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PT
1
22 tháng 12 2022
a: \(\dfrac{2x^4-x^3-x^2+7x-4}{x^2+x-1}\)
\(=\dfrac{2x^4+2x^3-2x^2-3x^3-3x^2+3x+4x^2+4x-4}{x^2+x-1}\)
=2x^2-3x+4
b: \(=\dfrac{y}{x\left(2x-y\right)}+\dfrac{4x}{y\left(y-2x\right)}\)
\(=\dfrac{y^2-4x^2}{xy\left(2x-y\right)}=\dfrac{-\left(2x-y\right)\left(2x+y\right)}{xy\left(2x-y\right)}=\dfrac{-2x-y}{xy}\)
c: \(=\dfrac{6\left(x+8\right)}{7\left(x-1\right)}\cdot\dfrac{\left(x-1\right)^2}{\left(x-8\right)\left(x+8\right)}=\dfrac{6\left(x-1\right)}{7\left(x-8\right)}\)
LN
1
1 tháng 1 2022
\(=\dfrac{6\left(x-2\right)^2}{x^3y}\cdot\dfrac{x^3y^2}{x-2}=6\left(x-2\right)\cdot y\)
LT
2
26 tháng 7 2017
Ta có \(\left(\frac{1}{x^2+4x+4}-\frac{1}{x^2-4x+4}\right):\left(\frac{1}{x+2}+\frac{1}{x-2}\right)\)
\(=\frac{\left(x-2\right)^2-\left(x+2\right)^2}{\left(x-2\right)^2\left(x+2\right)^2}:\frac{x-2+x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{\left(x-2+x+2\right)\left(x-2-x-2\right)}{\left(x-2\right)^2\left(x+2\right)^2}:\frac{2x}{\left(x+2\right)\left(x-2\right)}\)
\(\frac{-4.2x}{\left(x+2\right)^2\left(x-2\right)^2}.\frac{\left(x+2\right)\left(x-2\right)}{2x}=\frac{-4}{\left(x+2\right)\left(x-2\right)}\)
D
1
13 tháng 12 2023
ĐKXĐ: \(x\notin\left\{2;-3;-4\right\}\)
\(\dfrac{x^2-5x+6}{x^2+7x+12}\cdot\dfrac{x^2+3x}{x^2-4x+4}\)
\(=\dfrac{x^2-2x-3x+6}{x^2+3x+4x+12}\cdot\dfrac{x\left(x+3\right)}{\left(x-2\right)^2}\)
\(=\dfrac{x\left(x-2\right)-3\left(x-2\right)}{x\left(x+3\right)+4\left(x+3\right)}\cdot\dfrac{x\left(x+3\right)}{\left(x-2\right)^2}\)
\(=\dfrac{\left(x-3\right)\left(x-2\right)}{\left(x+3\right)\left(x+4\right)}\cdot\dfrac{x\left(x+3\right)}{\left(x-2\right)^2}\)
\(=\dfrac{x\left(x-3\right)}{\left(x-2\right)\left(x+4\right)}\)
TK
0
26 tháng 12 2019
ĐKx\(\ne\)2,x\(\ne\)0
\(=\)\(\frac{2(x+2)+2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\):\(\frac{4x}{\left(x+2\right)^2}\)
=\(\frac{2x+4+2x-4}{\left(x-2\right)\left(x+2\right)}\)\(\frac{(x+2)^2}{4x}\)
=\(\frac{x+2}{x-2}\)
28 tháng 12 2019
\(\left(\frac{2}{x-2}+\frac{2}{x+2}\right):\frac{4x}{x^2+4x+4}\)
\(=\left(\frac{2}{x-2}+\frac{2}{x+2}\right):\frac{4x}{\left(x+2\right)^2}\)
\(=\left(\frac{2}{x-2}+\frac{2}{x+2}\right).\frac{\left(x+2\right)^2}{4x}\)
\(=\frac{4x}{x^2-4}.\frac{\left(x+2\right)^2}{4x}\)
\(=\frac{4x.\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right).4}\)
\(=\frac{x+2}{x-2}\)
\(=(4x^2+16x-20x-80+32):(x+4)\\ =[4x(x+4)-20(x+4)+32]:(x+4)\\ =4x-20(\text{dư }32)\)
giải kiểu nì hay ghe ă:>