Toán Lớp 8: Giải các phương trình sau 1, 9x^2 -4-(3x+2)×(x-1)=0 2, (x+3)×(2x+3)=4x^2-9

Question

Toán Lớp 8:  Giải các phương trình sau 1,   9x^2 -4-(3x+2)×(x-1)=0 2,  (x+3)×(2x+3)=4x^2-9

thudan · Answer

Giải đáp:   1)S={-2/3;1/2}   2)S={-3/2;6}   Lời giải và giải thích chi tiết:   1)9x&sup2;-4-(3x+2)(x-1)=0   &hArr;(3x+2)(3x-2)-(3x+2)(x-1)=0   &hArr;(3x+2)[(3x-2)-(x-1)]=0   &hArr;(3x+2)(3x-2-x+1)=0   &hArr;(3x+2)(2x-1)=0   &hArr; ( left[  begin{array}{l}3x+2=0  2x-1=0 end{array}  right. )   &hArr; ( left[  begin{array}{l}3x=-2  2x=1 end{array}  right. )   &hArr; ( left[  begin{array}{l}x=- dfrac{2}{3}  x= dfrac{1}{2} end{array}  right. )   Vậy S={-2/3;1/2}   2)(x+3)(2x+3)=4x&sup2;-9   &hArr;(x+3)(2x+3)=(2x+3)(2x-3)   &hArr;(x+3)(2x+3)-(2x+3)(2x-3)=0   &hArr;(2x+3)[(x+3)-(2x-3)]=0   &hArr;(2x+3)(x+3-2x+3)=0   &hArr;(2x+3)(-x+6)=0   &hArr; ( left[  begin{array}{l}2x+3=0  -x+6=0 end{array}  right. )   &hArr; ( left[  begin{array}{l}2x=-3  -x=-6 end{array}  right. )   &hArr; ( left[  begin{array}{l}x=- dfrac{3}{2}  x=6 end{array}  right. )    Vậy S={-3/2;6}

thanhthuythu · Answer

1)     9x^2-4-(3x+2)(x-1)=0     =>9x^2-4-3x^2+x+2=0     =>6x^2+x-2=0     =>6x^2+4x-3x-2=0     =>2x(3x+2)-(3x+2)=0     =>(2x-1)(3x+2)=0     =>2x-1=0 hoặc 3x+2=0     =>2x=1 hoặc 3x=-2     =>x=1/2 hoặc x=-2/3     Vậy S={1/2;-2/3}     2)     (x+3)(2x+3)=4x^2-9     =>2x^2+9x+9-4x^2+9=0     =>-2x^2+9x+18=0     =>2x^2-9x-18=0     =>2x^2+3x-12x-18=0     =>x(2x+3)-6(2x+3)=0     =>(x-6)(2x+3)=0     =>x-6=0 hoặc 2x+3=0     =>x=6 hoặc 2x=-3     =>x=6 hoặc x=-3/2     Vậy S={6;-3/2}