Question

Giải phương trình

a) \(\frac{\left(x+2^{ }\right)^2}{2x-3}\)-1 = \(\frac{x^2+10}{2x-3x}\)

b) \(\frac{2}{x-1}\)+ \(\frac{2x+3}{x^2+x+1}\) = \(\frac{\left(2x-1\right)\left(2x+1\right)}{x^3-1}\)

c) \(\frac{x-3}{x-2}\) + \(\frac{x-2}{x-4}\) = 1

Answer 1

c) \(\frac{x-3}{x-2}+\frac{x-2}{x-4}=1\) đặt x-2 =t " cho bé hệ số lại

ĐK : \(\left\{\begin{matrix}x\ne2\\x\ne4\end{matrix}\right.\Rightarrow\left\{\begin{matrix}t\ne0\\t\ne-2\end{matrix}\right.\)

\(\frac{t-1}{t}=\frac{t}{t-2}\Leftrightarrow\left(t-1\right)\left(t-2\right)=t^2\Leftrightarrow t^2-3t+2=t^2\Rightarrow-3t=-2\)

\(t=\frac{2}{3}\Rightarrow x=2+\frac{2}{3}=\frac{8}{3}\)

Answer 2

a) \(A=\frac{\left(x+2\right)^2}{2x-3}-1=\frac{x^2+10}{2x-3x}\) xem lại đề thấy cái mẫu VP vô duyên thế!

b) \(B=\frac{2}{x-1}+\frac{2x+3}{x^2+x+1}=\frac{\left(2x-1\right)\left(2x+1\right)}{x^3-1}\) MSC=(x^3-1)

\(B=\frac{2\left(x^2+x+1\right)+\left(2x+3\right)\left(x-1\right)-\left(4x^2-1\right)}{MSC}=\frac{\left(2x^2+2x+2\right)+\left(2x^2+x-3\right)-4x^2+1}{MSC}=0\)

\(B=0\Leftrightarrow\frac{3x}{MSC}=0=>x=0\) thảo mãn đk x khác 1

Kết luận: x=0 là nghiệm duy nhất.

Answer 3

b) \(\frac{2}{x-1}\)+\(\frac{2x+3}{x^2+x+1}\)=\(\frac{\left(2x-1\right)\left(2x+1\right)}{x^3-1}\) ĐKXĐ: x \(\ne\) 1

<=>\(\frac{2\left(x^2+x+1\right)+\left(x-1\right)\left(2x+3\right)-\left(2x-1\right)\left(2x+1_{ }\right)}{x^3-1}\)=0

<=>\(\frac{2x^2+2x+2+2x^2+3x-2x-3-4x^2+1}{x^3-1}\)=0

<=> \(\frac{3x}{x^3-1}\)=0

<=> x = 3 (TM)

Vậy pt có nghiệm là x = 3