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31 tháng 7 2018
1,x^2-(x+1)(x-1)=0
x^2-x^2+1+0
1=0(vô lý)
2,5x^3+3x^2+3x+1=4x^2
x^3+3x^2+3x+1=0
(x+1)=0
x=-1
3,x^3+x^2=0
x^2(x+1)=0
x=0 or x=-1
4,2x^3-12x^2+18x=0
x^3-6x^2+9x=0
x(x^2-6x+9)=0
x(x-3)^2=0
x=0 or x=3
5,5x^2-4(x^2-2x+1)+20=0
5x^2-4x^2+8x-4+20=0
x^2+8x+16=0
(x+4)^2=0
x=-4
6,5x(x-3)+7x-21=0
5x(x-3)+7(x-3)=0
(5x+7)(x-3)=0
5x-7=0 or x-3=0
x=7/5 or x=3
7,2x^3-50x=0
2x(x^2-25)=0
2x(x-5)(x+5)=0
x=0 or x=5 or x=-5
8,(4x-1)^2-9(x+3)^2=0
(4x-1)^2-3^2*(x+3)^2=0
(4x-1)^2-(3x+9)^2=0
(4x-1-3x-9)(4x-1+3x+9)=0
(x-10)(7x+8)=0
x=10 or x=-8/7
9,3(x-2)^2-x+2=0
3*(x-2)*(x-2)-(x-2)=0
(3x-6)(x-2)-(x-2)=0
(x-2)(3x-6-1)=0
(x-2)(3x-7)=0
x=2 or x=7/3
10,9x^2+6x-8=0
9x^2+12x-6x-8=0
3x(3x-2)+4(3x-2)=0
(3x+4)(3x-2)=0
3x+4=0 or 3x-2=0
x=-4/3 or x=2/3
LA
21 tháng 6 2017
b)(2x - 1)^2 - (2x + 5) (2x - 5 ) = 18
4x 2 -4x+1-4x 2+25=18
26-4x=18
4x=8
x=2
21 tháng 6 2017
a,27x-18=2x-3x^2
<=> 3x^2-2x+27-18x=0
<=> 3x^2-20x+27=0
\(\Delta\)= 20^2-4-12.27
tính \(\Delta\)rồi tìm x1 ,x2
23 tháng 6 2017
\(\left(x+1\right)^2=x+1\)
\(\left(x+1\right)^2-\left(x+1\right)=0\)
\(\left(x+1\right)\left(x+1-1\right)=0\)
\(\left(x+1\right)x=0\)
\(\orbr{\begin{cases}x+1=0\\x=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=-1\\x=0\end{cases}}\)vậy.....
\(x\left(x-5\right)^2-4x+20=0\)
\(x\left(x-5\right)^2-4\left(x-5\right)=0\)
\(\left(x-5\right)\left[x\left(x-5\right)-4\right]=0\)
\(\left(x-5\right)\left(x^2-5x-4\right)=0\)
\(\orbr{\begin{cases}x-5=0\\x^2-5x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-0,7015621187\end{cases}}}\)vậy.........
\(x\left(x+6\right)-7x-42=0\)
\(x\left(x+6\right)-7\left(x+6\right)=0\)
\(\left(x+6\right)\left(x-7\right)=0\)
\(\orbr{\begin{cases}x+6=0\\x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-6\\x=7\end{cases}}}\) vậy....
\(x^3-5x^2+x-5=0\)
\(x^2\left(x-5\right)+\left(x-5\right)=0\)
\(\left(x-5\right)\left(x^2+1\right)=0\)
\(\orbr{\begin{cases}x-5=0\\x^2+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x^2=-1\Rightarrow x\in\Phi\end{cases}}}\)vậy........
\(x^4-2x^3+10x^2-20x=0\)
\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\left(x-2\right)\left(x^3+10x\right)=0\)
\(\orbr{\begin{cases}x-2=0\\x^3+10x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}}\)vậy..............
nhớ chọn mk nha
15 tháng 10 2018
a. \(\left(2x-1\right)^2-4x^2+1=0\)
\(\Leftrightarrow4x^2-4x+1-4x^2+1=0\)
\(\Leftrightarrow2-4x=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy ...
b/ \(6x^3-24x=0\)
\(\Leftrightarrow6x\left(x^2-4\right)=0\)
\(\Leftrightarrow6x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}6x=0\\x-2=0\\x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy ...
c/ \(2x\left(x-3\right)-4x+12=0\)
\(\Leftrightarrow2x\left(x-3\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow2\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy ...
d/ \(x^3-5x^2+x-5=0\)
\(\Leftrightarrow x^2\left(x-5\right)+\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2+1\right)=0\)
Mà \(x^2+1>0\)
\(\Leftrightarrow x-5=0\Leftrightarrow x=5\)
Vậy..
