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解方程:(1)(2x-1)2=2(1-2x)(2)4x2-3x-1=0...

(1)x2-2x-2=0, x2-2x=2,x2-2x+1=3,(x-1)2=3,∴x=1±3,∴x1=1+3,x2=1-3,(2)(x-3)2=4x(x-3),(x-3)2-4x(x-3)=0,(x-3)(x-3-4x)=0,∴x-3=0,x-3-4x=0,∴x1=3,x2=-1,(3)4x2-8x-1=0(用配方法) 4x2-8x=1,x2-2x=14,x2-2x...

(1)原方程可变形为(x+2)(x+2-3)=0,(x+2)(x-1)=0.x+2=0或x-1=0.∴x1=-2,x2=1.(2)原方程可变形为(3x+2-2x)(3x+2+2x)=0,即(x+2)(5x+2)=0.x+2=0或5x+2=0.∴x1=-2,x2=-25.(3)原方程可变形为(2x-1)(5+x+3)=0,即(2...

X1+2X2+3X3=1 X1-2X3=-1 2式-1式得 3X1+4X2+3X3=-1 X1+2X2+3X3=1 X1-2X3=-1 X1-3X3=-3 3式-2*1式得 X1+2X2+3X3=1 X3=2 2式-3式得 X1=3 X2=4

解: 系数矩阵A= 4 -2 -2 -3 4 -1 -2 -2 4 r1-2r3,r2-r3 0 -6 6 -1 6 -5 -2 -2 4 r1*(-1/6), r3-2r2 0 1 -1 -1 6 -5 0 -14 14 r2*(-1), r2+6r1, r3+14r1 0 1 -1 1 0 -1 0 0 0 r1r2 1 0 -1 0 1 -1 0 0 0 以通解为 c(1,1,1)'. c为任意常数. 满意请...

(1)(x+2)2-36=0,(x+2+6)(x+2-6)=0,x+2+6=0,x+2-6=0,x1=-8,x2=4;(2)4x2-3x-1=0,(4x+1)(x-1)=0,4x+1=0,x-1=0,x1=-14,x2=1;(3)2(x-3)2=x2-9,2(x-3)2-(x+3)(x-3)=0,(x-3)[2(x-3)-(x+3)]=0,(x-3)(x-...

3X1+2X2=5 X1 -4X2= -3 所以得到 系数行列式D=3*(-4) -2*1= -14 D1=5*(-4) -2*(-3)= -14 D2=3*(-3) -5*1= -14 于是解得 x1=D1/D= 1 x2=D2/D= 1

(1)(2x+1)2=3,2x+1=±3,解得:x1=?1+32,x2=?1?32;(2)x2+2x-24=0,(x+6)(x-4)=0,解得:x1=-6,x2=4;(3)2x2-4x+5=0∵b 2-4ac=16-4×2×5=-24<0,∴此方程没有实数根;(4)4x2+4x+10=1-8x,整理得:4x2+12x+9=0,(2x+3) 2=0,解得...

(1)分解因式得:(x+2)(x-4)=0,可得x+2=0或x-4=0,解得:x1=-2,x2=4;(2)方程移项得:4x(5x-3)-3(5x-3)=0,分解因式得:(4x-3)(5x-3)=0,解得:x1=34,x2=35;(3)分解因式得:(3x+2+2x)(3x+2-2x)=0,解得:x1=-25,x2=-2...

(1)196x2-1=0,整理,得(14x+1)(14x-1)=0,令14x+1=0或14x-1=0,解,得x1=-114,x2=114.(2)4x2+12x+9=81整理,得x2+3x-18=0,即:(x+6)(x-3)=0令x+6=0或x-3=0,解,得x1=-6,x2=3.(3)x2-7x-1=0即:x=7±72+42=7±532,∴x1=7+532,...

写出增广矩阵为 2 -4 5 3 1 3 -6 4 2 2 4 -8 3 1 K r2-1.5r1,r3-2r1 ~ 2 -4 5 3 1 0 0 -3.5 -2.5 0.5 0 0 -7 -5 K-2 r2*2, r3-r2 ~ 2 -4 5 3 1 0 0 -7 -5 1 0 0 0 0 K-3 若方程有解,显然k-3=0即k=3 此时增广矩阵为 2 -4 5 3 1 0 0 -7 -5 1 0 0...

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