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7 tháng 10 2017
bài 1
a)\(x^2+5x+6=\left(x+2\right)\left(x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x+3=0\\x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=-2\end{cases}}}\)
NM
2
17 tháng 8 2015
a) x^4 - 2x^2 + 1 = 0
=> ( x^2 - 1 )^2 = 0
=> x^2 - 1 = 0
=> x^2 = 1
=> x = 1 hoặc x = -1
CW
4 tháng 12 2016
a) x4-2x2+1=0
(thang Tran giải rồi nhé)
b) x4-2x2-8=0
<=> x^4 - 2x^2 +1 -9 =0
<=> (x^2 -1)^2 -9 =0
\(\Leftrightarrow\orbr{\begin{cases}x^2-1=-3\\x^2-1=3\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=-2\left(VN\right)\\x=+_-\sqrt{2}\end{cases}}}\)
Vậy x=+- căn 2
c) x4-4x2-60=0
\(\Leftrightarrow x^4-4x^2+4-64=0\)
\(\Leftrightarrow\left(x^2-2\right)-64=0\)
\(\Leftrightarrow\left(x^2+62\right)\left(x^2-66\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+62=0\\x^2-66=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=-62\left(VN\right)\\x^2=+_-\sqrt{66}\end{cases}}}\)
Vậy x=+- căn 66
d) x6-16x2+64=0
12 tháng 10 2020
a) 2x (x-5) -(x2-10x +25)=0
\(\Leftrightarrow\)2x(x-5)-(x-5)2=0
\(\Leftrightarrow\)(x-5)(2x-x+5)=0
\(\Leftrightarrow\)(x-5)(x+5)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-5=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
b) x2 - 9 +3x(x+3) = 0
\(\Leftrightarrow\)(x2 - 9) +3x(x+3) =0
\(\Leftrightarrow\)(x-3)(x+3)+3x(x+3)=0
\(\Leftrightarrow\)(x+3)(x-3+3x)=0
\(\Leftrightarrow\)(x+3)(4x-3)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x+3=0\\4x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-3\\4x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{3}{4}\end{matrix}\right.\)
c) x3 - 16x = 0
\(\Leftrightarrow\)x(x2-16)=0
\(\Leftrightarrow\)x(x-4)(x+4)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
d) (2x+3)(x-2) - (x2 -4x+4) = 0
\(\Leftrightarrow\)(2x+3)(x-2) -(x-2)2=0
\(\Leftrightarrow\)(x-2)(2x+3-x+2)=0
\(\Leftrightarrow\)(x-2)(x+5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
e) 9x2 -(x2 -2x +1)=0
\(\Leftrightarrow\)(3x)2-(x-1)2=0
\(\Leftrightarrow\)(3x-x+1)(3x+x-1)=0
\(\Leftrightarrow\)(2x+1)(4x-1)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x+1=0\\4x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x=-1\\4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{1}{4}\end{matrix}\right.\)
f)x3-4x2 -9x +36 = 0
\(\Leftrightarrow\)(x3-9x)-(4x2-36)=0
\(\Leftrightarrow\)x(x2-9)-4(x2-9)=0
\(\Leftrightarrow\)(x-4)(x2-9)=0
\(\Leftrightarrow\)(x-4)(x-3)(x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=4\\x=3\\x=-3\end{matrix}\right.\)
g) 3x - 6 = (x-1).(x-2)
\(\Leftrightarrow\)3(x-2)=(x-1)(x-2)
\(\Leftrightarrow\)x-1=3
\(\Leftrightarrow\)x=4
i) (x-2).(x+2) +(2x+1)2 =-5x.(x-3) =5 (?? đề sao vậy ??)
k) x2 -1 = (x-1).(2x+3)
\(\Leftrightarrow\)(x-1)(x+1)=(x-1)(2x+3)
\(\Leftrightarrow\)x+1=2x+3
\(\Leftrightarrow\)x-2x=3-1
\(\Leftrightarrow\)-x=2
\(\Leftrightarrow\)x=-2
l) (2x-1)2 +(x+3).(x-3) -5x(x-2)=6
\(\Leftrightarrow\)4x2-4x+1+x2-9-5x2+10x=6
\(\Leftrightarrow\)6x-8=6
\(\Leftrightarrow\)6x=14
\(\Leftrightarrow\)x=\(\frac{7}{3}\)
NC
Tìm x biết:
4x2 - 6x = 0
b) 4x2 + 4x = -1
c) 5x2 + x = 0
d) x3 - 5x = 4x2
3x(x-2) = x-2
x3 - 16x = 0
1
LT
6 tháng 9 2020
Tìm x biết:
4x2 - 6x = 0
\(\Leftrightarrow2x\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\2x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{3}{2}\end{matrix}\right.\)
Vậy \(x=\left\{0;\frac{3}{2}\right\}\)
b) 4x2 + 4x = -1
\(\Leftrightarrow4x^2+4x+1=0\)
\(\Leftrightarrow\left(2x+1\right)^2=0\)
\(\Leftrightarrow2x+1=0\)
\(\Leftrightarrow x=-\frac{1}{2}\)
