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24 tháng 12 2020

a, + b, \(A=\frac{x+2}{x-3}+\frac{2x-1}{x-1}-\frac{2x-1}{2x+1}\)DKXD : \(x\ne3;1;-\frac{1}{2}\)

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

\(=\frac{2x^3+3x^2-3x-2+4x^3-12x^2-x+4-2x^3+9x^2-10x+3}{MTC}\)

\(=\frac{4x^3-14x+2x^3+5}{MTC}\)

Đề sai ko kiểm tra lại hộ nhé !!! 

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 : \(\frac{9\left(-1\right)-4}{\left[2\left(-1\right)-1\right]\left[\left(-1\right)^2+2\right]}=-\frac{13}{-9}=\frac{13}{9}\)

21 tháng 12 2021

a: \(M=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)

21 tháng 12 2021

câu b c d e đâu anh ơi

 

a: Khi x=3 thì \(A=\dfrac{3+2}{3-1}=\dfrac{5}{2}\)

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

\(P=A:B=\dfrac{x+2}{x-1}\cdot\dfrac{x+1}{x+2}=\dfrac{x+1}{x-1}\)

3: Để P>1/3 thì \(P-\dfrac{1}{3}>0\)

=>\(\Leftrightarrow3\left(x+1\right)-x+1>0\)

=>3x+3-x+1>0

=>2x+4>0

hay x>-2

5 tháng 3 2022

a, ĐKXĐ:\(\left\{{}\begin{matrix}2x-2\ne0\\2-2x^2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1\ne0\\1-x^2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x^2\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne\pm1\end{matrix}\right.\Leftrightarrow x\ne\pm1\)

b, \(C=\dfrac{x}{2x-2}+\dfrac{x^2+1}{2-2x^2}\)

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

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

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

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

\(\Rightarrow C=\dfrac{1}{2\left(x+1\right)}\)

c, \(C=\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{2\left(x+1\right)}=\dfrac{1}{2}\\ \Rightarrow\dfrac{1}{x+1}=1\\ \Rightarrow x+1=1\\ \Rightarrow x=0\)

 

a: ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

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

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

c: Để C=1/2 thì 2x+2=2

hay x=0