec7m l\u00e0 hai s\u1ed1 ph\u1ee9c li\u00ean h\u1ee3p kh\u00f4ng th\u1ef1c ph\u1ea3i c\u00f3 c\u00e1c h\u1ec7 s\u1ed1 B, C l\u00e0 hai s\u1ed1 th\u1ef1c? v\u00ec sao? \u0110i\u1ec1u ng\u01b0\u1ee3c l\u1ea1i c\u00f3 \u0111\u00fang kh\u00f4ng?<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

a) \u0110\u1ecbnh l\u00ed Viet v\u1eabn \u0111\u00fang cho Ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai v\u1edbi h\u1ec7 s\u1ed1 ph\u1ee9c,<\/p>\n

Gi\u1ea3 s\u1eed Ph\u01b0\u01a1ng tr\u00ecnh: Az2<\/sup>+Bz+C=0 (A \u2260 0;A,B,C \u2208C) c\u00f3 hai nghi\u1ec7m:<\/p>\n

\"\"<\/p>\n

v\u1edbi \u03b1 l\u00e0 m\u1ed9t c\u0103n b\u1eadc hai c\u1ee7a bi\u1ec7t s\u1ed1 \u0394=B2<\/sup>-4AC<\/p>\n

\"\"<\/p>\n

b) Theo \u0111\u1ecbnh l\u00ed Viet th\u00ec hai s\u1ed1 ph\u1ee9c c\u00f3 nghi\u1ec7m c\u1ee7a Ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n

z2<\/sup>-(4-i)z+5(1-i)=0 (*)<\/p>\n

Ta c\u00f3 \u0394=(4-i)2<\/sup>-20(1-i)=-5+12i<\/p>\n

\u0394 l\u00e0 m\u1ed9t c\u0103n b\u1eadc hai l\u00e0 \u03b1=2+3i, n\u00ean (*) c\u00f3 hai nghi\u1ec7m l\u00e0:<\/p>\n

\"\"<\/p>\n

V\u1eady hai s\u1ed1 c\u1ea7n t\u00ecm l\u00e0: 3+i;1-2i<\/p>\n

c) Gi\u1ea3 s\u1eed ph\u01b0\u01a1ng tr\u00ecnh: z2<\/sup>+Bz+C=0 nh\u1eadn hai nghi\u1ec7m l\u00e0 hai s\u1ed1 ph\u1ee9c li\u00ean h\u1ee3p kh\u00f4ng th\u1ee9c sau \u0111\u00e2y: z1<\/sub>=a+bi;z2<\/sub>=a-bi v\u1edbi b \u2260 0;a.b \u2208R<\/p>\n

V\u00ec z1<\/sub>;z2<\/sub>\u00a0l\u00e0 nghi\u1ec7m Ph\u01b0\u01a1ng tr\u00ecnh: z2<\/sup>+Bz+C=0 n\u00ean ta c\u00f3:<\/p>\n

\"\"<\/p>\n

C\u1ed9ng v\u1ebf v\u1edbi v\u1ebf c\u1ee7a (1) v\u00e0 (2) ta \u0111\u01b0\u1ee3c:<\/p>\n

2a2<\/sup>+2aB+2C=0 <=> a2<\/sup>+aB+C=0 => c=-(a2<\/sup>+aB)<\/p>\n

Thay v\u00e0o (1) ta \u0111\u01b0\u1ee3c:<\/p>\n

(1) <=> a2<\/sup>-b2<\/sup>+2abi+B(a+bi)-(a2<\/sup>+aB)=0<\/p>\n

<=> -b2<\/sup>+2abi+bBi=0 (v\u00ec b \u2260 0)<\/p>\n

\"\"<\/p>\n

V\u00ec b \u2260 0 n\u00ean B = -2a \u2013 bi kh\u00f4ng th\u1ec3 l\u00e0 m\u1ed9t s\u1ed1 th\u1ef1c, v\u1eadt kh\u1eb3ng \u0111\u1ecbnh: B v\u00e0 C l\u00e0 hai s\u1ed1 th\u1ef1c l\u00e0 sai.<\/p>\n

– \u0110i\u1ec1u ng\u01b0\u1ee3c l\u1ea1i. n\u1ebfu B, C l\u00e0 hai s\u1ed1 th\u1ef1c th\u00ec Ph\u01b0\u01a1ng tr\u00ecnh z2<\/sup>+Bz+C=0 nh\u1eadn hai nghi\u1ec7m s\u1ed1 ph\u1ee9c li\u00ean h\u1ee3p l\u00e0 sai, ch\u1eb3ng h\u1ea1n Ph\u01b0\u01a1ng tr\u00ecnh z2<\/sup>+2z-3=0 c\u00f3 nghi\u1ec7m l\u00e0 z = 1; z =-3<\/p>\n