Vậy \(x=-\frac{1}{2}\)
c) 5x2 + x = 0
\(\Leftrightarrow x\left(5x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\5x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\frac{1}{5}\end{matrix}\right.\)
Vậy \(x=\left\{0;-\frac{1}{5}\right\}\)
d) x3 - 5x = 4x2
\(\Leftrightarrow x^3-4x^2-5x=0\)
\(\Leftrightarrow x^3+x^2-5x^2-5x=0\)
\(\Leftrightarrow x^2\left(x+1\right)-5x\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-5x\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\\x-5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=5\end{matrix}\right.\)
Vậy x ={0; - 1; 5}
3x(x-2) = x-2
\(\Leftrightarrow3x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy \(x=\left\{2;\frac{1}{3}\right\}\)
x3 - 16x = 0
\(\Leftrightarrow x\left(x^2-16\right)=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
Vậy x = {0; 4; -4}
NH
23 tháng 9 2018
1,=\(x^2-3x-2x^2+6x=-x^2+3x\)
2,=\(3x^2-x-5+15x=3x^2+14x-5\)
3,=\(5x+15-6x^2-6x=-6x^2-x+15\)
4,=\(4x^2+12x-x-3=4x^2+11x-3\)
4 tháng 9 2022
5: =>(x+5)^3=0
=>x+5=0
=>x=-5
6: =>(2x-3)^2=0
=>2x-3=0
=>x=3/2
7: =>(x-6)(x-10)=0
=>x=10 hoặc x=6
8: \(\Leftrightarrow x^3-12x^2+48x-64=0\)
=>(x-4)^3=0
=>x-4=0
=>x=4
28 tháng 7 2016
\(\left(x-2\right)^3+6\left(x+1\right)^2-x^3+12=0\)
\(x^3-3.x^2.2+3.x.2^2-2^3+6.x^2+2.x.1+1^2-x^3+12=0\)\(=x^3-6x^2+12x-8+6x^2+2x+1-x^3+12=0\)
\(14x+5=0\)
\(14x=0-5\)
\(14x=-5\)
\(x=-5:14\)
\(x=-\frac{5}{14}\)
KT
16 tháng 7 2018
a) \(x^3-x^2-5x+125\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
b) \(5x^2-5xy-3x+3y\)
\(=5x\left(x-y\right)-3\left(x-y\right)\)
\(=\left(x-y\right)\left(5x-3\right)\)
c) \(x^2-2x-4y^2+1\)
\(=\left(x-1\right)^2-4y^2\)
\(=\left(x-2y-1\right)\left(x+2y-1\right)\)
TD
24 tháng 9 2017
Tìm x
b) 16x - 5x2 - 3 = 0
\(\Leftrightarrow\) 5x2 - 16x + 3 = 0
\(\Leftrightarrow\) 5x2 - 15x - x + 3 = 0
\(\Leftrightarrow\) ( 5x2 - 15x ) - ( x - 3 ) = 0
\(\Leftrightarrow\) 5x ( x - 3 ) - ( x- 3 ) = 0
\(\Leftrightarrow\) ( x - 3 ) ( 5x - 1 ) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)
Vậy phương trình có nghiệm x = 3 hoặc x = \(\dfrac{1}{5}\)
O
4 tháng 11 2018
\(A.x^2-16x=0\)
\(x^2-\left(4x\right)^2=0\)
\(\left(x-4x\right)\left(x+4x\right)=0\)
\(\left(-3x\right)\left(5x\right)=0\)
\(\Rightarrow\) \(-3x=0\) hoặc \(5x=0\)
\(x=\dfrac{0}{-3}\) hoặc \(x=\dfrac{0}{5}\)
Vậy \(x=0\) hoặc \(x=0.\)
B. 4x2 - 4x + 1 = 0
(2x)2 - (2x)2 + 12 = 0
(2x - 2x + 1 ) (2x + 2x +1) = 0
1 (4x + 1) =0
=> 1 (4x + 1) =0
4x + 1 = 0
4x = 0-1
Vậy x = \(\dfrac{-1}{4}.\)
C. (3x-1)2 - (2x+3)2 = 0
(3x -1 -2x +3) (3x -1 +2x +3) = 0
(x + 2)(5x + 2) = 0
=> x + 2 =0 hoặc 5x + 2 =0
x = 0 - 2 hoặc 5x = 0 - 2
Vậy x = -2 hoặc x = \(\dfrac{-2}{5}.\)
Còn về câu d thì mình hơi phân vân, tại mình dốt toán lắm
4 tháng 11 2018
a/ \(x^2-16x=0\)
\(\Leftrightarrow x\left(x-16\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-16=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=16\end{matrix}\right.\)
Vậy...
b/ \(4x^2-4x+1=0\)
\(\Leftrightarrow\left(2x-1\right)^2=0\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy..
c/ \(\left(3x-1\right)^2-\left(2x+3\right)^2=0\)
\(\Leftrightarrow\left(3x-1-2x-3\right)\left(3x-1+2x+3\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{5}\end{matrix}\right.\)
Vậy...
d/ \(2013x^2-2014x+1=0\)
\(\Leftrightarrow2013x^2-x-2013x+1=0\)
\(\Leftrightarrow x\left(2013x-1\right)-\left(2013x-1\right)=0\)
\(\Leftrightarrow\left(2013x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2013x-1=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2013}\\x=1\end{matrix}\right.\)
Vậy..