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TK
6
26 tháng 8 2020
a) 5x( x - 1 ) = x - 1
<=> 5x2 - 5x = x - 1
<=> 5x2 - 5x - x + 1 = 0
<=> 5x2 - 6x + 1 = 0
<=> 5x2 - 5x - x + 1 = 0
<=> 5x( x - 1 ) - 1( x - 1 ) = 0
<=> ( x - 1 )( 5x - 1 ) = 0
<=> \(\orbr{\begin{cases}x-1=0\\5x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)
b) 2( x + 5 ) - x2 - 5x = 0
<=> 2x + 10 - x2 - 5x = 0
<=> -x2 - 3x + 10 = 0
<=> -x2 - 5x + 2x + 10 = 0
<=> -x( x + 5 ) + 2( x + 5 ) = 0
<=> ( x + 5 )( 2 - x ) = 0
<=> \(\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)
c) x2 - 2x - 3 = 0
<=> x2 + x - 3x - 3 = 0
<=> x( x + 1 ) - 3( x + 1 ) = 0
<=> ( x + 1 )( x - 3 ) = 0
<=> \(\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
d) 2x2 + 5x - 3 = 0
<=> 2x2 - x + 6x - 3 = 0
,<=> x( 2x - 1 ) + 3( 2x - 1 ) = 0
<=> ( 2x - 1 )( x + 3 ) = 0
<=> \(\orbr{\begin{cases}2x-1=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-3\end{cases}}\)
PN
26 tháng 8 2020
a) 5x ( x - 1 ) = x - 1 <=> 5x2 - 5x - x + 1 = 0
<=> 5x2 - 6x + 1 = 0 <=> 5x2 - x - ( 5x - 1 ) = 0
<=> x ( 5x - 1 ) - ( 5x - 1 ) = 0 <=> ( x - 1 )( 5x - 1 ) = 0
<=> x = 1 hoặc x = 1/5
b) 2 ( x + 5 ) - x2 - 5x = 0 <=> 2 ( x + 5 ) - x ( x + 5 ) = 0
<=> ( 2 - x ) ( x + 5 ) = 0 <=> x = 2 hoặc x = -5
c) x2 - 2x - 3 = 0 <=> x2 + x - 3x - 3 = 0
<=> x ( x + 1 ) - 3 ( x + 1 ) = 0 <=> ( x - 3 ) ( x + 1 ) = 0
<=> x = 3 hoặc x = -1
d) 2x2 + 5x - 3 = 0
Ta có : delta = 52 - 4.2.3 = 25 - 24 = 1
Khi đó : x = -1 hoặc x = 3/2
28 tháng 9 2018
\(2x\left(x-3\right)-x+3=0\)
<=> \(2x\left(x-3\right)-\left(x-3\right)=0\)
<=> \(\left(x-3\right)\left(2x-1\right)=0\)
<=> \(\orbr{\begin{cases}x=3\\x=\frac{1}{2}\end{cases}}\)
Vậy...
5 tháng 12 2017
1)⇔x2+1x-3x+3=0
⇔x(x+1)-3(x+1)=0
⇔(x+1)(x-3)=0
⇔x+1=0 hoặc x-3=0
⇔x=-1 hoặc x=3
5 tháng 12 2017
4)⇔x(1+5x)=0
⇔x=0 hoặc 1+5x=0
⇔x=0 hoặc 5x=-1
⇔x=0 hoặc x=-0.2
18 tháng 12 2016
a) \(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(\Leftrightarrow2x^2-10x-3x-2x^2=26\)
\(\Leftrightarrow-13x=26\Leftrightarrow x=-2\)
b) \(5x\left(x-1\right)=x-1\)
\(\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=\frac{1}{5}\end{array}\right.\)
c) \(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-5\\x=2\end{array}\right.\)
d) \(\left(2x-3\right)^2-\left(x+5\right)^2=0\)
\(\Leftrightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)
\(\Leftrightarrow\left(x-8\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=8\\x=-\frac{2}{3}\end{array}\right.\)
e) \(3x^3-48x=0\)
\(\Leftrightarrow3x\left(x^2-16\right)=0\)
\(\Leftrightarrow3x\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=4\\x=-4\end{array}\right.\)
f) \(x^3+x^2-4x=4\)
\(\Leftrightarrow x^2\left(x+1\right)-4\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=2\\x=-2\end{array}\right.\)
NN
8 tháng 12 2019
Ko viết lại đề
Câu 1: chia ra làm 3 trường hợp
Câu 2:
\(\left(x+2-x+2\right)\left(x+2\right)=0\)
\(4\left(x+2\right)=0\)
\(\Rightarrow x+2=0\)
\(x=-2\)
N
24 tháng 8 2017
a, \(2\left(x+5\right)-x^2-5x=0\)
\(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x=2\end{cases}}\)
11 tháng 4 2020
b)