 <\/p>\n

B\u00e0i 21 (trang 197 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b><\/p>\n

a) Gi\u1ea3i Ph\u01b0\u01a1ng tr\u00ecnh: (z2<\/sup>+i)(z2<\/sup>-2iz-1)=0<\/p>\n

b) T\u00ecm s\u1ed1 ph\u1ee9c B \u0111\u1ec3 Ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai z2<\/sup>\u00a0Bz+3i=0 c\u00f3 t\u1ed5ng b\u00ecnh Ph\u01b0\u01a1ng hai nghi\u1ec7m b\u1eb1ng 8.<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

a) Ph\u01b0\u01a1ng tr\u00ecnh: (z2<\/sup>+i)(z2<\/sup>-2iz-1)=0<\/p>\n

\"\"<\/p>\n

V\u1eady Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 3 nghi\u1ec7m:<\/p>\n

\"\"<\/p>\n

b) Theo b\u00e0i ra ta c\u00f3: z1<\/sub>2<\/sup>+z2<\/sub>2<\/sup>=8 <=> (z1<\/sub>+z2<\/sub>\u00a0)2<\/sup>-2z1<\/sub>\u00a0z2<\/sub>=8<\/p>\n

<=> B2<\/sup>-6i=8 (v\u00ec theo Viet z1<\/sub>+z2<\/sub>=-B;z1<\/sub>\u00a0z2<\/sub>=3i<\/p>\n

<=> B2<\/sup>=6i+8, v\u1eady B l\u00e0 c\u0103n b\u1eadc hai c\u1ee7a 6i + 8<\/p>\n

S\u1ed1 6i + 8 c\u00f3 c\u0103n b\u1eadc hai l\u00e0: 3+i v\u00e0-3-i<\/p>\n

V\u1eady B = 3 + i ho\u1eb7c B = -3 \u2013 i<\/p>\n

\u0110\u00e1p s\u1ed1: c\u00f3 hai s\u1ed1 B th\u1ecfa m\u00e3n b\u00e0i to\u00e1n.<\/p>\n

 <\/p>\n

B\u00e0i 22 (trang 197 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b><\/p>\n

\u0110\u1ed1 vui. M\u1ed9t h\u1ecdc sinh k\u00ed hi\u1ec7u m\u1ed9t c\u0103n b\u1eadc hai c\u1ee7a -1 l\u00e0 \u221a(-1) v\u00e0 t\u00ednh \u221a(-1).\u221a(-1) nh\u01b0 sau:<\/p>\n

a) T\u00ednh theo \u0111\u1ecbnh ngh\u0129a c\u1ee7a c\u0103n b\u1eadc hai l\u00e0 -1 th\u00ec \u221a(-1).\u221a(-1)=-1<\/p>\n

b) T\u00ednh theo t\u00ednh ch\u1ea5t c\u1ee7a c\u0103n b\u1eadc hai (t\u00edch c\u1ee7a hai c\u0103n b\u1eadc hai c\u1ee7a hai s\u1ed1 b\u1eb1ng c\u0103n b\u1eadc hai c\u1ee7a t\u00edch hai s\u1ed1 \u0111\u00f3) th\u00ec:<\/p>\n

\u221a(-1).\u221a(-1)=\u221a((-1)(-1) )=\u221a1=1<\/p>\n

T\u1eeb \u0111\u00f3, h\u1ecdc sinh suy ra \u2013 1 = 1. H\u00e3y t\u00ecm \u0111i\u1ec1u sai trong l\u1eadp lu\u1eadn tr\u00ean.<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

1. Tr\u01b0\u1edbc h\u1ebft kh\u00f4ng n\u00ean k\u00ed hi\u1ec7u \u221a(-1) l\u00e0 m\u1ed9t c\u0103n b\u1eadc hai c\u1ee7a -1, b\u1edfi v\u00ec trong ph\u1ea7n l\u00ed thuy\u1ebft ta \u0111\u00e3 bi\u1ebft s\u1ed1 -1 c\u00f3 d\u00f9ng c\u0103n b\u1eadc hai l\u00e0: \u221a(-(-1) ) i v\u00e0 -\u221a(-(-1) ) i. K\u00ed hi\u1ec7u \u221aa ch\u1ec9 n\u00ean ch\u1ec9: \u201cGi\u00e1 tr\u1ecb kh\u00f4ng \u00e2m c\u1ee7a c\u0103n b\u1eadc hai c\u1ee7a s\u1ed1 th\u1ef1c kh\u00f4ng \u00e2m a\u201d m\u00e0 th\u00f4i.<\/p>\n

