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23 tháng 12 2020

a) Điều kiện: \(x\ne\pm1\)

 \(B=\frac{x-1}{x+1}-\frac{x+1}{x-1}-\frac{4}{1-x^2}\)

\(B=\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}-\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}-\frac{-4}{\left(x-1\right).\left(x+1\right)}\)

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

\(B=\frac{-4x+4}{\left(x-1\right).\left(x+1\right)}=\frac{-4.\left(x-1\right)}{\left(x-1\right).\left(x+1\right)}=\frac{-4}{x+1}\)

b) \(x^2-x=0\Leftrightarrow x.\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

Khi  \(x=0\Leftrightarrow\frac{-4}{0-1}=\frac{-4}{-1}=4\)

Khi \(x=1\Leftrightarrow\frac{-4}{1-1}=0\)

c) \(\frac{-4}{x+1}=-3\Leftrightarrow-3.\left(x+1\right)=-4\Leftrightarrow x+1=\frac{4}{3}\Leftrightarrow x=\frac{1}{3}\)

9 tháng 6 2023

\(d,\) Để \(B< 0\) thì \(\dfrac{-4}{x+1}< 0\)

Vì \(-4< 0\Rightarrow x+1>0\) để \(\dfrac{-4}{x+1}< 0\)

Giải:

\(x+1>0\\ \Leftrightarrow x>-1\)

Vậy \(x>-1\) Thì \(B< 0\)

9 tháng 6 2023

P/s: tháo mác xanh ra đui mẹ:)

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

\(=\dfrac{x+4x+8+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)

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

\(=\dfrac{2x}{x-2}\)

24 tháng 12 2020

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

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

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

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

\(=\frac{8x-4}{\left(2x-1\right)\left(2x+1\right)}.\frac{2x+1}{x^2+2}=\frac{8x-4}{\left(2x-1\right)\left(x^2+2\right)}\)

b, Thay x = -1 ta được : \(\frac{9\left(-1\right)-4}{\left[2\left(-1\right)-1\right]\left[\left(-1\right)^2+2\right]}=-\frac{13}{-9}=\frac{13}{9}\)

5 tháng 2 2022

Đề bài là \(B=\dfrac{\left(x-1\right)^2-4}{\left(2x+1\right)^2-\left(x+2\right)^2}\) hay là \(B=\dfrac{\left(x-1\right)^2-4}{\left(2x+1\right)^2}-\left(x+2\right)^2?\)

5 tháng 2 2022

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

viết lại biểu thức 

a: ĐKXĐ: x<>1; x<>-1

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

c: Để A nguyên thì x+1-2 chia hết cho x+1

=>\(x+1\in\left\{1;-1;2;-2\right\}\)

=>\(x\in\left\{0;-2;-3\right\}\)

22 tháng 12 2020

\(P=\frac{3}{x+3}+\frac{1}{x-3}-\frac{18}{9-x^2}\)

a) Điều kiện: \(x\ne3;x\ne-3\)

b)  \(P=\frac{3}{x+3}+\frac{1}{x-3}-\frac{18}{9-x^2}\)

\(P=\frac{3.\left(x-3\right)}{\left(x+3\right).\left(x-3\right)}+\frac{x+3}{\left(x-3\right).\left(x+3\right)}-\frac{-18}{\left(x-3\right).\left(x+3\right)}\)

\(P=\frac{3x-9+x+3+18}{\left(x+3\right).\left(x-3\right)}=\frac{4x+12}{\left(x-3\right).\left(x+3\right)}=\frac{4.\left(x+3\right)}{\left(x-3\right).\left(x+3\right)}=\frac{4}{x-3}\)

c)  \(\frac{4}{x-3}=4\Leftrightarrow4=\left(x-3\right).4\Leftrightarrow4x-12=4\Leftrightarrow4x=16\Leftrightarrow x=4\)

a: ĐKXĐ: x<>1; x<>-1

b: \(P=\dfrac{x}{2\left(x-1\right)}-\dfrac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)

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

c: Để P=1/2 thì 1/2(x+1)=1/2

=>x+1=1

=>x=0