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NM
3
VN
23 tháng 12 2016
a) \(x^2-x+x-1-x^2+2x=3\)
(=) \(-1+2x=3\)
(=) \(2x=3+1\)
(=) \(2x=4\)
(=) \(x=4:2\)
(=) \(x=2\)
b) \(x^2-x-3x+3=0\)
(=) \(x\left(x-1\right)-3\left(x-1\right)=0\)
(=) \(\left(x-1\right)\left(x-3\right)=0\)
(=) \(\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}}\)(=) \(\orbr{\begin{cases}x=0+1\\x=0+3\end{cases}}\)(=) \(\orbr{\begin{cases}x=1\\x=3\end{cases}}\)
23 tháng 12 2016
a.)(x+1)(x-1)-x(x+2)=3
x^2-1-x^2-2x=3
-1-2x=3
x=-2
b.)x^2-4x+3=0
x^2-3x-x+3=0
x(x-3)-(x-3)=0
(x-3)(x-1)=0
suy ra x-3=0 hoặc x-1=0
x=3 hoặc x=1
NT
0
29 tháng 11 2023
a: \(x^3-4x^2-x+4=0\)
=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)
=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(x^2-1\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)
b: Sửa đề: \(x^3+3x^2+3x+1=0\)
=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)
=>\(\left(x+1\right)^3=0\)
=>x+1=0
=>x=-1
c: \(x^3+3x^2-4x-12=0\)
=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)
=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)
=>\(\left(x+3\right)\left(x^2-4\right)=0\)
=>\(\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)
d: \(\left(x-2\right)^2-4x+8=0\)
=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)
=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)
=>\(\left(x-2\right)\left(x-2-4\right)=0\)
=>(x-2)(x-6)=0
=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
NT
5
11 tháng 9 2017
a) \(x^3\)\(-\)\(\frac{1}{4}x\)\(=\)\(0\)
\(x\left(x^2-\frac{1}{4}\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x^2-\frac{1}{4}=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x^2=0,5^2\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=+-0,5\end{cases}}\)
Vậy .............................
b) \(\left(2x-1\right)^2\)\(-\)\(\left(x+3\right)^2\)\(=\)\(0\)
\(\left(2x-1+x+3\right)\left(2x-1-x-3\right)=0\)
\(\left(3x+2\right)\left(x-4\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}3x+2=0\\x-4=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}3x=-2\\x=4\end{cases}}\)\(\orbr{\begin{cases}x=\frac{-2}{3}\\x=4\end{cases}}\)
Vậy ................................
c) \(x^2\)\(\left(x-3\right)\)\(+\)\(12\)\(-\)\(4x\)\(=\)\(0\)
\(x^2\)\(\left(x-3\right)\)\(-\)\(4\)\(\left(x-3\right)\)\(=\)\(0\)
\(\left(x^2-4\right)\left(x-3\right)\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x^2\\x-3=0\end{cases}-4=0}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x^2\\x=3\end{cases}=2^2}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=+-2\\x=3\end{cases}}\)
11 tháng 9 2017
a)\(x^3-\frac{1}{4}x=0\)
\(\Leftrightarrow x\left(x^2-\frac{1}{4}\right)=0\)
\(\Leftrightarrow x\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x-\frac{1}{2}=0\\x+\frac{1}{2}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}}\)
N
1
4 tháng 8 2018
\(4x^2+4x-3=0\)
\(\left[\left(2x\right)^2+2.2x.1+1\right]-4=0\)
\(\left(2x+1\right)^2-2^2=0\)
\(\left(2x+1-2\right).\left(2x+1+2\right)=0\)
\(\left(2x-1\right).\left(2x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-1=0\\2x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{3}{2}\end{cases}}}\)
Vậy \(\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{3}{2}\end{cases}}\)
\(x^4-3x^3-x+3=0\)
\(x^3.\left(x-3\right)-\left(x-3\right)=0\)
\(\left(x-3\right).\left(x^3-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x^3-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)
Vậy \(\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
\(x^2.\left(x-1\right)-4x^2+8x-4=0\)
\(x^2.\left(x-1\right)-\left[\left(2x\right)^2-2.2x.2+2^2\right]=0\)
\(x^2.\left(x-1\right)-\left(2x-2\right)^2=0\)
\(x^2.\left(x-1\right)-4.\left(x-1\right)^2=0\)
\(\left(x-1\right).\left[x^2-4.\left(x-1\right)\right]=0\)
\(\left(x-1\right).\left[x^2-2.x.2+2^2\right]=0\)
\(\left(x-1\right).\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)
Vậy \(\begin{cases}x=1\\x=2\end{cases}\)
Tham khảo nhé~
N
0
14 tháng 10 2018
a) \(x^2-4x=0\)
\(x\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}}\)
b) \(4x^2-9=0\)
\(\left(2x\right)^2-3^2=0\)
