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1 tháng 2 2019

làm a thôi nha :D 

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

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

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

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

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

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

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

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

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

1 tháng 2 2019

a)\(C=\left[\frac{x.\left(x+1\right)}{\left(x-1\right)^2}\right]:\left[\frac{x+1}{x}-\frac{1}{-\left(x-1\right)}+\frac{-x^2+2}{x.\left(x-1\right)}\right]\)

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

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

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

b)\(\text{Để B nguyên }\Rightarrow x^2⋮x-1\)

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

\(\Rightarrow\text{Để }x^2⋮x-1\Rightarrow1⋮x-1\Rightarrow x-1\inƯ\left(1\right)=\left\{\pm1\right\}\Rightarrow x\in\left\{2;0\right\}\)

11 tháng 2 2019

a) ĐKXĐ: \(x\ne-1;0;1.\)Ta có:

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

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

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

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

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

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

Vậy \(A=\frac{x}{x-1}\)với \(x\ne-1;0;1.\)

b) A < 1 \(\Leftrightarrow\frac{x}{x-1}< 1\Leftrightarrow\frac{x}{x-1}-1< 0\Leftrightarrow\frac{x}{x-1}-\frac{x-1}{x-1}< 0\)\(\Leftrightarrow\frac{1}{x-1}< 0\)

\(\Leftrightarrow x-1< 0\)(do 1 > 0)\(\Leftrightarrow x< 1.\)

Kết hợp ĐKXĐ, A < 1 khi \(x< 1\)và \(x\ne-1;0.\)

c) \(A\inℤ\Leftrightarrow\frac{x}{x-1}\inℤ.\)Mà \(x\inℤ\)\(\Rightarrow x⋮\left(x-1\right)\Rightarrow\left(x-1+1\right)⋮\left(x-1\right)\Rightarrow1⋮\left(x-1\right)\Rightarrow\left(x-1\right)\inƯ\left(1\right)=\left\{1;-1\right\}.\)Ta lập bảng sau:

\(x-1\)1-1
\(x\)20
Kết luậnx thoả mãn ĐKXĐx không thoả mãn ĐKXĐ

Vậy để A nguyên thì x = 2.

9 tháng 5 2021

a) ĐKXĐ : x \(\ne-2;x\ne1;x\ne0\)

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

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

b) Để A > 1 

=> \(\frac{x-2}{x-1}>1\)

=> \(\frac{x-2}{x-1}-1>0\Rightarrow\frac{-1}{x-1}>0\Rightarrow x-1< 0\Rightarrow x< 1\)

Vậy để A > 1 thì x < 1 và x \(\ne-2;x\ne1;x\ne0\)

c) Ta có \(A=\frac{x-2}{x-1}=\frac{x-1-1}{x-1}=1-\frac{1}{x-1}\)

Để A \(\inℤ\Rightarrow\frac{1}{x-1}\inℤ\Rightarrow1⋮x-1\Rightarrow x-1\inƯ\left(1\right)\Rightarrow x-1\in\left\{1;-1\right\}\)

Khi x - 1 = 1 => x = 2( tm)

Khi x - 1 =-1 => x = 0 (loại) 

Vậy x = 2 thì A nguyên

21 tháng 4 2020

a) Ta có :A = \(\left(\frac{\left(x-1\right)^2}{3x+\left(x-1\right)^2}-\frac{1-2x^2+4x}{x^3-1}+\frac{1}{x-1}\right):\frac{x^2+x}{x^3+x}\)

ĐK: \(\hept{\begin{cases}x\ne0\\x\ne1\end{cases}}\)

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

    \(\frac{\left(x-1\right)^3-1+2x^2-4x+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{x+1}\)

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

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

b) Để A > - 1 <=> \(\frac{x^2+1}{x+1}>-1\)

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

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

Vì x2 + x + 2 >0 \(\forall x\)

=> A > 0 <=> x + 1 > 0 <=> x > -1