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1:
a: =>28x-8=9x+3
=>19x=11
=>x=11/19
b: =>(3x-1)(x-1)=(2x+1)(x+1)
=>3x^2-4x+1=2x^2+3x+1
=>x^2-7x=0
=>x=0 hoặc x=7
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a: \(\Leftrightarrow\dfrac{y+5}{y\left(y-5\right)}-\dfrac{y-5}{2y\left(y+5\right)}=\dfrac{y+25}{2\left(y-5\right)\left(y+5\right)}\)
\(\Leftrightarrow2\left(y+5\right)^2-\left(y-5\right)^2=y^2+25y\)
=>\(2y^2+20y+50-y^2+10y-25=y^2+25y\)
=>30y+25=25y
=>5y=-25
=>y=-5(loại)
b: \(\Leftrightarrow x\left(x+1\right)+x\left(x-3\right)=4x\)
=>x^2+x+x^2-3x-4x=0
=>2x^2-6x=0
=>2x(x-3)=0
=>x=0(nhận) hoặc x=3(loại)
c: =>x^2-9-6(2x+7)=-13(x+3)
=>x^2-9-12x-42+13x+39=0
=>x^2+x-6=0
=>(x+3)(x-2)=0
=>x=2(nhận) hoặc x=-3(loại)
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Ta có: 2xy - x - y = 1
=> x(2y-1) - y =1 => 2x(2y-1) - 2y = 2 => 2x(2y-1) - (2y-1) = 3
=> (2y-1)(2x-1) = 3
Ta có bảng:
2y-1 | 1 | -1 | 3 | -3 |
2x-1 | 3 | -3 | 1 | -1 |
x | 2 | -1 | 1 | -1 |
y | 1 | 0 | 2 | 0 |
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-2\right)\left(x^2-3x+5\right)=\left(2-x\right)\left(1-x^2\right)\\ \Leftrightarrow\left(x-2\right)\left(x^2-3x+5\right)=-\left(x-2\right)\left(1-x^2\right)\\ \Leftrightarrow\left(x-2\right)\left(x^2-3x+5\right)+\left(x-2\right)\left(1-x^2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x^2-3x+5+1-x^2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(-3x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\-3x+6=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=2\end{matrix}\right.\\ \Leftrightarrow x=2\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-3x+5\right)=-\left(x-2\right)\left(1-x^2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-3x+5\right)+\left(x-2\right)\left(1-x^2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-3x+5+1-x^2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(6-3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=2\end{matrix}\right.\)
Vậy \(S=\left\{2\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
dấu # là dấu gì
hả bn?
mk ko biết là dấu gì hết?
chúc bn học giỏi
ahjhj