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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: \(x^2-5x-6=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-1\end{matrix}\right.\)
22 tháng 9 2020
a) ( x - 1 )2 - ( x - 1 )( x + 1 ) = 0
<=> x2 - 2x + 1 - ( x2 - 1 ) = 0
<=> x2 - 2x + 1 - x2 + 1 = 0
<=> 2 - 2x = 0
<=> 2x = 2
<=> x = 1
b) ( 2x - 1 )2 - ( 2x + 1 )2 = 0
<=> [ ( 2x - 1 ) - ( 2x + 1 ) ][ ( 2x - 1 ) + ( 2x + 1 ) ] = 0
<=> ( 2x - 1 - 2x - 1 )( 2x - 1 + 2x + 1 ) = 0
<=> -2.4x = 0
<=> -8x = 0
<=> x = 0
c) 25( x + 3 )2 + ( 1 - 5x )( 1 + 5x ) = 8
<=> 52( x + 3 )2 + 12 - 25x2 = 8
<=> [ 5( x + 3 ) ]2 + 1 - 25x2 = 8
<=> ( 5x + 15 )2 + 1 - 25x2 = 8
<=> 25x2 + 150x + 225 + 1 - 25x2 = 8
<=> 150x + 226 = 8
<=> 150x = -218
<=> x = -218/150 = -109/75
d) 9( x + 1 )2 - ( 3x - 2 )( 3x + 2 ) = 10
<=> 32( x + 1 )2 - ( 9x2 - 4 ) = 10
<=> [ 3( x + 1 ) ]2 - 9x2 + 4 = 10
<=> ( 3x + 3 )2 - 9x2 + 4 = 10
<=> 9x2 + 18x + 9 - 9x2 + 4 = 10
<=> 18x + 13 = 10
<=> 18x = -3
<=> x = -3/18 = -1/6
22 tháng 9 2020
a) (x - 1)2 - (x - 1)(x + 1) = 0
=> (x - 1)2 - (x2 - 12) = 0
=> x2 - 2.x.1 + 12 - x2 + 1 = 0
=> x2 - 2x + 1 - x2 + 1 = 0
=> -2x + 1 + 1 = 0
=> -2x + 2 = 0
=> -2x = -2 => x = 1
b) (2x - 1)2 - (2x + 1)2 = 0
=> (2x - 1 - 2x + 1)(2x - 1 + 2x + 1) = 0
=> 0 = 0(đúng)
c) 25(x + 3)2 + (1 - 5x)(1 + 5x) = 8
=> 25(x2 + 2.x.3 + 32) + (12 - (5x)2) = 8
=> 25x2 + 150x + 225 + 1 - 25x2 = 8
=> 150x +225 + 1 = 8
=> 150x = -218
=> x = -109/75
d) 9(x + 1)2 - (3x - 2)(3x + 2) = 10
=> 9(x2 + 2x + 1) - [(3x)2 - 22 ] = 10
=> 9x2 + 18x + 9 - (9x2 - 4) = 10
=> 9x2 + 18x + 9 - 9x2 + 4 = 10
=> 18x + 9 + 4 = 10
=> 18x = -3
=> x = -1/6
28 tháng 7 2018
\(x\left(x-3\right)+x-3=0\)
\(\left(x-3\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}}\)
KL:......................
\(x^3-5x=0\)
\(x\left(x^2-5\right)=0\)
Làm tương tự như câu a
@_@ n...h..i......ề....u q...u.....................á!
GH
28 tháng 2 2018
a) \(\left(x-3\right)\left(2x+1\right)\left(4-5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\\4-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{-1}{2}\\x=\dfrac{4}{5}\end{matrix}\right.\)
Vậy ..................
b) \(2x^3-5x^2+3x=0\)
\(\Leftrightarrow x\left(2x^2-5x+3\right)=0\)
\(\Leftrightarrow x\left(2x^2-2x-3x+3\right)=0\)
\(\Leftrightarrow x\left[2x\left(x-1\right)-3\left(x-1\right)\right]=0\)
\(\Leftrightarrow x\left(x-1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy .................
c) \(\left(x-3\right)^2=\left(2x+1\right)^2\)
\(\Leftrightarrow\left(2x+1\right)^2-\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(2x+1-x+3\right)\left(2x+1+x-3\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy .......................
d) \(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-3x-4x+12\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left[x\left(x-3\right)-4\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=3\\x=4\end{matrix}\right.\)
Vậy ...................
TT
28 tháng 2 2018
a,\(\left(x-3\right)\left(2x+1\right)\left(4-5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\\4-5x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\\x=\dfrac{4}{5}\end{matrix}\right.\)
Vậy...
b,\(2x^3-5x^2+3x=0\)
\(\Leftrightarrow x\left(2x^2-5x+3\right)=0\)
\(\Leftrightarrow x\left(2x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-3=0\\x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\\x=1\end{matrix}\right.\)
Vậy...
c,Sửa đề:
\(\left(x-3\right)^2=\left(2x+1\right)^2\)
\(\Leftrightarrow\left(x-3\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left(x-3+2x+1\right)\left(x-3-2x-1\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(-x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\-x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-4\end{matrix}\right.\)
Vậy...
d,\(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x+4\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x+4=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-4\\x=3\end{matrix}\right.\)
Vậy...
18 tháng 6 2019
a,2x^2=x
=>2x^2 -x=0
=>x(2x-1)=0
th1 x=0
th2 2x-1=0 => 2x=1 =>x=1/2
18 tháng 6 2019
b,x^2-1/36=0
<=> x^2-(1/6)^2=0
<=> (x-1/6)(x+1/6)=0
th1 x-1/6=0 =>x=1/6
th2 x+1/6=0 =>x=-1/6
21 tháng 10 2020
a) \(7x-10=5x-6\)
\(7x-5x=-6+10\)
\(2x=4\)
\(x=2\)
b) \(3x\left(x-2\right)+x-2=0\)
\(\left(x-2\right)\left(3x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\3x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-\frac{1}{3}\end{cases}}\)
c) \(2x^2+7x-4=0\)
\(2x^2-x+8x-4=0\)
\(x\left(2x-1\right)+2\left(2x-1\right)=0\)
\(\left(2x-1\right)\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-1=0\\x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-2\end{cases}}\)
21 tháng 10 2020
7x-10=5x-6<=>7x-5x=-6+10<=>2x=4=>x=2
3x(x-2)+x-2=0<=>(x-2)(3x+1)=0<=>x-2=0=>x=2 HAY 3x+1=0=>x=-1/3
2x2+7x-4=0.
Câu cuối xem có lộn đề không nha bạn ơi!!!
3x( x - 2 ) + 10 - 5x = 0
<=> 3x( x - 2 ) - ( 5x - 10 ) = 0
<=> 3x( x - 2 ) - 5( x - 2 ) = 0
<=> ( x - 2 )( 3x - 5 ) = 0
<=> \(\orbr{\begin{cases}x-2=0\\3x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{5}{3}\end{cases}}\)