Câu hỏi của Kaijo

Question

1. giải phương trình (chú ý tim điểu kiện xác định).

a,  $\frac{2x-5}{x+5}$=3

b, $\frac{2}{x-1}$ =$\frac{6}{x+1}$

c,  $\frac{2x+1}{x-1}$ =$\frac{5\left(x-1\right)}{x+1}$

d,...

Nguyễn Lê Phước Thịnh · Accepted Answer

a) ĐKXĐ: x≠-5

Ta có: $\frac{2x-5}{x+5}=3$

$\Leftrightarrow\frac{2x-5}{x+5}-3=0$

$\Leftrightarrow\frac{2x-5}{x+5}-\frac{3\left(x+5\right)}{x+5}=0$

$\Leftrightarrow2x-5-3\left(x+5\right)=0$

$\Leftrightarrow2x-5-3x-15=0$

$\Leftrightarrow-x-20=0$

$\Leftrightarrow-x=20$

$\Leftrightarrow x=-20$(tmđk)

Vậy: x=-20

b) ĐKXĐ: x≠1;x≠-1

Ta có: $\frac{2}{x-1}=\frac{6}{x+1}$

$\Leftrightarrow\frac{2}{x-1}-\frac{6}{x+1}=0$

$\Leftrightarrow\frac{2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{6\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0$

$\Leftrightarrow2\left(x+1\right)-6\left(x-1\right)=0$

$\Leftrightarrow2x+2-6x+6=0$

$\Leftrightarrow-4x+8=0$

$\Leftrightarrow-4x=-8$

$\Leftrightarrow x=2$(tmđk)

Vậy: x=2

c) ĐKXĐ: x≠1;x≠-1

Ta có: $\frac{2x+1}{x-1}=\frac{5\left(x-1\right)}{x+1}$

$\Leftrightarrow\frac{2x+1}{x-1}-\frac{5\left(x-1\right)}{x+1}=0$

$\Leftrightarrow\frac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{5\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0$

$\Leftrightarrow\left(2x+1\right)\left(x+1\right)-5\left(x^2-2x+1\right)=0$

$\Leftrightarrow2x^2+2x+x+1-5x^2+10x-5=0$

$\Leftrightarrow-3x^2+13x-4=0$

$\Leftrightarrow-3x^2+x+12x-4=0$

$\Leftrightarrow x\left(-3x+1\right)+4\left(3x-1\right)=0$

$\Leftrightarrow x\left(1-3x\right)-4\left(1-3x\right)=0$

$\Leftrightarrow\left(1-3x\right)\left(x-4\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}1-3x=0\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\x=4\end{matrix}\right.$(thỏa mãn điều kiện)

Vậy: $x\in\left\{\frac{1}{3};4\right\}$

d) ĐKXĐ: x≠1;x≠-1

Ta có: $\frac{x}{x-1}-\frac{2x}{x^2-1}=0$

$\Leftrightarrow\frac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{2x}{\left(x-1\right)\left(x+1\right)}=0$

$\Leftrightarrow x\left(x+1\right)-2x=0$

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

$\Leftrightarrow x^2-x=0$

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

$\Leftrightarrow\left[{}\begin{matrix}x=0\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\x=1\left(ktm\right)\end{matrix}\right.$

Vậy: x=0

e) ĐKXĐ: x≠2

Ta có: $\frac{1}{x-2}+3=\frac{x-3}{2-x}$

⇔$\frac{1}{x-2}+3-\frac{x-3}{2-x}=0$

⇔$\frac{1}{x-2}+3+\frac{x-3}{x-2}=0$

⇔$\frac{1}{x-2}+\frac{3\left(x-2\right)}{x-2}+\frac{x-3}{x-2}=0$

$\Leftrightarrow1+3\left(x-2\right)+x-3=0$

$\Leftrightarrow1+3x-6+x-3=0$

$\Leftrightarrow4x-8=0$

$\Leftrightarrow4x=8$

$\Leftrightarrow x=2$(không thỏa mãn)

Vậy: x∈∅

f) ĐKXĐ: $x\ne\pm2$

Ta có: $\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}$

⇔$\frac{x+1}{x-2}+\frac{x-1}{x+2}-\frac{2\left(x^2+2\right)}{x^2-4}=0$

⇔$\frac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{2x^2+4}{\left(x+2\right)\left(x-2\right)}=0$

$\Leftrightarrow\left(x+1\right)\left(x+2\right)+\left(x-1\right)\left(x-2\right)-2x^2-4=0$

$\Leftrightarrow x^2+2x+x+2+x^2-2x-x+2-2x^2-4=0$

$\Leftrightarrow0=0$

Vậy: x∈R

g) ĐKXĐ: $x\ne\pm2$

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

⇔$\frac{x+2}{x-2}+\frac{1}{x+2}-\frac{x\left(x-5\right)}{\left(x-2\right)\left(x+2\right)}=0$

⇔$\frac{\left(x+2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x+2\right)\left(x-2\right)}-\frac{x^2-5x}{\left(x-2\right)\left(x+2\right)}=0$

⇔$\left(x+2\right)^2+x-2-x^2+5x=0$

$\Leftrightarrow x^2+4x+4+x-2-x^2+5x=0$

$\Leftrightarrow10x-2=0$

$\Leftrightarrow10x=2$

$\Leftrightarrow x=\frac{2}{10}=\frac{1}{5}$(thỏa mãn)

Vậy: $x=\frac{1}{5}$Câu trả lời: a) ĐKXĐ: x≠-5