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2 tháng 8 2021
\(ĐK:x\ne\pm3\)
\(P=\left[\frac{\left(2x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3-10x}{\left(x-3\right)\left(x+3\right)}\right]\cdot\frac{x-3}{x+2}\)
\(=\frac{2x^2-7x+3+x^2+3x-3+10x}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x-3}{x+2}\)
\(=\frac{3x^2+6x}{x+3}\cdot\frac{1}{x+2}=\frac{3x\left(x+2\right)}{\left(x+3\right)\left(x+2\right)}=\frac{3x}{x+3}\)
7 tháng 12 2014
1/(x^2+6x+9)-1/(x^2-6x+9)=(x-3)/(x-3)(x+3)-(x+3)/(x-3)(x+3)= -6/(x-3)(x+3)
1/(x+3)+1/(x-3)=
KG
2
6 tháng 5 2019
ĐKXĐ \(\hept{\begin{cases}x\ne3\\x\ne-3\\x\ne0\end{cases}}\)
\(A=\left(\frac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\right).\frac{x+3}{x}\)
\(=\frac{x^2+x}{x^2-3x}\)
6 tháng 5 2019
Ta có: A = \(\left(\frac{x^2-3}{x^2-9}+\frac{1}{x-3}\right):\frac{x}{x+3}\)
\(\Leftrightarrow\) A = \(\left(\frac{x^2-3}{\left(x-3\right)\left(x+3\right)}+\frac{1}{x-3}\right):\frac{x}{x+3}\)
\(\Leftrightarrow\) A = \(\left(\frac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\right).\frac{x+3}{x}\)
\(\Leftrightarrow\) A = \(\frac{x^2+x}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{x}\)
\(\Leftrightarrow\) A = \(\frac{x^2+x}{x^2-3x}\)
5 tháng 8 2019
Mk chỉ cần câu trả lời thôi nên đừng vào cmt bậy nha các bác ;)))
20 tháng 7 2020
\(B=\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}+\frac{x-1}{3+x}\right)\div\left(1-\frac{1}{x+3}\right)\)
\(B=\left(\frac{21}{x^2-9}+\frac{x-4}{x-3}+\frac{x-1}{x+3}\right)\div\left(\frac{x+3}{x+3}-\frac{1}{x+3}\right)\)
\(B=\left(\frac{21}{\left(x+3\right)\left(x-3\right)}+\frac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\right)\div\frac{x+2}{x+3}\)
\(B=\left(\frac{21}{\left(x+3\right)\left(x-3\right)}+\frac{x^2-x-12}{\left(x+3\right)\left(x-3\right)}+\frac{x^2-4x+3}{\left(x+3\right)\left(x-3\right)}\right)\cdot\frac{x+3}{x+2}\)
\(B=\left(\frac{21+x^2-x-12+x^2-4x+3}{\left(x+3\right)\left(x-3\right)}\right)\cdot\frac{x+3}{x+2}\)
\(B=\frac{2x^2-5x+12}{\left(x+3\right)\left(x-3\right)}\cdot\frac{x+3}{\left(x+2\right)}\)
\(B=\frac{2x^2-5x+12}{\left(x-3\right)\left(x+2\right)}\)
\(B=\frac{2x^2-5x+12}{x^2-x-6}\)
Đến đây là chịu ạ :(
KN
1
26 tháng 6 2017
\(C=\frac{1+x}{3-x}-\frac{1-2x}{3+x}-\frac{x\left(1-x\right)}{9-x^2}\)
\(C=\left(1+x\right)\left(3+x\right)-\left(1-2x\right)\left(3-x\right)-x\left(1-x\right)\)
\(C=3+4x+x^2-\left(3-5x+2x^2\right)-x+x^2\)
\(C=2x^2+3x+3-3+5x-2x^2\)
\(C=8x\)
T
4 tháng 1 2019
a) ĐKXĐ: \(x\ne\pm1\)
\(A=\left(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}\right):\left(\frac{1-x}{\left(1+x\right)\left(1-x\right)}-\frac{x\left(1+x\right)}{\left(1-x\right)\left(1+x\right)}+\frac{x}{x^2-1}\right)\)
\(=\frac{4x-1}{x^2-1}:\left(\frac{-x^2-2x+1}{1-x^2}-\frac{x}{1-x^2}\right)=\frac{4x-1}{x^2-1}:\frac{-x^2-3x+1}{1-x^2}\)
\(=\frac{1-4x}{1-x^2}:\frac{-x^2-3x+1}{1-x^2}=\frac{\left(1-4x\right)\left(1-x^2\right)}{\left(1-x^2\right)\left(-x^2-3x+1\right)}\)
\(=\frac{1-4x}{-x^2-3x+1}=\frac{4x-1}{x^2+3x-1}\) (chắc hết rút gọn được rồi)
30 tháng 4 2019
\(B=\left(\frac{21}{x^2-9}+\frac{\left(x-4\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right):\frac{x+2}{x+3}\)
\(B=\frac{2x^2-5x+12}{x^2-9}\cdot\frac{x+3}{x+2}\)
\(B=\frac{2x^2-5x-12}{\left(x-3\right)\left(x+2\right)}\)
\(B=\frac{2x^2-5x+12}{x^2-x-6}\)
Thik thì tách tiếp nha
\(A=\frac{x^2-x}{x^2-9}-\frac{1}{x-3}+\frac{1}{x+3}\)
\(=\frac{x\left(x-1\right)}{\left(x-3\right)\left(x+3\right)}-\frac{1}{x-3}+\frac{1}{x+3}\)
\(=\frac{x\left(x-1\right)-x-3+x-3}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x^2+x-6}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{\left(x-2\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x-2}{x-3}\)
\(A=\frac{x^2-x}{x^2-9}-\frac{1}{x-3}+\frac{1}{x+3}\)
\(A=\frac{x^2-x}{x^2-3^2}-\frac{1}{x-3}+\frac{1}{x+3}\)
\(A=\frac{x^2-x}{\left(x-3\right)\left(x+3\right)}-\frac{x+3}{\left(x-3\right)\left(x+3\right)}+\frac{x-3}{\left(x-3\right)\left(x-3\right)}\)
\(A=\frac{x^2-x-\left(x+3\right)+x-3}{\left(x-3\right)\left(x+3\right)}\)
\(A=\frac{x^2-x-x-3+x-3}{\left(x-3\right)\left(x+3\right)}\)
\(A=\frac{x^2-x}{\left(x-3\right)\left(x+3\right)}\)