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d. x+165+x+363=x+561+x+759
\(\Leftrightarrow2x+528=2x+1320\)
\(\Leftrightarrow2x-2x=1320-528\)
\(\Leftrightarrow0x=729\) (loại)
\(\Rightarrow\) PT vô N0
![](https://rs.olm.vn/images/avt/0.png?1311)
c: \(\Leftrightarrow2x+2x-6=12-2x\)
=>4x-6=12-2x
=>6x=18
hay x=3
b: \(\Leftrightarrow\left(x-1\right)\left(x+1\right)+x=2x-1\)
\(\Leftrightarrow x^2-1+x=2x-1\)
=>x2-x=0
=>x(x-1)=0
=>x=0(loại) hoặc x=1(nhận)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. \(x^2+2018x-2019=0\)
Ta có: \(a+b+c=1+2018+\left(-2019\right)=0\)\(\)
Nên \(x_1=1,x_2=-2019\)
Vậy . . .
b. \(\left\{{}\begin{matrix}3x-y=5\\x+3y=15\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-y=5\\3x+9y=45\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-10y=-40\\3x-y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=4\\x=3\end{matrix}\right.\)
Vậy hệ pt có nghiệm . . . .
![](https://rs.olm.vn/images/avt/0.png?1311)
A= [(2√x√x+3)+√x√x+3+3(√xx−9)]:(2√x−2√x−3−11)[(2xx+3)+xx+3+3(xx−9)]:(2x−2x−3−11)với x>= 0 , x #9
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a)\(\sqrt{5}\left(x+2\right)=\sqrt{10}\)
\(\Leftrightarrow x+2=\sqrt{2}\)
\(\Leftrightarrow x=\sqrt{2}-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)