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2
25 tháng 10 2021
\(a,x^2-5x\)
\(=x\left(x-5\right)\)
\(b,5x\left(x+5\right)+4x+20\)
\(=5x\left(x+5\right)+4\left(x+5\right)\)
\(=\left(5x+4\right)\left(x+5\right)\)
\(c,7x\left(2x-1\right)-4x+2\)
\(=7x\left(2x-1\right)-2\left(2x-1\right)\)
\(=\left(7x-2\right)-\left(2x-1\right)\)
2
25 tháng 10 2021
\(d,x^2-16+2\left(x+4\right)\)
\(=x^2-16+2x+8\)
\(=x\left(x-2\right)-8\) ( Ý này thì k chắc lắm, sai thông cảm :)) )
\(e,x^2-10x+9\)
\(=x^2-x-9x+9\)
\(=x\left(x-1\right)-9\left(x-1\right)\)
\(=\left(x-9\right)\left(x-1\right)\)
\(f,\left(2x-1\right)^2-\left(x-3\right)^2=0\) ( mk đoán bài này là tìm x, sai thì bảo mk để mk sửa nhé )
\(\Rightarrow\left(2x-1\right)^2=\left(x-3\right)^2\)
\(\Leftrightarrow\pm\left(2x-1\right)=\pm\left(x-3\right)\)
\(\Rightarrow\hept{\begin{cases}2x-1=x-3\\-\left(2x-1\right)=-\left(x-3\right)\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2x-1-x+3=0\\-2x+1-x+3=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x+2=0\\-3x+4=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\left(-2\right)\\x=\frac{4}{3}\end{cases}}\)
Vậy ...
23 tháng 7 2019
1. a) Ta có: 2x2 - x + 1 = x(2x + 1) - 2x + 1 = x(2x + 1) - (2x + 1) + 2 = (x - 1)(2x + 1) + 2
Do (x - 1)(2x + 1) \(⋮\)2x + 1
=> 2 \(⋮\)2x + 1
=> 2x + 1 \(\in\)Ư(2) = {1; -1; 2; -2}
Do : 2x + 1 là số lẻ => 2x + 1 \(\in\){1; -1}
+) 2x + 1 = 1 => 2x = 0 => x = 0
+) 2x + 1 = -1 => 2x = -2 => x = -1
b) 2x + y + 2xy - 3 = 0
=> 2x(1 + y) + (1 + y) = 4
=> (2x + 1)(1 + y) = 4
=> 2x + 1;1 + y \(\in\)Ư(4) = {1; -1;2 ;-2; 4; -4}
Do: 2x + 1 là số lẻ => 2x + 1 \(\in\){1; -1}
=> 1 + y \(\in\){4; -4}
Lập bảng :
| 2x + 1 | 1 | -1 |
| 1 + y | 4 | -4 |
| x | 0 | -1 |
| y | 3 | -5 |
Vậy ....
c) x2 + 2xy = 0
=> x(x + 2y) = 0
=> \(\hept{\begin{cases}x=0\\x+2y=0\end{cases}}\)
=> \(\hept{\begin{cases}x=0\\2y=0\end{cases}}\)
=> \(\hept{\begin{cases}x=0\\y=0\end{cases}}\)
Vậy x = y = 0
27 tháng 5 2022
a: \(\left(3x-1\right)^2-\left(x+3\right)^3=\left(2-x\right)\left(x^2+2x+4\right)\)
\(\Leftrightarrow9x^2-6x+1-x^3-9x^2-27x-27=8-x^3\)
\(\Leftrightarrow-x^3-33x-26-8+x^3=0\)
=>-33x=34
hay x=-34/33
b: \(\left(x+1\right)\left(x-1\right)\left(x^2+1\right)-\left(x^2-1\right)^2=2\)
\(\Leftrightarrow\left(x^2+1\right)\left(x^2-1\right)-\left(x^2-1\right)^2=2\)
\(\Leftrightarrow x^4-1-x^4+2x^2-1=2\)
\(\Leftrightarrow2x^2=4\)
hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)
c: \(x^2-2\sqrt{3}x+3=0\)
\(\Leftrightarrow\left(x-\sqrt{3}\right)^2=0\)
hay \(x=\sqrt{3}\)
d: \(\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)-\left(x-\sqrt{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\sqrt{2}\right)\left(x+\sqrt{2}-x+\sqrt{2}\right)=0\)
\(\Leftrightarrow x-\sqrt{2}=0\)
hay \(x=\sqrt{2}\)
A
19 tháng 10 2020
a, \(x\left(x+1\right)-x\left(x-5\right)=6\Leftrightarrow x^2+x-x^2+5x=6\)
\(\Leftrightarrow x=1\)
b, \(4x^2-4x+1=0\Leftrightarrow\left(2x-1\right)^2=0\Leftrightarrow x=\frac{1}{2}\)
c, \(x^2-\frac{1}{4}=0\Leftrightarrow\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)=0\Leftrightarrow x=\pm\frac{1}{2}\)
