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Em ơi em đăng câu hỏi thì dùng công thức toán học để viết chữ viết thế này ai hiểu đúng được em?
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a: \(\dfrac{5x+y^2}{x^2y}-\dfrac{5y-x^2}{xy^2}\)
\(=\dfrac{5xy+y^3-x\left(5y-x^2\right)}{x^2y^2}\)
\(=\dfrac{5xy+y^3-5xy+x^3}{x^2y^2}=\dfrac{x^3+y^3}{x^2y^2}\)
b: \(\dfrac{x+9}{\left(x-3\right)\left(x+3\right)}-\dfrac{3}{x\left(x+3\right)}\)
\(=\dfrac{x^2+9x-3x+9}{x\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x+3\right)^2}{x\left(x-3\right)\left(x+3\right)}=\dfrac{x+3}{x^2-3x}\)
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1: =>x^2+4x-21=0
=>(x+7)(x-3)=0
=>x=3 hoặc x=-7
2: =>(2x-5-4)(2x-5+4)=0
=>(2x-9)(2x-1)=0
=>x=9/2 hoặc x=1/2
3: =>x^3-9x^2+27x-27-x^3+27+9(x^2+2x+1)=15
=>-9x^2+27x+9x^2+18x+9=15
=>18x=15-9-27=-21
=>x=-7/6
6: =>4x^2+4x+1-4x^2-16x-16=9
=>-12x-15=9
=>-12x=24
=>x=-2
7: =>x^2+6x+9-x^2-4x+32=1
=>2x+41=1
=>2x=-40
=>x=-20
![](https://rs.olm.vn/images/avt/0.png?1311)
`a,3(x-2)^2+9(x-1)=3(x^2+x-3)`
`<=>3(x^2-4x+4)+9x-9=3x^2+3x-9`
`<=>3x^2-12x+12+9x-9=3x^2+3x-9`
`<=>3x^2-3x+3=3x^2+3x-9`
`<=>6x=12`
`<=>x=12`
`b,(x+3)^2-(x-3)=6x+18`
`<=>(x+3-x+3)(x+3+x-3)+6x+18`
`<=>6.2x=6(x+3)`
`<=>2x=x+3`
`<=>x=3`
`c,(2x+7)^2=9(x+2)^2`
`<=>(2x+7)^2=(3x+6)^2`
`<=>(3x+6-2x-7)(3x+6+2x+7)=0`
`<=>(x-1)(5x+13)=0`
`<=>` $\left[ \begin{array}{l}x-1=0\\5x+13=0\end{array} \right.$
`<=>` $\left[ \begin{array}{l}x=1\\5x=-13\end{array} \right.$
`<=>` $\left[ \begin{array}{l}x=1\\x=-\dfrac{13}{5}\end{array} \right.$
a) Ta có: \(3\left(x-2\right)^2+9\left(x-1\right)=3\left(x^2+x-3\right)\)
\(\Leftrightarrow3\left(x^2-4x+4\right)+9x-9=3x^2+3x-9\)
\(\Leftrightarrow3x^2-12x+12+9x-9-3x^2-3x+9=0\)
\(\Leftrightarrow-6x+12=0\)
\(\Leftrightarrow-6x=-12\)
hay x=2
Vậy: x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-3\right)\left(x+3\right)\left(x^2+9\right)-\left(x^2-9\right)^2\)
\(=\left(x^2-9\right)\left(x^2+9\right)-\left(x^2-9\right)^2\)
\(=\left(x^2-9\right)\left(x^2+9-x^2+9\right)\)
\(=\left(x^2-3^2\right).18\)
\(=18\left(x-3\right)\left(x+3\right)\)
(x-3)(x+3)(x^2+9)- (x^2-9)^2
=(x^2-9)[(x-3)(x+3)-(x^2-9)]
=(x^2-9)[x^2-9-x^2+9]
=(x^2-9)*0
=0
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+9x^2+18x+9=15\)
\(\Leftrightarrow45x=6\)
hay \(x=\dfrac{2}{15}\)
(x2 - 9)2 - 9(x - 3)2
= [(x - 3)(x + 3)]2 - 9(x - 3)2
= (x - 3)2(x + 3)2 - 9(x - 3)2
= (x - 3)2(x2 + 6x + 9 - 9)
= (x - 3)2(x2 + 6x)
= x(x + 6)(x - 3)2