\(2x\cdot\left(2x-3\right)=\left(3-2x\right)\cdot\left(2-5x\right)\\ \Leftrightarrow-2x\cdot\left(3-2x\right)-\left(3-2x\right)\cdot\left(2-5x\right)=0\\ \Leftrightarrow\left(3-2x\right)\cdot\left(-2x-2+5x\right)=0\\ \Leftrightarrow\left(3-2x\right)\cdot\left(3x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3-2x=0\\3x-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{3}{2}\\x=\frac{2}{3}\end{matrix}\right.\)
c)
\(2x^3+6x^2=x^2+3x\\ \Leftrightarrow2x^3+6x^2-x^2-3x=0\\ \Leftrightarrow x\cdot\left(2x^2+6x-x-3\right)=0\\ \Leftrightarrow x\cdot\left(-3+6x-x+2x^2\right)=0\\ \Leftrightarrow x\cdot\left[-3\cdot\left(1-2x\right)-x\cdot\left(1-2x\right)\right]=0\\ \Leftrightarrow x\cdot\left(-3-x\right)\cdot\left(1-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\-3-x=0\\1-2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=-3\\x=\frac{1}{2}\end{matrix}\right.\)
d)
\(x^2-5x+6=0\\ \Leftrightarrow x^2-3x-2x+6=0\\ \Leftrightarrow6-2x-3x+x^2=0\\ \Leftrightarrow2\cdot\left(3-x\right)-x\cdot\left(3-x\right)=0\\ \Leftrightarrow\left(2-x\right)\cdot\left(3-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2-x=0\\3-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
e)
\(\left(2x+5\right)^2=\left(x+2\right)^2\\ \Leftrightarrow\left(2x+5\right)^2-\left(x+2\right)^2=0\\ \Leftrightarrow\left(2x+5+x+2\right)\cdot\left(2x+5-x-2\right)=0\\ \Leftrightarrow\left(3x+7\right)\cdot\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x+7=0\\x+3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\frac{7}{3}\\x=-3\end{matrix}\right.\)
S
11 tháng 4 2020
a) \(\left(x+3\right)\left(x+5\right)+\left(x+3\right)\left(3x-4\right)=0\)
➜\(\left(x+3\right)\left(x+5+1+3x-4\right)=0\)
➜\(\left[{}\begin{matrix}x+3=0\\x+3x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=\frac{1}{2}\end{matrix}\right.\)
Mk đang hok zoom sorry nha!!!
VM
25 tháng 10 2019
\(2x^2-6x=0\)
\(\Rightarrow2x.\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0:2\\x=0+3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{0;3\right\}.\)
\(2x.\left(x+2\right)-3.\left(x+2\right)=0\)
\(\Rightarrow\left(x+2\right).\left(2x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+2=0\\2x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0-2\\2x=3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3:2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=\frac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{-2;\frac{3}{2}\right\}.\)
\(x^3-16x=0\)
\(\Rightarrow x.\left(x^2-16\right)=0\)
\(\Rightarrow x.\left(x^2-4^2\right)=0\)
\(\Rightarrow x.\left(x-4\right).\left(x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=0+4\\x=0-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
Vậy \(x\in\left\{0;4;-4\right\}.\)
Chúc bạn học tốt!
A
1 tháng 9 2020
Bài 1 :
a, \(\left(x-3\right)^2-4=0\Leftrightarrow\left(x-3\right)^2=4\Leftrightarrow\left(x-3\right)^2=\left(\pm2\right)^2\)
TH1 : \(x-3=2\Leftrightarrow x=5\)
TH2 : \(x-3=-2\Leftrightarrow x=1\)
b, \(x^2-2x=24\Leftrightarrow x^2-2x-24=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+4\right)=0\)
TH1 : \(x-6=0\Leftrightarrow x=6\)
TH2 : \(x+4=0\Leftrightarrow x=-4\)
c, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-4\right)=0\)
\(\Leftrightarrow2x+30=0\Leftrightarrow x=-15\)
d, tương tự
\(\left(x-5\right)^2+\left(2x+3\right)^2-\left(5x-3\right)\left(x+2\right)=0\)
\(x^2-10x+25+4x^2+12x+9-5x^2-10x+3x+6=0\)
\(-5x+40=0\)
\(-5\left(x-8\right)=0\)
\(x-8=0\)
\(x=8\)