\(\Leftrightarrow6x^2-9x-32x+48=0\)

\(\Leftrightarrow3x\left(2x-3\right)-16\left(2x-3\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(3x-16\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{16}{3}\end{matrix}\right.\)

\(x\) = 1,5; \(x\) = \(\dfrac{16}{3}\)

`a)2x^2+3(x-1)(x+1)=5x(x+1)`

`<=>2x^2+3x^2-3=5x^2+5x`

`<=>5x=-3`

`<=>x=-3/5`

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`b)(x-3)^3+3-x=0` nhỉ?

`<=>(x-3)^3-(x-3)=0`

`<=>(x-3)(x^2-1)=0`

`<=>[(x=3),(x^2=1<=>x=+-1):}`

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`c)5x(x-2000)-x+2000=0`

`<=>5x(x-2000)-(x-2000)=0`

`<=>(x-2000)(5x-1)=0`

`<=>[(x=2000),(x=1/5):}`

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`d)3(2x-3)+2(2-x)=-3`

`<=>6x-9+4-2x=-3`

`<=>4x=2`

`<=>x=1/2`

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`e)x+6x^2=0`

`<=>x(1+6x)=0`

`<=>[(x=0),(x=-1/6):}`

\(\left(x-2\right)^5-\left(x-2\right)^3=0\)

\(\Rightarrow\left(x-2\right)^3\left(\left(x-2\right)^2-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\\left(x-2\right)^2-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\\left(x-2\right)^2=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x-2=1\\x-2=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=3\\x=1\end{matrix}\right.\)

Vậy \(x\in\left\{1;2;3\right\}\)

⇒ ( x - 2)³ . (x - 2)² - (x - 2)³  . 1 = 0                                                          ⇒ ( x - 2)³ . [( x - 2)² - 1] = 0 

1)

a) \(=15x^3-20x^2+10x\)

b) \(=3x^4-x^3+4x^2-9x^3+3x-12x=3x^4-10x^3+4x^2-9x\)

2) 

a) \(\Rightarrow x\left(x^2-6x+12\right)=0\)

\(\Rightarrow x=0\)(do \(x^2-6x+12=\left(x^2-6x+\dfrac{36}{4}\right)+3=\left(x-\dfrac{6}{2}\right)^2+3\ge3>0\))

b) \(\Rightarrow\left(x+3\right)^3=0\Rightarrow x=-3\)

(3x²-5x+2)+(3x²+5x)= bao nhiêu ạ

Giúp em vs ạ . Em cảm ơn

1) \(\left(x-3\right)^2-4=0\)

\(\Leftrightarrow\left(x-3-2\right)\left(x-3+2\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)

2) \(x^2-2x=24\)

\(\Leftrightarrow x^2-2x-24=0\)

\(\Leftrightarrow x^2+4x-6x-24=0\)

\(\Leftrightarrow x\left(x+4\right)-6\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

Câu 3 số xấu rồi e

\(A=\frac{3}{1-x}+\frac{4}{x}\ge\frac{\left(\sqrt{3}+2\right)^2}{1-x+x}=7+4\sqrt{3}\)

Dấu = xảy ra khi: \(x=\frac{2}{\sqrt{3}+2}\)

\(\left(2x-4\right)\left(3x+1\right)< 0\)

=> TH1:  \(\begin{matrix}2x-4< 0\\3x+1>0\end{matrix}\)\(\Leftrightarrow\left\{{}\begin{matrix}2x< 4\\3x>-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 2\\x>-\dfrac{1}{3}\end{matrix}\right.\) (tm)

TH2: \(\begin{matrix}2x-4>0\\3x+1< 0\end{matrix}\)\(\Leftrightarrow\left\{{}\begin{matrix}2x>4\\3x< -1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>2\\x< -\dfrac{1}{3}\end{matrix}\right.\) (vô lí)

=> \(2>x>-\dfrac{1}{3}\)

x∈(-1/3, 2)