Câu hỏi của thaonguyen

Question

1) (x+5)(x+2)-3(4x-3)=(5-x)2

2) (x+2)3-(x-2)3=12x(x-1)-8

3) 3x(12x-4)-9x(4x-3)=30

4) (12x-5)(4x-1)+(3x-7)(1-16x)=81

Nguyễn Lê Phước Thịnh · Accepted Answer

1) Ta có: $\left(x+5\right)\left(x+2\right)-3\left(4x-3\right)=\left(5-x\right)^2$

$\Leftrightarrow x^2+2x+5x+10-12x+9=25-10x+x^2$

$\Leftrightarrow x^2-5x+19-25+10x-x^2=0$

$\Leftrightarrow5x-6=0$

$\Leftrightarrow5x=6$

$\Leftrightarrow x=\frac{6}{5}$

Vậy: $x=\frac{6}{5}$

2) Ta có: $\left(x+2\right)^3-\left(x-2\right)^3=12x\left(x-1\right)-8$

$\Leftrightarrow x^3+6x^2+12x+8-\left(x^3-6x^2+12x-8\right)=12x^2-12x-8$

$\Leftrightarrow x^3+6x^2+12x+8-x^3+6x^2-12x+8-12x^2+12x+8=0$

$\Leftrightarrow12x+24=0$

$\Leftrightarrow12x=-24$

$\Leftrightarrow x=-2$

Vậy: x=-2

3) Ta có: $3x\left(12x-4\right)-9x\left(4x-3\right)=30$

$\Leftrightarrow36x^2-12x-36x^2+27x-30=0$

$\Leftrightarrow15x-30=0$

$\Leftrightarrow15x=30$

$\Leftrightarrow x=2$

Vậy: x=2

4) Ta có: $\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81$

$\Leftrightarrow48x^2-12x-20x+5+3x-48x^2-7+112x-81=0$

$\Leftrightarrow83x-83=0$

$\Leftrightarrow83x=83$

$\Leftrightarrow x=1$

Vậy: x=1Câu trả lời: 1) Ta có: $\left(x+5\right)\left(x+2\right)-3\left(4x-3\right)=\left(5-x\right)^2$