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nguyenthihuyentrang

( 2x  - 5) - ( 4x - 1 ) ( x + 3 ) = 5 

=> ( 2x )   - 2 . 2x. 5 + 52 - 4x2 + 12x - 3 - x = 5 

=> 4x2    - 20x + 15 - 4x + 11x - 3 = 5 

=> -20x + 11x = 5 + 3 - 15 

=> -9x = -7  =>  x = 7/9 

^^ Học tốt! 

12 tháng 6 2017

M.n giup mk vs ak!! mk se tag k cho m.n, camon ah!

Bài 2:

a: ĐKXĐ: \(x\notin\left\{0;2;-2;3\right\}\)\(A=\left(\dfrac{-\left(x+2\right)}{x-2}-\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\right):\dfrac{x\left(x-3\right)}{x^2\left(2-x\right)}\)

\(=\dfrac{-x^2-4x-4-4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-x\left(x-2\right)}{x-3}\)

\(=\dfrac{-4x^2-8x}{\left(x+2\right)}\cdot\dfrac{-x}{x-3}\)

\(=\dfrac{-4x\left(x+2\right)}{x+2}\cdot\dfrac{-x}{x-3}=\dfrac{4x^2}{x-3}\)

b: Để A>0 thì x-3>0

hay x>3

 

a) Đặt A(x)=0

\(\Leftrightarrow-4x-5=0\)

\(\Leftrightarrow-4x=5\)

hay \(x=-\dfrac{5}{4}\)

b) Đặt B(x)=0

\(\Leftrightarrow3\left(2x-1\right)-2\left(x+1\right)=0\)

\(\Leftrightarrow6x-3-2x-2=0\)

\(\Leftrightarrow4x=5\)

hay \(x=\dfrac{5}{4}\)

14 tháng 12 2021

\(a,\Leftrightarrow x^2\left(x+5\right)-9\left(x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\\x=-3\end{matrix}\right.\\ b,\Leftrightarrow\left[{}\begin{matrix}1-2x=3x-2\\2x-1=3x-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=1\end{matrix}\right.\)

a) \(\Leftrightarrow x^2\left(x+5\right)-9x-45=0\)
    \(\Leftrightarrow x^2\left(x+5\right)-9x\left(x+5\right)=0\)
    \(\Leftrightarrow\left(x+5\right)\left(x^2-9\right)=0\)
    \(\Leftrightarrow\left(x+5\right)\left(x-3\right)\left(x+3\right)=0\)
    \(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-3=0\\x+3=0\end{matrix}\right.\)
    \(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=3\\x=-3\end{matrix}\right.\)
Vậy...
b) \(\Leftrightarrow\left(1-2x\right)^2-\left(3x-2\right)^2=0\)
    \(\Leftrightarrow\left(1-2x-3x+2\right)\left(1-2x+3x-2\right)=0\)
    \(\Leftrightarrow\left(3-5x\right)\left(x-1\right)=0\)
    \(\Leftrightarrow\left[{}\begin{matrix}3-5x=0\\x-1=0\end{matrix}\right.\)
    \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=1\end{matrix}\right.\)
Vậy...

6 tháng 3 2018

a.

\(3-2x=3\left(x+1\right)-x-2\)

\(\Leftrightarrow3-2x=3x+3-x-2\)

\(\Leftrightarrow-2x-3x+x=3-2-3\)

\(\Leftrightarrow-4x=-2\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy.........

6 tháng 3 2018

b.

\(\left(3x+2\right)\left(4x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\4x-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2}{3}\\x=\dfrac{5}{4}\end{matrix}\right.\)

Vậy........