\(\left(2x+3\right)\left(2x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+3=0\\2x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-3}{2}\\x=\frac{3}{2}\end{cases}}}\)
c) \(2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\left(x-3\right)\left(2x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\2x+5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{2}\end{cases}}}\)
d) \(x\left(2x+9\right)-4x-18=0\)
\(x\left(2x+9\right)-2\left(2x+9\right)=0\)
\(\left(2x+9\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+9=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-9}{2}\\x=2\end{cases}}}\)
e) \(\left(2x-1\right)^2-\left(x+2\right)^2=0\)
\(\left(2x-1-x-2\right)\left(2x-1+x+2\right)=0\)
\(\left(x-3\right)\left(3x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\3x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{-1}{3}\end{cases}}}\)
T
14 tháng 10 2018
\(x^2-4x=0\)
\(x.\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-4=0\Leftrightarrow x=4\end{cases}}\)
\(4x^2-9=0\)
\(2^2x^2-9=0\)
\(\left(2x\right)^2-9=0\)
\(\left(2x\right)^2-3^2=0\)
\(\Rightarrow\orbr{\begin{cases}\left(2x\right)^2=\left(-3\right)^2\\\left(2x\right)^2=3^2\end{cases}\Rightarrow\orbr{\begin{cases}2x=-3\\2x=3\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-3}{2}\\x=\frac{3}{2}\end{cases}}}}\)
\(2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\left(x-3\right)\cdot\left(2x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-3\right)=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0+3\\2x=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{2}\end{cases}}}\)
\(x\left(2x+9\right)-4x-18=0\)
\(x\left(2x+9\right)-\left(4x+18\right)=0\)
\(x\left(2x+9\right)-\left(2\cdot2x+2\cdot9\right)=0\)
\(x\left(2x+9\right)-2.\left(2x+9\right)=0\)
\(\left(2x+9\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}2x+9=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-9\\x=0+2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-9}{2}\\x=2\end{cases}}}\)
\(\left(2x-1\right)^2-\left(x+2\right)^2=0\)
\(\Rightarrow\left(2x-1\right)^2=\left(x+2\right)^2\)
\(\Rightarrow\orbr{\begin{cases}2x-1=x+2\\2x-1=-x+2\end{cases}\Rightarrow\orbr{\begin{cases}2x=3+x\\2x=-x+3\end{cases}\Rightarrow\orbr{\begin{cases}2x-x=3\\2x+x=3\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}}}\)
\(\)
14 tháng 11 2021
\(4x\left(3-\dfrac{1}{4}x\right)+\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow12x-x^2+x^2-4=0\Rightarrow12x=4\Rightarrow x=\dfrac{1}{3}\)
NH
14 tháng 11 2021
\(12x-x^2+x^2-2^2=0\)
\(12x-2=0\)
\(12x=2\)
\(x=\dfrac{1}{6}\)
Vậy x=1/6
E
15 tháng 9 2021
a)\(3x^2-4x=0<=>x(3x-4)=0\)
TH1: x=0
TH2 3x-4=0 <=>x=4/3
KL:.....
b) (x+3)(x−1)+2x(x+3)=0.
<=> (x+3)(x-1+2x)=0
TH1: x+3=0 <=> x=-3
TH2 x-1=0 <=> x=1
KL:.....
c) \(9x^2+6x+1=0. <=>(3x+1)^2=0<=>3x+1=0<=>x=-1/3 \)
KL:......
d) \(x^2−4x=4.<=>(x-2)^2=0<=>x-2=0<=>x=2\)
KL:....
15 tháng 9 2021
a) \(3x^2-4x=0\)
\(\Leftrightarrow x\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{3}\end{matrix}\right.\)
b) \(\left(x+3\right)\left(x-1\right)+2x\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{3}\end{matrix}\right.\)
c) \(9x^2+6x+1=0\)
\(\Leftrightarrow\left(3x+1\right)^2=0\)
\(\Leftrightarrow3x+1=0\Leftrightarrow x=-\dfrac{1}{3}\)
d) \(x^2-4x=4\)
\(\Leftrightarrow\left(x-2\right)^2=8\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=2\sqrt{2}\\x-2=-2\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}+2\\x=-2\sqrt{2}+2\end{matrix}\right.\)
Ta có : ( x2 - 1 )( x2 + 4x + 3 ) = 0
\(\Leftrightarrow\)( x - 1 )( x + 1 )[ ( x2 + 4x + 4 ) - 1 ] = 0
\(\Rightarrow\)( x - 1 )( x + 1 )[ ( x + 2 )2 - 12 ] = 0
\(\Rightarrow\)( x - 1 )( x + 1 )( x + 2 - 1 )( x + 2 + 1 ) = 0
\(\Rightarrow\)( x - 1 )( x + 1 )( x + 1 )( x + 3 ) = 0
\(\Rightarrow\)( x - 1 )( x + 1 )2( x + 3 ) = 0
\(\Rightarrow\)x - 1 = 0 hoặc ( x + 1 )2 = 0 hoặc x + 3 = 0
\(\Rightarrow\)x = 1 hoặc x = - 1 hoặc x = - 3
Vậy : x = 1 hoặc x = - 1 hoặc x = - 3
(x^2-1)(x^2+4x+3)=0
(x-1)(x+1)(x^2+3x+x+3)=0
(x-1)(x+1)[ x(x+1)+3(x+1)]=0
(x-1)(x+1)(x+3)(x+1)=0
x-1=0 hoặc x+1=0 hoặc x+3=0
=> x=1;-1;-3