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DL
4
25 tháng 11 2018
\(\frac{x^2+x-6}{x^3-4x^2-18x+9}=\frac{x^2+3x-2x-6}{x^3+3x^2-7x^2-21x+3x+9}\)
\(=\frac{x\left(x+3\right)-2\left(x+3\right)}{x^2\left(x+3\right)-7x\left(x+3\right)+3\left(x+3\right)}\)
\(=\frac{\left(x+3\right)\left(x-2\right)}{\left(x+3\right)\left(x^2-7x+3\right)}=\frac{x-2}{x^2-7x+3}\) (điều kiện: x khác -3)
HN
25 tháng 11 2018
t phân tích \(x^2-7x+3\) được như này =))
\(x^2-7x+3=x^2-2.x.\frac{7}{2}+\left(\frac{7}{2}\right)^2-\frac{49}{4}+3\)
\(=\left(x-\frac{7}{2}\right)^2-\frac{37}{4}\)
\(=\left(x-\frac{7}{2}\right)^2-\left(\frac{\sqrt{37}}{2}\right)^2\)
\(=\left(x-\frac{7}{2}-\frac{\sqrt{37}}{2}\right)\left(x-\frac{7}{2}+\frac{\sqrt{37}}{2}\right)\)
\(=\left(x-\frac{7+\sqrt{37}}{2}\right)\left(x-\frac{7-\sqrt{37}}{2}\right)\)
ND
1
3 tháng 4 2018
Ta có :
\(\dfrac{x^2+x-6}{x^3-4x^2-18x+9}=\dfrac{\left(x-2\right)\left(x+3\right)}{\left(x+3\right)\left(x^2-7x+3\right)}=\dfrac{x-2}{x^2-7x+3}\)
NH
0
H
15 tháng 12 2018
\(\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}\)
\(=\frac{x\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{-4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{x^2+2xy+x^2-2xy-4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2x^2-4xy}{\left(x-2y\right)\left(x+2y\right)}\)
29 tháng 9 2019
\(A=\frac{4x}{x^2-2x}+\frac{3}{2-x}+\frac{12x}{x^3-4x}\)
\(A=\frac{4x}{x\left(x-2\right)}-\frac{3}{x-2}+\frac{12x}{x\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{4x\left(x+2\right)-3x\left(x+2\right)+12x}{x\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{x\left(x+2\right)+12x}{x\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{x^2+2x+12x}{x\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{x^2+14x}{x\left(x-2\right)\left(x+2\right)}\)
25 tháng 1 2018
a)\(\text{ĐKXĐ:}\hept{\begin{cases}x^3-4x\ne0\\6-3x\ne0\\x+2\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne\mp2\end{cases}}\)
\(M=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(=\left[\frac{x^2}{x\left(x-2\right)\left(x+2\right)}-\frac{6}{3\left(x-2\right)}+\frac{1}{x+2}\right]:\left[\frac{x^2-4+10-x^2}{x+2}\right]\)
\(=\left[\frac{x^2}{x\left(x-2\right)\left(x+2\right)}-\frac{2x\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}+\frac{x\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}\right].\frac{x+2}{6}\)
\(=\frac{x^2-2x^2-4x+x^2-2x}{x\left(x-2\right)\left(x+2\right)}\cdot\frac{x+2}{6}\)
\(=\frac{1}{x+2}\)
b) /x/= \(\frac{1}{2}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)
*\(\text{Với }x=\frac{1}{2}\text{ta có pt:}\)
\(M=\frac{1}{x+2}=\frac{1}{\frac{1}{2}+2}=\frac{2}{5}\)
*\(\text{Với x= -1/2 ta có pt:}\)
\(M=\frac{1}{x+2}=\frac{1}{-\frac{1}{2}+2}=\frac{2}{3}\)
27 tháng 1 2018
a) = (\(\frac{x^2}{x\left(x^2\right)-4}+\frac{6}{3\left(2-x\right)}+\frac{1}{x+2}\)):(x-2+\(\frac{10-x^2}{x+2}\))
=(\(\frac{x^2}{x\left(x-2\right)\left(x+2\right)}+\frac{-6}{3\left(x-2\right)}+\frac{1}{x+2}\)) :(x-2+\(\frac{10-x^2}{x+2}\))
=(\(\frac{3x^2-6x\left(x+2\right)+\left(x-2\right)3x}{3x\left(x-2\right)\left(x+2\right)}\)) :(\(\frac{\left(x-2\right)\left(x+2\right)+10-x^2}{x+2}\))
=(\(\frac{3x^2-6x^2-12x+3x^2-6x}{3x\left(x-2\right)\left(x+2\right)}\)):(\(\frac{x^2-4+10-x^2}{x+2}\))
=\(\frac{-18x}{3x\left(x-2\right)\left(x+2\right)}\):\(\frac{6}{x+2}\)
=\(\frac{-6}{\left(x-2\right)\left(x+2\right)}\):\(\frac{6}{x+2}\)
=\(\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}\)
=\(\frac{-1}{x-2}\)
Vậy M=\(\frac{-1}{x-2}\)
b)Vì /x/ =1/2 nên x=1/2 hoặc x=-1/2Thay x=1/2 vào M ta được;
\(\frac{-1}{\frac{1}{2}-2}\)=\(\frac{2}{3}\)
Thay x=-1/2 vào M ta được:
\(\frac{-1}{-\frac{1}{2}-2}\)=\(\frac{2}{5}\)
Vậy \(M\in\)\(\hept{\begin{cases}\\\end{cases}\frac{2}{5};\frac{2}{3}}\)khi /x/=1/2
Đặt \(A=\frac{x^2+x-6}{x^3-4x^2-18x+9}\)
\(A=\frac{x^2+3x-2x-6}{x^3+3x^2-7x^2-21x+3x+9}\)
\(A=\frac{x\left(x+3\right)-2\left(x+3\right)}{x^2\left(x+3\right)-7x\left(x+3\right)+3\left(x+3\right)}\)
\(A=\frac{\left(x-2\right)\left(x+3\right)}{\left(x^2-7x+3\right)\left(x+3\right)}\)
\(A=\frac{x-2}{x^2-7x+3}\)