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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?
![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa đề thành rút gọn phân số nhé :
\(\frac{\left(\frac{x}{x+3+3}-\frac{x}{3.x^2+3x}+\frac{9}{x^2-9}\right):3}{x+3}\)
\(=\frac{\frac{x}{3\left(x+6\right)}-\frac{x}{9x^2+9x}+\frac{9}{\left(x^2-9\right).3}}{x+3}\)
\(=\frac{\frac{x}{3x+18}-9x-9+\frac{3}{x^2-9}}{x+3}\)
\(=\frac{\frac{x-9x\left(3x+18\right)}{3x+18}+\frac{3-9\left(x^2-9\right)}{x^2-9}}{x+3}\)
\(=\frac{\frac{x-27x^2-162x}{3x+18}+\frac{3-9x^2+81}{x^2-9}}{x+3}\)
\(=\frac{\frac{27x^2-161x}{3x+18}+\frac{-9x^2+84}{x^2-9}}{x+3}\)
\(=\frac{27x^2-161x}{\left(3x+18\right):\left(x+3\right)}+\frac{-9x^2+84}{\left(x^2-9\right):\left(x+3\right)}\)
Đến đây thì dễ r ha
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{37\cdot5^4}{25^2}=\dfrac{37\cdot5^4}{5^4}=37\\ \dfrac{2^4\cdot2^6\cdot3^8\cdot9^2}{4^4\cdot3^{11}}=\dfrac{2^{10}\cdot3^8\cdot3^4}{2^8\cdot3^{11}}=2^2\cdot3=12\\ \dfrac{3\cdot9^4\cdot9^3}{3^2\cdot9}=\dfrac{3\cdot3^8\cdot3^6}{3^2\cdot3^2}=3^{11}\\ \dfrac{125\cdot5\cdot64-25^3\cdot10\cdot4}{5^7\cdot8}=\dfrac{5^3\cdot5\cdot2^6-5^6\cdot2\cdot5\cdot2^2}{5^7\cdot2^3}=\dfrac{5^4\cdot2^3\left(2^3-5^3\right)}{5^7\cdot2^3}=\dfrac{8-125}{5^3}=\dfrac{-117}{125}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a. $x(3x+1)+(x-1)^2-(2x+1)(2x-1)=0$
$\Leftrightarrow (3x^2+x)+(x^2-2x+1)-(4x^2-1)=0$
$\Leftrightarrow 3x^2+x+x^2-2x+1-4x^2+1=0$
$\Leftrightarrow (3x^2+x^2-4x^2)+(x-2x)+(1+1)=0$
$\Leftrightarrow -x+2=0$
$\Leftrightarrow x=2$
b.
$(x+1)^3+(2-x)^3-9(x-3)(x+3)=0$
$\Leftrightarrow [(x+1)+(2-x)][(x+1)^2-(x+1)(2-x)+(2-x)^2]-9(x-3)(x+3)=0$
$\Leftrightarrow 3[x^2+2x+1-(x-x^2+2)+(x^2-4x+4)]-9(x-3)(x+3)=0$
$\Leftrightarrow 3(3x^2-3x+3)-9(x^2-9)=0$
$\Leftrightarrow 9(x^2-x+1)-9(x^2-9)=0$
$\Leftrightarrow 9(x^2-x+1-x^2+9)=0$
$\Leftrightarrow 9(-x+10)=0$
$\Leftrightarrow -x+10=0\Leftrightarrow x=10$
c.
$(x-1)^3-(x+3)(x^2-3x+9)+3x^2=25$
$\Leftrightarrow (x^3-3x^2+3x-1)-(x^3+3^3)+3x^2=25$
$\Leftrightarrow x^3-3x^2+3x-1-x^3-27+3x^2=25$
$\Leftrightarrow (x^3-x^3)+(-3x^2+3x^2)+3x-28=25$
$\Leftrightarrow 3x-28=25$
$\Leftrightarrow x=\frac{53}{3}$
d.
$(x+2)^3-(x+1)(x^2-x+1)-6(x-1)^2=23$
$\Leftrightarrow (x^3+6x^2+12x+8)-(x^3+1)-6(x^2-2x+1)=23$
$\Leftrightarrow x^3+6x^2+12x+8-x^3-1-6x^2+12x-6=23$
$\Leftrightarrow (x^3-x^3)+(6x^2-6x^2)+(12x+12x)+(8-1-6)=23$
$\Leftrightarrow 24x+1=23$
$\Leftrgihtarrow 24x=22$
$\Leftrightarrow x=\frac{11}{12}$
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
Đề như này hả bạn? :V
\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-9x^2+27x-27-\left(x^3-27\right)+9\left(x+1\right)^2=15\)
\(\Leftrightarrow-9x^2+27x+9\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow45x=6\)\(\Leftrightarrow x=\dfrac{2}{15}\)
Vậy...
![](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)
(\(3+\dfrac{x}{3-x}+\dfrac{2x}{3+x}-\dfrac{4x^2-3x-9}{x^2-9}\) ):\(\left(\dfrac{2}{3-x}-\dfrac{x-1}{3x-x^2}\right)\)\(=\left(\dfrac{3x^2-27}{\left(x-3\right)\left(x+3\right)}+\dfrac{-x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{2x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{4x^2-3x-9}{\left(x-3\right)\left(x+3\right)}\right)\)\(:\left(\dfrac{2x}{x\left(3-x\right)}-\dfrac{x-1}{x\left(3-x\right)}\right)\)
\(=\dfrac{3x^2-27-x^2-3x+2x^2-6x-4x^2+3x+9}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+1}{x\left(3-x\right)}\)
\(=\dfrac{-6x-18}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+1}{x\left(3-x\right)}\) \(=\dfrac{-6\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+1}{x\left(3-x\right)}\)
\(=\dfrac{6}{3-x}.\dfrac{x\left(x-3\right)}{x+1}\) \(=\dfrac{6x}{x+1}\)
3(x+3) - x2 + 9
= 3(x +3) - ( x2 - 9 )
=3(x+3) - (x2 - 32)
= 3(x + 3) - (x - 3)(x+3)
= (x+3)(3 - x + 3)
= x(x+3)
Chúc học tốt^^