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Bài 5.5:
\(\left(2x-3\right)\left(x+1\right)+\left(4x^3-6x^2-6x\right):\left(-2x\right)=18\)
\(\Leftrightarrow\left(2x^2+2x-3x-3\right)+2x\cdot\left(2x^2-3x-3\right):\left(-2x\right)=18\)
\(\Leftrightarrow2x^2-x-3-2x^2+3x+3=18\)
\(\Leftrightarrow2x=18\)
\(\Leftrightarrow x=\dfrac{18}{2}\)
\(\Leftrightarrow x=9\)
B1:
\(x^n\left(x+1\right)-x^n-x^{n-1}=0\)
\(\Rightarrow x^{n-1}\left(x^2+x\right)-x^{n-1}.x-x^{n-1}=0\)
\(\Rightarrow x^{n-1}\left(x^2+x-x-1\right)=0\)
\(\Rightarrow x^{n-1}\left(x^2-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^{n-1}=0\\x^2-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
Bài 2:
a: ĐKXĐ: \(x\notin\left\{0;-1;\dfrac{1}{2}\right\}\)
b: \(D=\left(\dfrac{x+2}{3x}+\dfrac{2}{x+1}-3\right):\dfrac{2-4x}{x+1}-\dfrac{3x-x^2+1}{3x}\)
\(=\dfrac{\left(x+2\right)\left(x+1\right)+6x-3\cdot3x\left(x+1\right)}{3x\left(x+1\right)}\cdot\dfrac{x+1}{2-4x}+\dfrac{x^2-3x-1}{3x}\)
\(=\dfrac{x^2+3x+2+6x-9x^2-9x}{3x}\cdot\dfrac{1}{2-4x}+\dfrac{x^2-3x-1}{3x}\)
\(=\dfrac{-8x^2+2}{3x}\cdot\dfrac{1}{-4x+2}+\dfrac{x^2-3x-1}{3x}\)
\(=\dfrac{-2\left(2x-1\right)\left(2x+1\right)}{3x\cdot\left(-2\right)\left(2x-1\right)}+\dfrac{x^2-3x-1}{3x}\)
\(=\dfrac{2x+1}{3x}+\dfrac{x^2-3x-1}{3x}\)
\(=\dfrac{2x+1+x^2-3x-1}{3x}=\dfrac{x^2-x}{3x}=\dfrac{x-1}{3}\)
c: Khi x=1 thì \(D=\dfrac{1-1}{3}=0\)
1: =>(x+2018)(6x-3)=0
=>x+2018=0 hoặc 6x-3=0
=>x=1/2 hoặc x=-2018
2: x(x-11)+3(11-x)=0
=>(x-11)(x-3)=0
=>x=11 hoặc x=3
4: =>(x+5)(2x-4)=0
=>2x-4=0 hoặc x+5=0
=>x=2 hoặc x=-5
3: =>(x-3)(x+2)=0
=>x=3 hoặc x=-2
Bài 1:
\(6x\left(x+2018\right)-3\left(x+2018\right)=0\)
\(\Leftrightarrow\left(x+2018\right)\left(6x-3\right)=0\)
\(\Leftrightarrow3\left(x+2018\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2018\\2x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2018\\x=\dfrac{1}{2}\end{matrix}\right.\)
Bài 2:
\(x\left(x-11\right)+3\left(11-x\right)=0\)
\(\Leftrightarrow x\left(x-11\right)-3\left(x-11\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=11\end{matrix}\right.\)
Câu 3:
\(x\left(x-3\right)-2\left(3-x\right)=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Câu 4:
\(2x\left(x+5\right)-4\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(2x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\2x=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Bài 1:
a: \(2x^2-8x=0\)
=>\(x^2-4x=0\)
=>x(x-4)=0
=>\(\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
b: \(\left(x+2\right)^2-x\left(x-1\right)=10\)
=>\(x^2+4x+4-x^2+x=10\)
=>5x+4=10
=>5x=6
=>\(x=\dfrac{6}{5}\)
c: \(x^3-6x^2+9x=0\)
=>\(x\left(x^2-6x+9\right)=0\)
=>\(x\left(x-3\right)^2=0\)
=>\(\left[{}\begin{matrix}x=0\\\left(x-3\right)^2=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Bài 1. Đề khó nhìn quá mình không làm được ._.
Bài 2.
12x2 + 24x - 15 = ( 2x - a )( 6x - 3 )
<=> 12x2 + 24x - 15 = 12x2 - 6x - 6ax + 3a
<=> 12x2 + 24x - 15 = 12x2 + ( -6 - 6a )x + 3a
Đồng nhất hệ số ta được :
\(\hept{\begin{cases}-6-6a=24\\-15=3a\end{cases}}\Leftrightarrow a=-5\)