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a.x-\(\dfrac{5x+2}{6}=\dfrac{7-3x}{4}\)
⇔\(x=\dfrac{7-3x}{4}+\dfrac{5x+2}{6}\)
⇔\(x=\dfrac{21-9x+10x+4}{12}\)
⇔x=\(\dfrac{x+25}{12}\)
⇔12x=x+25
⇔x=\(\dfrac{25}{11}\)
Vậy pt đã cho có n0 là S=\(\left\{\dfrac{25}{11}\right\}\)
b.ĐKXĐ:x≠-2;x≠2
\(\dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\)
⇔\(\dfrac{\left(x-2\right)\cdot\left(x-2\right)-3\cdot\left(x+2\right)}{\left(x-2\right)\cdot\left(x+2\right)}\)=\(\dfrac{2x-22}{\left(x-2\right)\cdot\left(x+2\right)}\)
⇔\(\dfrac{x^2-7x-2}{\left(x-2\right)\cdot\left(x+2\right)}=\dfrac{2x-22}{\left(x-2\right)\cdot\left(x+2\right)}\)
⇒\(\left(x^2-7x-2\right)\cdot\left(x-2\right)\cdot\left(x+2\right)=\left(2x-22\right)\cdot\left(x-2\right)\cdot\left(x+2\right)\)
⇔x2-7x-2=2x-22
⇔x2-9x+20=0
⇔(x-4)(x-5)=0
⇔\(\left\{{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
Vậy pt đã cho có n0 là S={4;5}
a) (2x+12x−1−2x−12x+1):4x10x−5=(2x+1)2−(2x−1)2(2x−1)(2x+1).10x+54x(2x+12x−1−2x−12x+1):4x10x−5=(2x+1)2−(2x−1)2(2x−1)(2x+1).10x+54x
=4x2+4x+1−4x2+4x−1(2x−1)(2x+1).5(2x+1)4x4x2+4x+1−4x2+4x−1(2x−1)(2x+1).5(2x+1)4x
=8x.5(2x+1)(2x−1)(2
b) \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)=\left(\dfrac{1}{x\left(x+1\right)}-\dfrac{x\left(2-x\right)}{x\left(x+1\right)}\right):\left(\dfrac{1}{x}+\dfrac{x^2}{x}-\dfrac{2x}{x}\right)=\left(\dfrac{1-2x+x^2}{x\left(x+1\right)}\right):\left(\dfrac{1+x^2-2x}{x}\right)=\left(\dfrac{\left(x-1\right)^2}{x\left(x+1\right)}\right)\cdot\left(\dfrac{x}{\left(x-1\right)^2}\right)=\dfrac{\left(x-1\right)^2\cdot x}{\left(x-1\right)^2\cdot x\cdot\left(x+1\right)}=\dfrac{1}{x+1}\)
a: \(=\dfrac{x^3-1}{x+2}\cdot\dfrac{x^2+x+1-x^2+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x+2}{x+2}=1\)
b: \(=\dfrac{\left(x+2\right)\left(x-1\right)\left(x+1\right)}{2\left(x+5\right)}\cdot\left(\dfrac{x+1-2x+2}{\left(x-1\right)\left(x+1\right)}+\dfrac{1}{x+2}\right)\)
\(=\dfrac{\left(x+2\right)\left(x-1\right)\left(x+1\right)}{2\left(x+5\right)}\cdot\left(\dfrac{-\left(x-3\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{1}{x+2}\right)\)
\(=\dfrac{\left(x+2\right)\left(x-1\right)\left(x+1\right)}{2\left(x+5\right)}\cdot\dfrac{-\left(x^2-x-6\right)+x^2-1}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}\)
\(=\dfrac{-x^2+x+6+x^2-1}{2\left(x+5\right)}=\dfrac{x+5}{2\left(x+5\right)}=\dfrac{1}{2}\)
4)a)\(\dfrac{x+5}{x-5}-\dfrac{x-5}{x+5}=\dfrac{20}{x^2-25}\)(1)
ĐKXĐ:\(\left\{{}\begin{matrix}x-5\ne0\\x+5\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne5\\x\ne-5\end{matrix}\right.\)
(1)\(\Rightarrow\left(x+5\right)\left(x+5\right)-\left(x-5\right)\left(x-5\right)=20\)
\(\Leftrightarrow x^2+10x+25-\left(x^2-10x+25\right)=20\)
\(\Leftrightarrow x^2+10x+25-x^2+10x-25=20\)
\(\Leftrightarrow x^2-x^2+10x+10x=-25+25=20\)
\(\Leftrightarrow20x=20\)
