(5x+15)(9-18x)=0
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1: \(\dfrac{2x^3+11x^2+18x-3}{2x+3}\)
\(=\dfrac{2x^3+3x^2+8x^2+12x+6x+9-12}{2x+3}\)
\(=x^2+4x+3-\dfrac{12}{2x+3}\)
a, \(x^4-5x^3+2x^2+10x+2=0\)
\(\Rightarrow x^4+x^3-6x^3-6x^2+8x^2+8x+2x+2=0\)
\(\Rightarrow x^3\left(x+1\right)-6x^2\left(x+1\right)+8x\left(x+1\right)+2\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x^3-6x^2+8x+2\right)=0\)
Vì \(x^3-6x^2+8x+2>0\) nên \(x+1=0\Rightarrow x=-1\)
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a. \(3x\left(2x+1\right)=6x^2+3x\)
b. \(\left(12x^3-18x^2+6x\right):6x=2x^2-3x+1\)
c. \(\dfrac{7x+6}{5x-1}+\dfrac{8x-9}{5x-1}=\dfrac{15x-3}{5x-1}=\dfrac{3\left(5x-1\right)}{5x-1}=3\)
\(15x^2-34x+15\)
\(=15x^2-25x-9x+15\)
\(=5x\left(3x-5\right)-3\left(3x-5\right)\)
\(=\left(5x-3\right)\left(3x-5\right)\)
\(x^2-25x=0\)
\(\Rightarrow x\left(x-25\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-25=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=25\end{cases}}}\)
vậy_
\(\left(4x-1\right)^2-9=0\)
\(\Rightarrow\left(4x-1\right)^2-3^2=0\)
\(\Rightarrow\left(4x-1+3\right)\left(4x-1-3\right)=0\)
\(\Rightarrow\left(4x+2\right)\left(4x-4\right)=0\)
\(\Rightarrow2\cdot\left(2x+1\right)\cdot4\cdot\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+1=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}}\)
vậy_
Bài 2 :
a) \(3x^2-18x+27\)
\(=3\left(x^2-6x+9\right)\)
\(=3\left(x^2-2\cdot x\cdot3+3^2\right)\)
\(=3\left(x+3\right)^2\)
b) \(xy-y^2-x+y\)
\(=y\left(x-y\right)-\left(x-y\right)\)
\(=\left(x-y\right)\left(y-1\right)\)
c) \(x^2-5x-6\)
\(=x^2+x-6x-6\)
\(=x\left(x+1\right)-6\left(x+1\right)\)
\(=\left(x+1\right)\left(x-6\right)\)
5x.(9-x) - 15.(9-x)= 0
(9-x).(5x-15) = 0
=> 9-x = 0 => x = 9
5x -15 = 0 => 5x = 15 => x = 3
KL:...
TH1: 5x + 15 = 0
5x = -15
x = -3
TH2: 9 - 18x = 0
18x = 9
x = 1/2