2. Sai l\u1ea7m ch\u00ednh \u1edf \u0111i\u1ec3m b). h\u1ecdc sinh \u0111\u00f3 \u0111\u00e3 xem k\u00ed hi\u1ec7u m\u1edbi c\u1ee7a m\u00ecnh \u221a(-1) nh\u01b0 l\u00e0 c\u0103n b\u1eadc hai s\u1ed1 h\u1ecdc c\u1ee7a m\u1ed9t s\u1ed1 th\u1ef1c kh\u00f4ng \u00e2m, m\u1eb7c d\u00f9 r\u1eb1ng \u221a(-1) kh\u00f4ng ph\u1ea3i l\u00e0 m\u1ed9t s\u1ed1 th\u1ef1c. (h\u1ecdc sinh \u0111\u00f3 d\u00f9ng \u221a(-1) \u0111\u1ec3 ch\u1ec9 s\u1ed1 \u1ea3o i ho\u1eb7c s\u1ed1 \u1ea3o -i) v\u00e0 k\u00ed hi\u1ec7u m\u1edbi \u221a(-1) c\u1ee7a h\u1ecdc sinh \u0111\u00f3 c\u0169ng kh\u00f4ng c\u00f3 t\u00ednh ch\u1ea5t t\u01b0\u01a1ng t\u1ef1 nh\u01b0 t\u00ednh ch\u1ea5t c\u1ee7a \u221aa (V\u1edbi a l\u00e0 s\u1ed1 th\u1ef1c kh\u00f4ng \u00e2m) m\u00e0 b\u1eb1ng ch\u1ee9ng l\u00e0 ch\u00ednh m\u00e2u thu\u1eabn t\u00ecm \u0111\u01b0\u1ee3c trong b)<\/p>\n

3. M\u1ed9t sai l\u1ea7m n\u1eefa ph\u1ea3i nh\u1eafc \u0111\u1ebfn \u0111\u00f3 l\u00e0: t\u00ednh ch\u1ea5t trong b) \u201ct\u00edch c\u1ee7a hai c\u0103n b\u1eadc hai c\u1ee7a hai s\u1ed1 b\u1eb1ng c\u0103n b\u1eadc hai c\u1ee7a hai s\u1ed1 \u0111\u00f3\u201d l\u00e0 ph\u00e1t bi\u1ec3u sai, ch\u1eb3ng h\u1ea1n.<\/p>\n

V\u00ed d\u1ee5: s\u1ed1 2 l\u00e0 m\u1ed9t c\u0103n b\u1eadc hai c\u1ee7a 4.<\/p>\n

S\u1ed1 -3 l\u00e0 m\u1ed9t c\u0103n b\u1eadc hai c\u1ee7a s\u1ed1 9<\/p>\n

S\u1ed1 6 l\u00e0 m\u1ed9t c\u0103n b\u1eadc hai c\u1ee7a s\u1ed1 4.9<\/p>\n

Theo t\u00ednh ch\u1ea5t tr\u00ean th\u00ec:2(-3) = 6, \u0111\u01b0\u1eddng nhi\u00ean sai.<\/p>\n

V\u00ed d\u1ee5 2. S\u1ed1 I l\u00e0 m\u1ed9t c\u0103n b\u1eadc hai c\u1ee7a s\u1ed1 -1;<\/p>\n

S\u1ed1 I +1 l\u00e0 m\u1ed9t c\u0103n b\u1eadc hai c\u1ee7a 2i<\/p>\n

S\u1ed1 I \u2013 1 l\u00e0 m\u1ed9t c\u0103n b\u1eadc hai c\u1ee7a s\u1ed1 -1.2i<\/p>\n

Theo t\u00ednh ch\u1ea5t tr\u00ean th\u00ec:<\/p>\n

i(i+1)=1-i <=> -1+i=1-i. Sai b\u1ea3n ch\u1ea5t c\u1ee7a sai l\u1ea7m c\u1ee7a bi\u1ebfn \u0111\u1ed5i trong b) kh\u00f4ng ph\u1ea3i sai do sai trong 3) m\u00e0 do sai trong 2). Nh\u01b0ng sai l\u1ea7m trong 3) c\u0169ng c\u1ea7n tr\u00e1ch.<\/p>\n

4. C\u1ea7n gi\u1ea3i th\u00edch th\u00eam s\u1ef1 ph\u00e2n t\u00edch trong 2) nh\u01b0 sau:<\/p>\n

T\u00ednh ch\u1ea5t. n\u1ebfu x, y l\u00e0 c\u00e1c s\u1ed1 th\u1ef1c kh\u00f4ng \u00e2m th\u00ec: \u221ax \u221ay=\u221a(x.y) (1)<\/p>\n

Khi k\u00ed hi\u1ec7u: \u221a(-1).\u221a(-1)=\u221a((-1)(-1) )=1, ngh\u0129a l\u00e0 \u0111\u00e3 xem s\u1ed1 -1 th\u00f5a m\u00e3n t\u00edch ch\u1ea5t -1 \u2265 0<\/p>\n

Con \u0111\u01b0\u1eddng d\u1eabn \u0111\u1ebfn sai l\u1ea7m c\u1ee7a h\u1ecdc sinh \u0111\u00f3 (c\u00f3 l\u1ebd) di\u1ec5n ta nh\u01b0 s\u1ef1 ph\u00e2n t\u00edch trong 4).<\/p>\n

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