Câu hỏi của Pokiwar!!

Question

Giải phương trình sau :

a,(x+1)(x-5)-x(x-6)=3x+7

b,$\dfrac{x-2}{x+3}-\dfrac{2x-1}{x}=\dfrac{11-2x^2}{x^2+3x}$

Cheewin · Accepted Answer

a) (x2-5x+x-5)-x2+6x-3x-7=0 <=> -2x-12=0 <=> x=-6 vậy S=(-6)Câu trả lời: a) (x2-5x+x-5)-x2+6x-3x-7=0 <=> -2x-12=0 <=> x=-6 vậy S=(-6)

Cheewin · Answer

b) $\dfrac{x\left(x-2\right)-\left(2x-1\right)\left(x+3\right)}{x\left(x+3\right)}=\dfrac{11-2x^2}{x\left(x+3\right)}$ Đk: x $\ne$ 0,-3 => $x^2-2x-\left(2x^2+6x-x-3\right)=11-2x^2$ <=> $x^2-2x-2x^2-6x+x+3-11+2x^2=0$ $x^2-7x-8=0$ $\left(x-8\right)\left(x+1\right)=0$ => x=8 hoặc x=-1Câu trả lời: b) $\dfrac{x\left(x-2\right)-\left(2x-1\right)\left(x+3\right)}{x\left(x+3\right)}=\dfrac{11-2x^2}{x\left(x+3\right)}$ Đk: x $\ne$ 0,-3 => $x^2-2x-\left(2x^2+6x-x-3\right)=11-2x^2$ <=> $x^2-2x-2x^2-6x+x+3-11+2x^2=0$ $x^2-7x-8=0$ $\left(x-8\right)\left(x+1\right)=0$ => x=8 hoặc x=-1

Trần Hương Giang · Answer

a) $\left(x+1\right)\left(x-5\right)-x\left(x-6\right)=3x+7$

$\Leftrightarrow x^2+x-5x-5-x^2+6x=3x+7$

$\Leftrightarrow x^2-x^2+x-5x+6x-3x=7+5$

$\Leftrightarrow-x=12$

$\Leftrightarrow x=-12$

Vậy phương trình có tập nghiệm $S=\left\{-12\right\}$

b) $\dfrac{x-2}{x+3}-\dfrac{2x-1}{x}=\dfrac{11-2x^2}{x^2+3x}$

$\Leftrightarrow\dfrac{\left(x-2\right)x}{\left(x+3\right)x}-\dfrac{\left(2x-1\right)\left(x+3\right)}{x\left(x+3\right)}=\dfrac{11-2x^2}{x^2+3x}$

$\Leftrightarrow\dfrac{x^2-2x}{x^2+3x}-\dfrac{2x^2-x+6x-3}{x^2+3x}=\dfrac{11-2x^2}{x^2+3x}$

$\Leftrightarrow x^2-2x-2x^2-5x+3=11-2x^2$

$\Leftrightarrow x^2-2x^2+2x^2-2x-5x+3-11=0$

$\Leftrightarrow x^2-7x-8=0$

$\Leftrightarrow x^2-8x+x-8=0$

$\Leftrightarrow x\left(x-8\right)+\left(x-8\right)=0$

$\Leftrightarrow\left(x+1\right)\left(x-8\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x+1=0\x-8=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}-1\8\end{matrix}\right.$

Vậy phương trình có tập nghiệm $S=\left\{-1;8\right\}$Câu trả lời: a) $\left(x+1\right)\left(x-5\right)-x\left(x-6\right)=3x+7$