Cho bt A=(x/x^2-4+1/x+2-2/x-2):(2-x+6/x+2) a)rut goc A b) tinh gt cua A khi x+-4
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BT
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29 tháng 6 2022
Bài 2:
a: ĐKXĐ: \(x\notin\left\{0;2;-2;3\right\}\)\(A=\left(\dfrac{-\left(x+2\right)}{x-2}-\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\right):\dfrac{x\left(x-3\right)}{x^2\left(2-x\right)}\)
\(=\dfrac{-x^2-4x-4-4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-x\left(x-2\right)}{x-3}\)
\(=\dfrac{-4x^2-8x}{\left(x+2\right)}\cdot\dfrac{-x}{x-3}\)
\(=\dfrac{-4x\left(x+2\right)}{x+2}\cdot\dfrac{-x}{x-3}=\dfrac{4x^2}{x-3}\)
b: Để A>0 thì x-3>0
hay x>3
20 tháng 7 2022
a: \(B=\dfrac{10x}{\left(x+4\right)\left(x-1\right)}-\dfrac{2x-3}{x+4}-\dfrac{x+1}{x-1}\)
\(=\dfrac{10x-\left(2x^2-2x-3x+3\right)-\left(x^2+5x+4\right)}{\left(x+4\right)\left(x-1\right)}\)
\(=\dfrac{10x-2x^2+5x-3-x^2-5x-4}{\left(x+4\right)\left(x-1\right)}\)
\(=\dfrac{-3x^2+10x-7}{\left(x+4\right)\left(x-1\right)}\)
\(=\dfrac{-\left(3x^2-10x+7\right)}{\left(x-1\right)\left(x+4\right)}=-\dfrac{\left(x-1\right)\left(3x-7\right)}{\left(x-1\right)\left(x+4\right)}\)
\(=\dfrac{-3x+7}{x+4}\)
b: \(B+3=\dfrac{-3x+7+3x+12}{x+4}=\dfrac{19}{x+4}>0\)
=>B>-3
9 tháng 1 2018
dkxd \(\hept{\begin{cases}\\\end{cases}}x-2=0;x+2=0\Leftrightarrow\hept{\begin{cases}\\\end{cases}x=+2;x=-2}\)
b/ \(\frac{x^2}{x^2-4}-\frac{x}{x+2}-\frac{2}{x-2}=\frac{x^2}{\left(x-2\right).\left(x+2\right)}-\frac{x.\left(x-2\right)}{\left(x+2\right).\left(x-2\right)}-\frac{2.\left(x+2\right)}{\left(x-2\right).\left(x+2\right)}\)
\(\frac{x^2-x^2-2x-2x+4}{\left(x-2\right).\left(x+2\right)}=\frac{4}{\left(x-2\right)\left(x+2\right)}\)
tới khúc này bí rồi ^^
9 tháng 1 2018
a,ĐKXĐ của A là:\(x\ne+2;-2\)
b,\(\frac{x^2-x^2+2x-2x+4}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{4}{\left(x+2\right)\left(x-2\right)}\)
c,Để A\(\in\)Z=> (x+2)(x-2)\(\inƯ\)(4) hay \(x^2-4\inƯ\)(4)=\(\left(4;-4;2;-2;1;-1\right)\)
Ta có bảng
| \(x^2-4\) | x |
| 4 | \(\sqrt{8}\) |
| -4 | 0 |
| 2 | \(\sqrt{6}\) |
| -2 | \(\sqrt{2}\) |
| 1 | \(\sqrt{5}\) |
Vậy A\(Z=>x\in\)( 0;\(\sqrt{8};\sqrt{6};\sqrt{2};\sqrt{5}\))
10 tháng 12 2022
a: ĐKXĐ: x<>2; x<>-2
b: \(A=\dfrac{3x\left(x-2\right)+2x+6}{2\left(x-2\right)\left(x+2\right)}=\dfrac{3x^2-6x+2x+6}{2\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{3x^2+4x+6}{2\left(x-2\right)\left(x+2\right)}\)
c: Khi x=-3 thì \(A=\dfrac{3\cdot\left(-3\right)^2-4\cdot3+6}{2\left(-3-2\right)\left(-3+2\right)}=\dfrac{21}{10}\)
a,\(A=\left(\frac{x}{x^2-4}+\frac{1}{x+2}-\frac{2}{x-2}\right):\left(2-x+\frac{6}{x+2}\right)\)
\(=\left(\frac{x}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x+2\right)\left(x-2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\right):\left(\frac{-\left(x-2\right)\left(x+2\right)}{x+2}+\frac{6}{x+2}\right)\)
\(=\left(\frac{2x-2-2x+4}{\left(x+2\right)\left(x-2\right)}\right):\left(\frac{-\left(x^2-4\right)+6}{x+2}\right)\)
\(=\frac{2}{\left(x+2\right)\left(x-2\right)}.\frac{x-2}{-\left(x^2-4\right)+6}=\frac{2}{-\left(x+2\right)^2\left(x-2\right)+6}\)
Thay x = 4 ta được :
\(\frac{2}{-\left(4+2\right)^2\left(4-2\right)+6}=\frac{2}{-26}=-\frac{1}{13}\)
Tương tự với x = -4