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8 tháng 7 2019

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

A=\(\frac{1}{x\left(x+1\right)}\)x \(\frac{\left(x-1\right)\left(x+1\right)}{2\left(x-1\right)\left(x-1\right)}\)

A=\(\frac{1}{2x^2-2x}\)

B=\(\frac{x+1}{x-2}\)-\(\frac{2x}{x+2}\)-\(\frac{2+5x}{x^2-4}\)ĐKXĐ : x\(\ne\)2, -2

B=\(\frac{x+1}{x-2_{ }}\)-\(\frac{2x}{x+2}\)-\(\frac{2+5x}{\left(x-2\right)\left(x+2\right)}\)

B=\(\frac{x^2+3x+2}{\left(x-2\right)\left(x+2\right)}\)-\(\frac{2x^2-4x}{\left(x-2\right)\left(x+2\right)}\)-\(\frac{2+5x}{\left(x-2\right)\left(x+2\right)}\)

B=\(\frac{-x^2+2x}{\left(x-2\right)\left(x+2\right)}\)

B=\(\frac{-x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

B=\(\frac{-x}{x+2}\)

b: \(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)

\(=\dfrac{\left(x+2\right)\left(x+3\right)+\left(x+1\right)\left(x+3\right)+\left(x+2\right)\left(x+1\right)}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)

\(=\dfrac{x^2+5x+6+x^2+4x+3+x^2+3x+2}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)

\(=\dfrac{3x^2+12x+11}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)

13 tháng 12 2016

ban nen tu tinh se tot hon

 

14 tháng 12 2016

mk tính mãi k ra mới hỏi chứ khocroi sắp thi hkì r chỉ jùm vs ik

31 tháng 10 2019

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

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

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

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

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

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

31 tháng 10 2019

b) \(Q=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{5x-5x}{2x\left(x+5\right)}\)

\(=\frac{x\left(x^2+2x\right)}{2x\left(x+5\right)}+\frac{2\left(x-5\right)\left(x+5\right)}{2x\left(x+5\right)}+\frac{50-5x}{2x\left(x+5\right)}\)

\(=\frac{x^3+2x^2+2\left(x^2-25\right)+50-5x}{2x\left(x+5\right)}\)

\(=\frac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)

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

\(=\frac{x^3-x^2+5x^2-5x}{2x\left(x+5\right)}\)

\(=\frac{x^2\left(x-1\right)+5x\left(x-1\right)}{2x\left(x+5\right)}\)

\(=\frac{\left(x-1\right)\left(x^2+5x\right)}{2x\left(x+5\right)}\)

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

\(=\frac{x-1}{2}\)

19 tháng 4 2020
https://i.imgur.com/wgXaoMx.jpg
16 tháng 8 2019

\(a,\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)\(\Leftrightarrow\frac{x^2+3x+2+x^2-3x+2}{x^2-4}=\frac{2\left(x^2+2\right)}{x^2-4}\)

\(\Leftrightarrow2\left(x^2+2\right)=2\left(x^2+2\right)\)(luôn đúng)

Vậy pt có vô số nghiệm

\(b,\Leftrightarrow\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\)

\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left(2x+3-x+5\right)=0\)\(\Leftrightarrow\left(\frac{-4x+10}{2-7x}\right)\left(x+8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}-4x+10=0\\x+8=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-8\end{cases}}\)

Mấy câu rút gọn bạn quy đồng nha

16 tháng 8 2019

bạn có thể giải ra giúp mik đc ko?