3x=5x=0
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CH
1
IM
8 tháng 8 2016
a)
\(\Rightarrow x\left(x-5\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x-5=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=5\end{array}\right.\)
b)
\(\Rightarrow3x\left(x-2\right)-2\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(3x-2\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x-2=0\\3x-2=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=2\\x=\frac{2}{3}\end{array}\right.\)
c)
\(\Rightarrow\left(3x-1\right)\left(5x+x-2\right)=0\)
\(\Rightarrow\left(3x-2\right)^2.2=0\)
\(\Rightarrow3x-2=0\)
\(\Rightarrow x=\frac{2}{3}\)
NT
2
28 tháng 6 2018
\(a)\) \(\left(5x-1\right)^2-5x\left(5x-1\right)=0\)
\(\Leftrightarrow\)\(\left(5x-1\right)\left(5x-1-5x\right)=0\)
\(\Leftrightarrow\)\(\left(5x-1\right).\left(-1\right)=0\)
\(\Leftrightarrow\)\(5x-1=0\)
\(\Leftrightarrow\)\(5x=1\)
\(\Leftrightarrow\)\(x=\frac{1}{5}\)
Vậy \(x=\frac{1}{5}\)
\(b)\) \(x\left(x+1\right)\left(x+2\right)=0\)
Suy ra \(x=0\) hoặc \(x+1=0\) hoặc \(x+2=0\)
\(\Leftrightarrow\)\(x=0\) hoặc \(x=-1\) hoặc \(x=-2\)
Vậy \(x=0\) hoặc \(x=-1\) hoặc \(x=-2\)
\(c)\) \(\left(3x+2\right)x-3\left(3x+2\right)=0\)
\(\Leftrightarrow\)\(\left(3x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}3x+2=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=0-2\\x=0+3\end{cases}}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}3x=-2\\x=3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-2}{3}\\x=3\end{cases}}}\)
Vậy \(x=\frac{-2}{3}\) hoặc \(x=3\)
Chúc bạn học tốt ~
HH
28 tháng 6 2018
a/ \(\left(5x-1\right)^2-5x\left(5x-1\right)=0\)
<=> \(\left(5x-1\right)\left(5x-1-5x\right)=0\)
<=> \(-1\left(5x-1\right)=0\)
<=> \(5x-1=0\)
<=> \(5x=1\)
<=> \(x=\frac{1}{5}\)
b/ \(x\left(x+1\right)\left(x+2\right)=0\)
<=> \(x=0\) hoặc \(\orbr{\begin{cases}x+1=0\\x+2=0\end{cases}}\)
<=> \(x=0\)hoặc \(\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)
c/ \(\left(3x+2\right)x-3\left(3x+2\right)=0\)
<=> \(\left(3x+2\right)\left(x-3\right)=0\)
<=> \(\orbr{\begin{cases}3x+2=0\\x-3=0\end{cases}}\)
<=> \(\orbr{\begin{cases}3x=-2\\x=3\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-\frac{2}{3}\\x=3\end{cases}}\)
HN
1
28 tháng 9 2021
a: Ta có: \(x^2+3x-10=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
b: Ta có: \(x^2-5x-6=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-1\end{matrix}\right.\)
BT
14 tháng 7 2016
a)5x ( 1 - 2x ) - 3x ( x + 18 ) = 0
5x - 10x^2 - 3x^2 - 54x =0
-13x^2 - 49x =0
(-13x - 49)x =0
\(\Rightarrow\orbr{\begin{cases}-13x-49=0\\x=0\end{cases}\Rightarrow\orbr{\begin{cases}-13x=49\\x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{49}{13}\\x=0\end{cases}}}\)
Vậy x= -49/13 hoặc x=0
b)5x - 10x^2 - 3x^2 - 54 = 0
(câu b giống câu a)
20 tháng 11 2021
a) \(\Rightarrow\left(x-2\right)\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
b) \(\Rightarrow\left(x-3\right)\left(5x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)
c) \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)
d) \(\Rightarrow\left(x-7\right)\left(3x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{2}{3}\end{matrix}\right.\)
N
20 tháng 11 2021
\(a,\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ b,\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\\ c,\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\\ d,\Leftrightarrow\left(x-7\right)\left(3x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{2}{3}\end{matrix}\right.\)
20 tháng 8 2017
Ta có : x(x - 3) - 2x + 6 = 0
<=> x(x - 3) - (2x - 6) = 0
=> x(x - 3) - 2(x - 3) = 0
=> (x - 2)(x - 3) = 0
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
NG
0
ND
2
TN
16 tháng 6 2017
a)\(3x\left(x-2\right)+2\left(2-x\right)=0\)
\(\Leftrightarrow3x\left(x-2\right)-2\left(x-2\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-2=0\\x-2=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=2\end{cases}}\)
b)\(5x\left(3x-1\right)+x\left(3x-1\right)-2\left(3x-1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(5x+x-2\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(6x-2\right)=0\)
\(\Leftrightarrow2\left(3x-1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)^2=0\Rightarrow3x-1=0\Rightarrow x=\frac{1}{3}\)
16 tháng 6 2017
a/3x(x-2)+2(2-x)=0
=>(2-3x)(2-x)=0
=>\(\orbr{\begin{cases}2-3x=0\\2-x=0\end{cases}}\)=>\(\orbr{\begin{cases}3x=2\\x=2\end{cases}}\)=>\(\orbr{\begin{cases}x=\frac{2}{3}\\x=2\end{cases}}\)
b/5x(3x-1)+x(3x-1)-2(3x-1)=0
=>(5x+x-2)(3x-1)=0
=>(6x-2)(3x-1)=0
=>\(\orbr{\begin{cases}6x-2=0\\3x-1=0\end{cases}}\)=>\(\orbr{\begin{cases}6x=2\\3x=1\end{cases}}\)=>x=\(\frac{1}{3}\)
CW
7 tháng 7 2016
1) x (x-2016) + 2015 (2016-x) = 0
x (x-2016) - 2015 (x- 2016) = 0
(x-2015)(x-2016) =0
\(\Rightarrow\orbr{\begin{cases}x-2015=0\\x-2016=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2015\\x=2016\end{cases}}}\)
Vậy x= 2015; 2016
2) -5x (x-15) + (15-x) = 0
-5x (x-15) - (x-15) =0
(-5x -1) (x-15) =0
\(\Rightarrow\orbr{\begin{cases}-5x-1=0\\x-15=0\end{cases}\Rightarrow\orbr{\begin{cases}-5x=1\\x=15\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{5}\\x=15\end{cases}}}\)
Vậy x= -1/5; 15
3) 3x (3x-7) - (7-3x) =0
3x(3x-7) + (3x -7) =0
(3x+1) (3x-7) =0
\(\Rightarrow\orbr{\begin{cases}3x+1=0\\3x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=-1\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{3}\\x=\frac{7}{3}\end{cases}}}\)
Vậy x= -1/3 ; 7/3
x=0 tick cho mình nha miko hậu đậu