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MH
1 tháng 11 2021
1.
a) \(2x^4-4x^3+2x^2\)
\(=2x^2\left(x^2-2x+1\right)\)
\(=2x^2\left(x-1\right)^2\)
b) \(2x^2-2xy+5x-5y\)
\(=\left(2x^2-2xy\right)+\left(5x-5y\right)\)
\(=2x\left(x-y\right)+5\left(x-y\right)\)
\(=\left(x-y\right)\cdot\left(2x+5\right)\)
KJ
1 tháng 11 2021
2 .
a,
\(4x\left(x-3\right)-x+3=0\)
⇒\(4x\left(x-3\right)-\left(x-3\right)=0\)
⇒\(\left(x-3\right)\left(4x-1\right)=0\)
⇒\(\left[{}\begin{matrix}x-3=0\\4x-1=0\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=3\\4x=1\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{4}\end{matrix}\right.\)
vậy \(x\in\left\{3;\dfrac{1}{4}\right\}\)
b,
\(\)\(\left(2x-3\right)^2-\left(x+1\right)^2=0\)
⇒\(\left(2x-3-x-1\right)\left(2x-3+x+1\right)\) = 0
⇒\(\left(x-4\right)\left(3x-2\right)=0\)
⇔\(\left[{}\begin{matrix}x-4=0\\3x-2=0\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=4\\3x=2\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=4\\x=\dfrac{2}{3}\end{matrix}\right.\)
vậy \(x\in\left\{4;\dfrac{2}{3}\right\}\)
HP
1
NH
1
3 tháng 8 2023
a: =>2x^2=4
=>x^2=2
=>\(x=\pm\sqrt{2}\)
b: =>(x+1)^2-4=0
=>(x+1+2)(x+1-2)=0
=>(x+3)(x-1)=0
=>x=1 hoặc x=-3
c: =>(2x-1)^2-3^2=0
=>(2x-1-3)(2x-1+3)=0
=>(2x-4)(2x+2)=0
=>x=2 hoặc x=-1
d: x^2-x=0
=>x(x-1)=0
=>x=0 hoặc x=1
27 tháng 7 2021
Bài 1:
a) Ta có: \(P=1+\dfrac{3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{4}{x-2}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\dfrac{4\left(x+2\right)-x-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{4x+8-x-x+2}\)
\(=1+3\cdot\dfrac{\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=1+\dfrac{3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{\left(x+3\right)\left(2x+10\right)+3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+10x+6x+30+3x-6}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+19x-6}{\left(x+3\right)\left(2x+10\right)}\)
N
10 tháng 10 2021
\(a.\left(x^2+4x+4\right)+\left(x^2-6x+9\right)=2x^2+14x\)
\(x^2+4x+4+x^2-6x+9-2x^2-14x=0\)
\(-18x+13=0\)
\(x=\dfrac{13}{18}\)
Vậy \(S=\left\{\dfrac{13}{18}\right\}\)
\(b.\left(x-1\right)^3-125=0\)
\(\left(x-1\right)^3=125\)
\(x-1=5\)
\(x=6\)
Vậy \(S=\left\{6\right\}\)
\(c.\left(x-1\right)^2+\left(y +2\right)^2=0\)
\(Do\left(x-1\right)^2\ge0\forall x;\left(y+2\right)^2\ge0\forall y\)
\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x,y\)
Mà \(\left(x-1\right)^2+\left(y+2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Vậy \(S=\left\{1;-2\right\}\)
\(d.x^2-4x+4+x^2-2xy+y^2=0\)
\(\left(x-2\right)^2+\left(x-y\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-2\right)^2=0\\\left(x-y\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-y=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
Vậy \(S=\left\{2;2\right\}\)
7 tháng 11 2021
a) \(\Leftrightarrow x^2-x-x^2+2x=5\)
\(\Leftrightarrow x=5\)
b) \(\Leftrightarrow4x\left(x^2-9\right)=0\)
\(\Leftrightarrow4x\left(x-3\right)\left(x+3\right)=0
\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}4x=0\\x-3=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
Vậy x = 0 , x = 3 hoặc x = -3
7 tháng 11 2021
\(a,\Leftrightarrow x^2-x-x^2+2x=5\\ \Leftrightarrow x=5\\ b,\Leftrightarrow4x\left(x^2-9\right)=0\\ \Leftrightarrow4x\left(x-3\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow2x\left(x-1\right)-\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(2x-x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\\ d,\Leftrightarrow\left(x^2-9x+14\right)\left(x^2-9x+20\right)-72=0\\ \Leftrightarrow\left(x^2-9x+17\right)^2-3^2-72=0\\ \Leftrightarrow\left(x^2-9x+17\right)^2-81=0\\ \Leftrightarrow\left(x^2-9x+17-9\right)\left(x^2-9x+17+9\right)=0\\ \Leftrightarrow\left(x-8\right)\left(x-1\right)\left(x^2-9x+26\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=1\\\left(x-\dfrac{9}{2}\right)^2+\dfrac{23}{4}=0\left(vô.n_0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=8\end{matrix}\right.\)
2x2 + 2x +\(\frac{1}{2}\)= 0
<=> 2 ( x2 + x +\(\frac{1}{4}\)) = 0
<=> ( x +\(\frac{1}{2}\))2 = 0
<=> x +\(\frac{1}{2}\)=0
<=> x =\(-\frac{1}{2}\)
\(2x^2+2x+\frac{1}{2}=2\left(x^2+x+\frac{1}{4}\right)=\)\(2\left(x+\frac{1}{2}\right)^2=0\)
=>\(\left(x+\frac{1}{2}\right)^2=0\)
=> \(x+\frac{1}{2}=0\)
=> \(x=\frac{-1}{2}\)