Question

gải pt:

A = \(\frac{2x}{x-1}+\frac{4}{x^2+2x-3}=\frac{2x-5}{x+3}\)

Answer 1

\(ĐKXĐ:\hept{\begin{cases}x\ne1\\x\ne-3\end{cases}}\)

\(A=\frac{2x}{x-1}+\frac{4}{x^2+2x-3}=\frac{2x-5}{x+3}\)

\(\Leftrightarrow\frac{2x}{x-1}+\frac{4}{\left(x-1\right)\left(x+3\right)}-\frac{2x-5}{x+3}=0\)

\(\Leftrightarrow\frac{2x\left(x+3\right)+4-\left(2x-5\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}=0\)

\(\Leftrightarrow2x^2+6x+4-2x^2+7x-5=0\)

\(\Leftrightarrow13x-1=0\)

\(\Leftrightarrow x=\frac{1}{13}\)

Vậy tập nghiệm của phương trình là \(S=\left\{\frac{1}{13}\right\}\)

Answer 2

S=1/3 nha!

Answer 3

trả lời

cậu chỉ cần dùng phương trình bậc 2 

hoặc  tìm x  ẩn 

hok tốt

Answer 4

ĐKXĐ: \(\hept{\begin{cases}x\ne1\\x\ne-3\end{cases}}\)

\(A=\frac{2x}{x-1}+\frac{4}{x^2+2x-3}=\frac{2x-5}{x+3}\)

\(\Leftrightarrow A=\frac{2x}{x-1}+\frac{4}{x^2+2x-3}-\frac{2x-5}{x+3}=0\)

\(\Leftrightarrow A=\frac{2x}{x-1}+\frac{4}{x^2-x+3x-3}-\frac{2x-5}{x+3}=0\)

\(\Leftrightarrow A=\frac{2x}{x-1}+\frac{4}{x\left(x-1\right)+3\left(x-1\right)}-\frac{2x-5}{x+3}=0\)

\(\Leftrightarrow A=\frac{2x}{x-1}+\frac{4}{\left(x-1\right)\left(x+3\right)}-\frac{2x-5}{x+3}=0\)

\(\Leftrightarrow A=\frac{2x\left(x+3\right)+4-\left(x-1\right)\left(2x-5\right)}{\left(x-1\right)\left(x+3\right)}=0\)

\(\Leftrightarrow A=\frac{2x^2+6x+4-\left(2x^2-5x-2x+5\right)}{\left(x-1\right)\left(x+3\right)}=0\)

\(\Leftrightarrow A=\frac{13x-1}{\left(x-1\right)\left(x+3\right)}=0\)

\(\Leftrightarrow13x-1=0\)

\(\Leftrightarrow x=\frac{1}{13}\)( Tm ĐKXĐ )