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

(1)由原方程,得4x2=25,∴x2=254,∴x=±254,即x=±52;(2)由原方程,得(x-1)(2x+1)=0,∴x-1=0或2x+1=0,∴x=1,或x=-12;(3)由原方程,得x2-4x+3=0,∴(x-1)(x-3)=0,∴x-1=0或x-3=0,∴x=1或x=3;(4)由原方程,得2x2-6x+1=0,∵该方...

(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)去分母得:x-1+2(x+1)=4,去括号得:x-1+2x+2=4,移项合并得:3x=3,解得:x=1,经检验x=1是增根,原分式方程无解;(2)去分母得:2-x-1=x-3,移项合并得:-2x=-4,解得:x=2,经检验x=2是原分式方程的解.

原式=(x?2)2x(x+1)÷?(x2?4)x+1+1x+2=-(x?2)2x(x+1)×x+1(x+2)(x?2)+1x+2=-x?2x(x+1)+1x+2=2x2+2x∵x为方程x2+2x-1=0的解,∴x2+2x=1,∴原式=2.

(1)根据平方差公式得,4x2-9=0,(2x-3)(2x+3)=0,∴2x-3=0或2x+3=0,∴x1=32或x2=-32;(2)x(x-1)=2(x-1),移项得,x(x-1)-2(x-1)=0,提公因式得,(x-2)(x-1)=0,x-2=0或x-1=0,∴x1=2或x2=-1;(3)根据十字相乘法得,x2-4x+3=...

(1)4x2-25=0,x2=254,∴x1=52,x2=-52;(2)(x-2)2=3x(x-2),(x-2)2-3x(x-2)=0,(x-2)(x-2-3x)=0,(x-2)(2x+2)=0,∴x-2=0,2x+2=0,解得x1=2,x2=-1;(3)x2+3=4x,x2-4x+3=0,(x-1)(x-3)=0,∴x-1=0,x-3=0,解得x1=1,...

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

(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...

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(x+1)=0,x=0或x+1=0,∴x1=0,x2=-1;(2)(2x+11)(2x-11)=0,2x+11=0或2x-11=0,∴x1=-112,x2=112;(3)(2x+1)(3x-2)=0,2x+1=0或3x-2=0,∴x1=-12,x2=23;(4)(x-4+5-2x)(x-4-5+2x)=0,(1-x)(3x-9)=0,1-x=0或3x-9=...

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