\(\Leftrightarrow x=1\left(nh\text{ậ}n\right)\)
S=\(\left\{1\right\}\)
câu nào cũng ghi lại đề nha
a) \(x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
b)\(x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
c) \(\left(x+1\right)\left(x+2\right)+\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+1+x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{2}\end{matrix}\right.\)
d) \(\dfrac{1}{x-2}+3-\dfrac{3-x}{x-2}=0\)
\(\Leftrightarrow\dfrac{1+3\left(x-2\right)-\left(3-x\right)}{x-2}=0\)
\(\Leftrightarrow\dfrac{1+3x-6-3+x}{x-2}=0\) ( đk \(x\ne2\) )
\(\Leftrightarrow4x-8=0\Rightarrow x=2\)
đ) \(\dfrac{8-x}{x-7}-8-\dfrac{1}{x-7}=0\)
\(\Leftrightarrow\dfrac{8-x-8\left(x-7\right)-1}{x-7}=0\) (đk \(x\ne7\))
\(\Leftrightarrow8-x-8x+56-1=0\)
\(\Leftrightarrow-9x+63=0\)
\(\Leftrightarrow x=7\)
1) \(\dfrac{x^2}{x+1}+\dfrac{2x}{x^2-1}-\dfrac{1}{1-x}+1\)
\(=\dfrac{x^2}{x+1}+\dfrac{2x}{x^2-1}+\dfrac{1}{x-1}+1\)
\(=\dfrac{x^2}{x+1}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}+\dfrac{1}{x-1}+1\) MTC: \(\left(x-1\right)\left(x+1\right)\)
\(=\dfrac{x^2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}+\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}+\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2\left(x-1\right)+2x+\left(x+1\right)+\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^3-x^2+2x+x+1+x^2-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x\left(x^2+3\right)}{\left(x-1\right)\left(x+1\right)}\)
b) \(\dfrac{1}{x^3-x}-\dfrac{1}{\left(x-1\right)x}+\dfrac{2}{x^2-1}\)
\(=\dfrac{1}{x\left(x^2-1\right)}-\dfrac{1}{\left(x-1\right)x}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1}{x\left(x-1\right)\left(x+1\right)}-\dfrac{1}{\left(x-1\right)x}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\) MTC: \(x\left(x-1\right)\left(x+1\right)\)
\(=\dfrac{1}{x\left(x-1\right)\left(x+1\right)}-\dfrac{x+1}{x\left(x-1\right)\left(x+1\right)}+\dfrac{2x}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1-\left(x+1\right)+2x}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1-x-1+2x}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1}{\left(x-1\right)\left(x+1\right)}\)
1) \(\dfrac{x^2-4}{x^2+2x+1}:\dfrac{4-2x}{2x+2}=\dfrac{\left(x-2\right)\left(x+2\right)2\left(x+1\right)}{\left(x+1\right)^22\left(2-x\right)}=\dfrac{2\left(x-2\right)\left(x+2\right)\left(x+1\right)}{-2\left(x-2\right)\left(x+1\right)\left(x+1\right)}=\dfrac{-\left(x+2\right)}{x+1}=\dfrac{-x-2}{x+1}\)
2) \(\dfrac{x+1}{x+2}:\left(\dfrac{x+2}{x+3}:\dfrac{x+3}{x+1}\right)=\dfrac{x+1}{x+2}:\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x+3\right)\left(x+3\right)}=\dfrac{\left(x+1\right)\left(x+3\right)\left(x+3\right)}{\left(x+2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{\left(x+3\right)^2}{\left(x+2\right)^2}=\dfrac{x^2+6x+9}{x^2+4x+4}\)