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Bài 2 .
a) \(\dfrac{2x}{x^2+2xy}+\dfrac{y}{xy-2y^2}+\dfrac{4}{x^2-4y^2}\)
\(=\dfrac{2x}{x\left(x+2y\right)}+\dfrac{y}{y\left(x-2y\right)}+\dfrac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{2xy\left(x-2y\right)+xy\left(x+2y\right)+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{2x^2y-2xy^2+x^2y+2xy^2+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{3x^2y+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
b) Sai đề hay sao ý
c) \(\dfrac{2x+y}{2x^2-xy}+\dfrac{16x}{y^2-4x^2}+\dfrac{2x-y}{2x^2+xy}\)
\(=\dfrac{2x+y}{x\left(2x-y\right)}+\dfrac{-16x}{\left(2x-y\right)\left(2x+y\right)}+\dfrac{2x-y}{x\left(2x+y\right)}\)
\(=\dfrac{\left(2x+y\right)^2-16x^2+\left(2x-y\right)^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{4x^2+4xy+y^2-16x^2+4x^2-4xy+y^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{-8x^2}{x\left(2x-y\right)\left(2x+y\right)}\)
d) \(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
.....
\(=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{32}{1-x^{32}}\)
a: \(Q=\left(\dfrac{-x\left(x-2\right)}{2\left(x^2+4\right)}-\dfrac{2x^2}{\left(x^2+4\right)\left(x-2\right)}\right)\cdot\dfrac{2+x-x^2}{x^2}\)
\(=\dfrac{-x\left(x^2-4x+4\right)-4x^2}{2\left(x^2+4\right)\left(x-2\right)}\cdot\dfrac{-\left(x^2-x-2\right)}{x^2}\)
\(=\dfrac{-x^3+4x^2-4x-4x^2}{2\left(x^2+4\right)}\cdot\dfrac{-\left(x+1\right)}{x^2}\)
\(=\dfrac{-x\left(x^2+4\right)}{2\left(x^2+4\right)}\cdot\dfrac{-\left(x+1\right)}{x^2}=\dfrac{x+1}{x}\)
b: Để Q là số nguyên thì \(x+1⋮x\)
hay \(x=1\)
b: \(\Leftrightarrow\dfrac{2}{\left(x+7\right)\left(x-3\right)}=\dfrac{3x+21}{\left(x-3\right)\left(x+7\right)}\)
=>3x+21=2
=>x=-19/3
d: \(\Leftrightarrow\left(2x+1\right)^2-\left(2x-1\right)^2=8\)
\(\Leftrightarrow4x^2+4x+1-4x^2+4x-1=8\)
=>8x=8
hay x=1
a: \(=\dfrac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}:\left(\dfrac{1}{x+1}+\dfrac{x}{x-1}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\right)\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}:\dfrac{x-1+x^2+x+2}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{x^2+2x+1}=\dfrac{4x}{x^2+2x+1}\)
b: \(=\dfrac{x+2}{-\left(x-2\right)}\cdot\dfrac{\left(x-2\right)^2}{4x^2}\cdot\left(\dfrac{2}{2-x}-\dfrac{4}{\left(x+2\right)\left(x^2-2x+4\right)}\cdot\dfrac{x^2-2x+4}{2-x}\right)\)
\(=\dfrac{-\left(x+2\right)\left(x-2\right)}{4x^2}\cdot\left(\dfrac{2}{2-x}-\dfrac{4}{\left(x+2\right)\left(2-x\right)}\right)\)
\(=\dfrac{-\left(x+2\right)\left(x-2\right)}{4x^2}\cdot\dfrac{2x+4-4}{\left(2-x\right)\left(x+2\right)}\)
\(=\dfrac{2x}{4x^2}=\dfrac{1}{2x}\)
2)
a) \(\dfrac{1}{x}.\dfrac{6x}{y}\)
\(=\dfrac{6x}{xy}\)
\(=\dfrac{6}{y}\)
b) \(\dfrac{2x^2}{y}.3xy^2\)
\(=\dfrac{2x^2.3xy^2}{y}\)
\(=\dfrac{6x^3y^2}{y}\)
\(=6x^3y\)
c) \(\dfrac{15x}{7y^3}.\dfrac{2y^2}{x^2}\)
\(=\dfrac{15x.2y^2}{7y^3.x^2}\)
\(=\dfrac{30xy^2}{7x^2y^3}\)
\(=\dfrac{30}{7xy}\)
d) \(\dfrac{2x^2}{x-y}.\dfrac{y}{5x^3}\)
\(=\dfrac{2x^2.y}{\left(x-y\right).5x^3}\)
\(=\dfrac{2y}{5x\left(x-y\right)}\)
1.
a) \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
b) \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Bài 1:
a, \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
Vậy \(x=-4\) hoặc \(x=-1\)
b, \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x=3\) hoặc \(x=-2\)
a) \(A=\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(\Leftrightarrow A=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}-\dfrac{2x^2}{4\left(2-x\right)+x^2\left(2-x\right)}\right)\left(\dfrac{x^2-x-2}{x^2}\right)\)
\(\Leftrightarrow A=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}-\dfrac{2x^2}{\left(2-x\right)\left(4+x^2\right)}\right)\cdot\dfrac{x^2-x-2}{x^2}\)
ĐKXĐ: \(x\ne0;x\ne2\)
\(\Leftrightarrow A=\left(\dfrac{-x\left(2-x\right)^2-4x^2}{2\left(x^2+4\right)\left(2-x\right)}\right)\cdot\dfrac{x^2-x-2}{x^2}\)
\(\Leftrightarrow A=\dfrac{-x^3-2x^2-4x}{2\left(x^2+4\right)\left(2-x\right)}\cdot\dfrac{x^2-x-2}{x^2}\)
\(\Leftrightarrow A=-\dfrac{\left(x^2+2x+4\right)\left(x+1\right)\left(x+2\right)}{2\left(x^2+4\right)\left(2-x\right)}\)