d, \(5x^2=20x\Leftrightarrow5x^2-20x=0\Leftrightarrow5x\left(x-4\right)=0\Leftrightarrow x=0;4\)
e, \(4x^2-9-x\left(2x-3\right)=0\Leftrightarrow4x^2-9-2x^2=3x\Leftrightarrow2x^2-9-3x=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x-3\right)=0\Leftrightarrow x=-\frac{3}{2};3\)
f, \(4x^2-25=\left(2x-5\right)\left(2x+7\right)\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow-2\left(2x+5\right)=0\Leftrightarrow x=-\frac{5}{2}\)
19 tháng 10 2020
a) x( x + 1 ) - x( x - 5 ) = 6
⇔ x2 + x - x2 + 5x = 6
⇔ 6x = 6
⇔ x = 1
b) 4x2 - 4x + 1 = 0
⇔ ( 2x - 1 )2 = 0
⇔ 2x - 1 = 0
⇔ x = 1/2
c) x2 - 1/4 = 0
⇔ ( x - 1/2 )( x + 1/2 ) = 0
⇔ \(\orbr{\begin{cases}x-\frac{1}{2}=0\\x+\frac{1}{2}=0\end{cases}}\Leftrightarrow x=\pm\frac{1}{2}\)
d) 5x2 = 20x
⇔ 5x2 - 20x = 0
⇔ 5x( x - 4 ) = 0
⇔ \(\orbr{\begin{cases}5x=0\\x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
e) 4x2 - 9 - x( 2x - 3 ) = 0
⇔ ( 2x - 3 )( 2x + 3 ) - x( 2x - 3 ) = 0
⇔ ( 2x - 3 )( 2x + 3 - x ) = 0
⇔ ( 2x - 3 )( x + 3 ) = 0
⇔ \(\orbr{\begin{cases}2x-3=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-3\end{cases}}\)
f) 4x2 - 25 = ( 2x - 5 )( 2x + 7 )
⇔ ( 2x - 5 )( 2x + 5 ) - ( 2x - 5 )( 2x + 7 ) = 0
⇔ ( 2x - 5 )( 2x + 5 - 2x - 7 ) = 0
⇔ ( 2x - 5 )(-2) = 0
⇔ 2x - 5 = 0
⇔ x = 5/2
PT
10 tháng 9 2018
a) \(x^2-2x=24\)
\(\Rightarrow x^2-2x-24=0\)
\(\Rightarrow x^2-6x+4x-24=0\)
\(\Rightarrow x\left(x-6\right)+4\left(x-6\right)=0\)
\(\Rightarrow\left(x-6\right)\left(x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-6=0\\x+4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
b) \(\left(5-2x\right)^2-16=0\)
\(\Rightarrow\left(5-2x\right)^2-4^2=0\)
\(\Rightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\)
\(\Rightarrow\left(1-2x\right)\left(9-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}1-2x=0\\9-2x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=1\\2x=9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\)
c)Sửa đề
\(x^2-4x+4-9x^2+6x-1=0\)
\(\Rightarrow\left(x^2-4x+4\right)-\left(9x^2-6x+1\right)=0\)
\(\Rightarrow\left(x-2\right)^2-\left(3x-1\right)^2=0\)
\(\Rightarrow\left(x-2-3x+1\right)\left(x-2+3x-1\right)=0\)
\(\Rightarrow\left(-2x-1\right)\left(4x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-2x-1=0\\4x-3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-2x=1\\4x=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\)
d) \(2x^2+y^2+2xy-4x+4=0\)
\(\Rightarrow\left(x^2+2xy+y^2\right)+\left(x^2-4x+4\right)=0\)
\(\Rightarrow\left(x+y\right)^2+\left(x-2\right)^2=0\)
Vì \(\left(x+y\right)^2\ge0\) với mọi x và y
\(\left(x-2\right)^2\ge0\) với mọi x
\(\Rightarrow\left(x+y\right)^2+\left(x-2\right)^2\ge0\) với mọi x và y
Mà \(\left(x+y\right)^2+\left(x-2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x+y=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}y=-x\\x=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}y=-2\\x=2\end{matrix}\right.\)
23 tháng 8 2017
a, \(2x^2-4x=0\)
\(\Leftrightarrow2x\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Vậy...
