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\(\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}\)
\(=\frac{x\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{-4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{x^2+2xy+x^2-2xy-4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2x^2-4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(M=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)
a) ĐKXĐ : x ≠ -3 , x ≠ 2
\(=\frac{x+2}{x+3}-\frac{5}{x^2-2x+3x-6}-\frac{1}{x-2}\)
\(=\frac{x+2}{x+3}-\frac{5}{x\left(x-2\right)+3\left(x-2\right)}-\frac{1}{x-2}\)
\(=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{x+3}{\left(x+3\right)\left(x-2\right)}\)
\(=\frac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)
\(=\frac{x^2-4x+3x-12}{\left(x+3\right)\left(x-2\right)}=\frac{x\left(x-4\right)+3\left(x-4\right)}{\left(x+3\right)\left(x-2\right)}\)
\(=\frac{\left(x+3\right)\left(x-4\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x-4}{x-2}\)
b) Để M = 1/3
=> \(\frac{x-4}{x-2}=\frac{1}{3}\)( x ≠ -3 , x ≠ 2 )
=> 3( x - 4 ) = x - 2
=> 3x - 12 - x + 2 = 0
=> 2x - 10 = 0
=> 2x = 10
=> x = 5 ( tm )
Vậy x = 5 thì M = 1/3
đk: \(x\ne2,x\ne-3\)
a) Ta có: \(M=\frac{-4+x^2}{x^2+x-6}-\frac{5}{x^2+x-6}-\frac{x+3}{x^2+x-6}\)
\(=\frac{x^2-x-12}{x^2+x-6}=\frac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\frac{x-4}{x-2}\)
b) \(M=\frac{1}{3}\Rightarrow\frac{x-4}{x-2}=\frac{1}{3}\Leftrightarrow3x-12=x-2\Leftrightarrow x=5\)
a. A=\(1+\left(\frac{x+1}{x^3+1}-\frac{1}{x-x^2-1}-\frac{2}{x+1}\right):\frac{x^3-2x^2}{x^3-x^2+x}\)
\(=1+\left(\frac{x+1+x+1-2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right).\frac{x\left(x^2-x+1\right)}{x^2\left(x-2\right)}\)
\(=1+\frac{-2x^2+4x}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{x^2-x+1}{x\left(x-2\right)}\)
\(=1+\frac{-2x\left(x-2\right)}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{x^2-x+1}{x\left(x-2\right)}\)
\(=1-\frac{2}{x+1}=\frac{x-1}{x+1}\)
b.\(\left|x-\frac{3}{4}\right|=\frac{5}{4}\Rightarrow\orbr{\begin{cases}x-\frac{3}{4}=\frac{5}{4}\\x-\frac{3}{4}=-\frac{5}{4}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=-\frac{1}{2}\end{cases}}\)
Với \(x=2\Rightarrow A=\frac{2-1}{2+1}=\frac{1}{3}\)
Với \(x=-\frac{1}{2}\Rightarrow A=\frac{-\frac{1}{2}-1}{-\frac{1}{2}+1}=-3\)
a,\(A=\frac{6x+12}{\left(x+2\right)\left(2x-6\right)}=\frac{6\left(x+2\right)}{2\left(x+2\right)\left(x-3\right)}=\frac{3}{x-3}\)
b, Giá trị của x để phân thức có giá trị bằng (-2) :
\(\frac{3}{x-3}=-2\Rightarrow x=1,5\)
Đề chỗ \(x^{2-2}\) là sao bạn ??? Tức là mũ là 2-2 hay là \(x^2-2?\)
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/= \(\frac{1}{2}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)
*\(\text{Với }x=\frac{1}{2}\text{ta có pt:}\)
\(M=\frac{1}{x+2}=\frac{1}{\frac{1}{2}+2}=\frac{2}{5}\)
*\(\text{Với x= -1/2 ta có pt:}\)
\(M=\frac{1}{x+2}=\frac{1}{-\frac{1}{2}+2}=\frac{2}{3}\)
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
ĐKXĐ: x khác 2 và -3
\(P=\frac{x+2}{x+3}-\frac{5}{x+2x-3x-6}-\frac{1}{x-2}\)
\(P=\frac{\left(x+2\right).\left(x-2\right)}{\left(x+3\right).\left(x-2\right)}-\frac{5}{\left(x-2\right).\left(x+3\right)}-\frac{x+3}{\left(x-2\right).\left(x+3\right)}\)
\(P=\frac{x^2-4-5-x-3}{\left(x-2\right).\left(x+3\right)}=\frac{x^2-x-12}{\left(x+2\right).\left(x+3\right)}=\frac{\left(x-4\right).\left(x+3\right)}{\left(x+2\right).\left(x+3\right)}=\frac{x-4}{x+2}\)