Tìm x, biết:
a) ( x-1) x ( x-2) = 0
b) ( x+1) + ( x+2) + ... + ( x+100) = 9050
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CM
8
17 tháng 10 2021
\(a,\Leftrightarrow3\left(x+3\right)=0\Leftrightarrow x=-3\\ b,\Leftrightarrow\left(x^2-2\right)\left(6x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=2\\6x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\\x=-\dfrac{1}{6}\end{matrix}\right.\\ c,\Leftrightarrow\left(x-2013\right)\left(4x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2013\\x=\dfrac{1}{4}\end{matrix}\right.\\ d,\Leftrightarrow\left(x+1\right)^2-\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x+1-1\right)=0\\ \Leftrightarrow x\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
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 8 2021
b: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x^2+1\right)=4\)
\(\Leftrightarrow x^3-1-x^3-x=4\)
\(\Leftrightarrow-x=5\)
hay x=-5
c: Ta có: \(\left(2x-1\right)^3+\left(x+2\right)^3-9x\left(x+1\right)\left(x-1\right)=7\)
\(\Leftrightarrow8x^3-12x^2+6x-1+x^3+6x^2+12x+8-9x^3+9x=7\)
\(\Leftrightarrow-6x^2+27x=0\)
\(\Leftrightarrow-3x\left(2x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{9}{2}\end{matrix}\right.\)
20 tháng 10 2021
a: \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
b: \(\left(x+1\right)^2-4\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
23 tháng 10 2021
\(a,\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-2\right)^3=0\Leftrightarrow x-2=0\Leftrightarrow x=2\\ c,\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
13 tháng 9 2021
a) \(\left(x+3\right)\left(2x-1\right)-\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow2x^2+5x-3-x^2+2x+3=0\)
\(\Leftrightarrow x^2+7x=0\Leftrightarrow x\left(x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\end{matrix}\right.\)
b) \(\left(x+4\right)\left(2x-3\right)-3\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow2x^2+5x-12-3x^2+12=0\)
\(\Leftrightarrow x^2-5x=0\Leftrightarrow x\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
TP
25 tháng 12 2022
\((2x-1)^2+(x+3)^2-5(x+7)(x-7)=0\)
\(< =>4x^2-4x+1+x^2+6x+9-5\left(x^2-7^2\right)=0\\ < =>4x^2-4x+1+x^2+6x+9-5x^2+245=0\\ < =>2x+255=0\\ < =>2x=-255=>x=\dfrac{-255}{2}\)
Vậy \(x=\dfrac{-255}{2}\)
DX
25 tháng 12 2022
\(\Rightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)
\(\Rightarrow2x+255=0\Rightarrow2x=-255\Rightarrow x=-\dfrac{255}{2}\)
22 tháng 9 2021
\(\left|x-3\right|+\left|x-\dfrac{1}{2}\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\)( vô lý)
Vậy \(S=\varnothing\)
22 tháng 9 2021
b: \(\left|x-3\right|+\left|x-\dfrac{1}{2}\right|\ge0\forall x\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
a) ( x - 1 ) ( x - 2 ) = 0
=> x - 1 = 0 hoặc x - 2 = 0
x - 1 = 0 x - 2 = 0
x = 0 + 1 = 1 x = 0 + 2 = 2
Vậy x = 1 hoặc x = 2
b) ( x + 1 ) + ( x + 2 ) + ... + ( x + 100 ) = 9050
( x + x + ... + x ) + ( 1 + 2 + ... + 100 ) = 9050
=> 100x + 5050 = 9050
100x = 9050 - 5050
100x = 4000
x = 4000 : 100
x = 40