b, \(x^3+2x^2+x=0\)
\(\Leftrightarrow x\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\\left(x+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy...
c, \(x^2\left(x+1\right)+2x\left(x+1\right)=0\)
\(\Leftrightarrow\left(x^2+2x\right)\left(x+1\right)=0\)
\(\Leftrightarrow x\left(x+2\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x+2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=-1\end{matrix}\right.\)
Vậy...
d, \(x\left(2x-3\right)-2\left(3-2x\right)=0\)
\(\Leftrightarrow x\left(2x-3\right)+2\left(2x-3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+2=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy...
DT
23 tháng 8 2017
a) \(2x^2-4x=0\\ \Leftrightarrow2x\left(x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
b) \(x^3+2x^2+x=0\\\Leftrightarrow x\left(x^2+2x+1\right)=0\\ \Leftrightarrow x\left(x+1\right)^2=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
c) \(x^2\left(x+1\right)+2x\left(x+1\right)=0\\ \Leftrightarrow x\left(x+1\right)\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)
d) \(x\left(2x-3\right)-2\left(2x-3\right)=0\\ \Leftrightarrow\left(x-2\right)\left(2x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{3}{2}\end{matrix}\right.\)
TT
0
NT
1
28 tháng 2 2022
a: \(x^2\left(2x-3\right)+8x-12=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x^2+4\right)=0\)
=>2x-3=0
hay x=3/2
b: \(\Leftrightarrow\left(2x-5\right)\left(2x+10\right)-\left(2x-5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+10-x+1\right)=0\)
=>(2x-5)(x+11)=0
=>x=5/2 hoặc x=-11
c: \(\Leftrightarrow2x\left(x^2-16\right)=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\)
hay \(x\in\left\{0;4;-4\right\}\)
HN
1
1 tháng 9 2020
Sửa đề :
(x - 2)2 - 16 = 0
=> (x - 2)2 = 16
=> (x - 2)2 = (\(\pm\)4)2
=> \(\orbr{\begin{cases}x-2=4\\x-2=-4\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=-2\end{cases}}\)
(x - 2)2 - x2 + 4 = 0
=> x2 - 4x + 4 - x2 + 4 = 0
=> (x2 - x2) - 4x + (4 + 4) = 0
=> -4x + 8 = 0
=> -4x = -8
=> x = 2
(2x + 3)2 - (\(\frac{1}{3}-2x\))2 = -2/3x + 5
=> \(\left(2x+3\right)\left(2x+3\right)-\left(\frac{1}{3}-2x\right)\left(\frac{1}{3}-2x\right)=-\frac{2}{3}x+5\)
=> \(2x\left(2x+3\right)+3\left(2x+3\right)-\frac{1}{3}\left(\frac{1}{3}-2x\right)+2x\left(\frac{1}{3}-2x\right)=-\frac{2}{3}x+5\)
=> \(4x^2+6x+6x+9-\frac{1}{9}+\frac{2}{3}x+\frac{2}{3}x-4x^2=-\frac{2}{3}x+5\)
=> \(\left(4x^2-4x^2\right)+\left(6x+6x+\frac{2}{3}x+\frac{2}{3}x\right)+\left(9-\frac{1}{9}\right)+\frac{2}{3}x-5=0\)
=> \(\frac{40}{3}x+\frac{80}{9}+\frac{2}{3}x-5=0\)
=> \(\frac{40}{3}x+\frac{2}{3}x+\frac{80}{9}-5=0\)
=> 14x + 80/9 - 5 = 0
=> x = -5/18
Trả lời:
\(\left(x-1\right)^2-\left(2x\right)^2=0\)
\(\Leftrightarrow\left(x-1-2x\right)\left(x-1+2x\right)=0\)
\(\Leftrightarrow\left(-x-1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-x-1=0\\3x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{1}{3}\end{cases}}}\)
Vậy x = - 1; x = 1/3 là nghiệm của pt.
\(\left(x-1\right)^2-\left(2x\right)^2=0\)
\(\Leftrightarrow\left(x-1-2x\right)\left(x-1+2x\right)=0\)
\(\Leftrightarrow\left(-x-1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-x-1=0\\3x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\3x=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{1}{3}\end{cases}}\)
Vậy \(S=\left\{-1;\frac{1}{3}\right\}\)