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\(=\dfrac{a+x+1}{a+x}:\dfrac{a+x-1}{a+x}\cdot\left(\dfrac{2ax-1+a^2+x^2}{2ax}\right)\)

\(=\dfrac{a+x+1}{a+x-1}\cdot\dfrac{\left(a+x\right)^2-1}{2ax}\)

\(=\dfrac{a+x+1}{a+x-1}\cdot\dfrac{\left(a+x+1\right)\left(a+x-1\right)}{2ax}\)

\(=\dfrac{\left(a+x+1\right)^2}{2ax}\)

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

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

\(A=\left(\dfrac{x^2-2x+1}{x^2+x+1}-\dfrac{-2x^2+4x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x-1}\right):\dfrac{2x}{x^3+x}\)

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

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

8 tháng 9 2016

\(\left[\left(1+\frac{1}{x^2}\right)\div\left(1+2x+x^2\right)+\frac{2}{\left(x+1\right)^3}\times\left(1+\frac{1}{x}\right)\right]\div\frac{x-1}{x^3}\)

\(=\left[\frac{x^2+1}{x^2}\times\frac{1}{\left(x+1\right)^2}+\frac{2}{\left(x+1\right)^3}\times\frac{x+1}{x}\right]\div\frac{x-1}{x^3}\)

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

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

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

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

\(=\frac{1}{x^2}\times\frac{x^3}{x-1}\)

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

8 tháng 9 2016

e cảm ơn cj nhug bài này thầy chữa tối wa òi hehe

3 tháng 9 2016

\(A=\left(\frac{3}{\sqrt{x}-1}-\frac{\sqrt{x}-3}{x-1}\right):\left(\frac{x+2}{x+\sqrt{x}-2}-\frac{\sqrt{x}}{\sqrt{x}+2}\right)\left(ĐK:x\ge0;\ne1\right)\)

\(=\left[\frac{3}{\sqrt{x}-1}-\frac{\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]:\left[\frac{x+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}}{\sqrt{x}+2}\right]\)

\(=\frac{3\left(\sqrt{x}+1\right)-\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{3\sqrt{x}+3-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{x+2-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{2\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\)

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

15 tháng 1 2017

a/ Đặt: \(x+\frac{1}{x}=a\)

Ta có: \(x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)=a^3-3a\)

\(x^6+\frac{1}{x^6}=\left(x^3+\frac{1}{x^3}\right)^2-2=\left(\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)\right)^2-2\)

\(=\left(a^3-3a\right)^2-2\)

\(\Rightarrow M=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2}{\left(x+\frac{1}{x}\right)^3+x^3+\frac{1}{x^3}}\)

\(=\frac{a^6-\left(a^3-3a\right)^2+2-2}{a^3+a^3-3a}\)

\(=\frac{\left(a^3+a^3-3a\right)\left(a^3-a^3+3a\right)}{\left(a^3+a^3-3a\right)}=3a\)

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

b/ \(\frac{3x^2+3}{x}=3x+\frac{3}{x}\ge2.3=6\)

Đấu =  xảy ra khi \(x=\frac{1}{x}\Leftrightarrow x=1\)

16 tháng 8 2016

quy đồng H lên rồi rút gọn

sau ko rút gọn xong thì tìm x nguyên khi H=6

17 tháng 8 2016

bạn giải rõ ra giúp mình với 

Mình ngu lắm khocroi