27 tháng 6 2017
a) ... \(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\hept{\begin{cases}x=1\\x=2\\x=-2\end{cases}}\)Vậy.....
b) ... \(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+10x\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x^2+10\right)=0\Leftrightarrow\hept{\begin{cases}x=0\\x=2\\x^2=-10\Rightarrow x\in\theta\end{cases}}\)(\(\theta\)là rỗng) Vậy.........
c) ... \(\Leftrightarrow2x-3=x+5\Leftrightarrow x=8\)Vậy.......
d) ... \(\Leftrightarrow x\left(x^2-16\right)=0\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\Leftrightarrow\hept{\begin{cases}x=0\\x=4\\x=-4\end{cases}}\)Vậy......
21 tháng 10 2018
a) \(x^2\left(x-3\right)+12-4x=0\)
\(x^2\left(x-3\right)-4\left(x-3\right)=0\)
\(\left(x-3\right)\left(x^2-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x^2-4=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x\in\left\{\pm2\right\}\end{cases}}\)
b) \(x\left(2x-7\right)-3\left(7-2x\right)=0\)
\(x\left(2x-7\right)+3\left(2x-7\right)=0\)
\(\left(2x-7\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-7=0\\x+3=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-3\end{cases}}\)
c) \(\left(2x-1\right)^2-25=0\)
\(\left(2x-1\right)^2-5^2=0\)
\(\left(2x-1-5\right)\left(2x-1+5\right)=0\)
\(\left(2x-6\right)\left(2x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-6=0\\2x+4=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
d) \(\left(3x-5\right)^2-\left(2x-3\right)^2=0\)
\(\left(3x-5-2x+3\right)\left(3x-5+2x-3\right)=0\)
\(\left(x-2\right)\left(5x-8\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\5x-8=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{8}{5}\end{cases}}\)
29 tháng 9 2017
Câu a : Mình ko biết làm .
Câu b : Bạn làm rồi khỏi làm nữa
Câu c :
\(x\left(2x-7\right)-4x+14=0\)
\(x\left(2x-7\right)-\left(4x-14\right)=0\)
\(x\left(2x-7\right)-2\left(2x-7\right)=0\)
\(\left(2x-7\right)\left(x-2\right)=0\)
\(\left[{}\begin{matrix}2x-7=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\end{matrix}\right.\)
Vậy \(x=\dfrac{7}{2}\) và \(x=2\)
Câu d :
\(\left(2x-3\right)^2-\left(x+5\right)^2=0\)
\(\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)
\(\left(x-8\right)\left(3x+2\right)=0\)
\(\left[{}\begin{matrix}x-8=0\\3x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(x=8\) và \(x=-\dfrac{2}{3}\)
DH
29 tháng 9 2017
Vậy em xin câu a ^^
a, \(6x^3+x^2+x+1=0\)
\(\Rightarrow6x^3+3x^2-2x^2-x+2x+1=0\)
\(\Rightarrow3x^2\left(2x+1\right)-x\left(2x+1\right)+\left(2x+1\right)=0\)
\(\Rightarrow\left(2x+1\right)\left(3x^2-x+1\right)=0\) (1)
Ta có: \(3x^2-x+1=3x^2-\dfrac{1}{2}x-\dfrac{1}{2}x+\dfrac{1}{12}+\dfrac{11}{12}\)
\(=3x\left(x-\dfrac{1}{6}\right)-\dfrac{1}{2}\left(x-\dfrac{1}{6}\right)+\dfrac{11}{12}\)
\(=3\left(x-\dfrac{1}{6}\right)^2+\dfrac{11}{12}>0\) (2)
Từ (1) và (2) suy ra \(2x+1=0\Rightarrow x=-\dfrac{1}{2}\)
Chúc bạn học tốt!!!
1 ) 2x2 - 5x + 4x - 10 = 0
=> 2x2 + 4x - 5x - 10 = 0
=> 2x ( x + 2 ) - 5. ( x + 2 ) = 0
=> ( x + 2 ) . ( 2x - 5 ) = 0
=> \(\orbr{\begin{cases}x+2=0\\2x-5=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-2\\x=\frac{5}{2}\end{cases}}\)
Vậy \(x\in\left\{-2;\frac{5}{2}\right\}\)
2 ) x2 ( 2x - 3 ) + 3 - 2x = 0
=> x2 ( 2x - 3 ) - ( 2x - 3 ) = 0
=> ( 2x - 3 ) . ( x2 - 1 ) = 0
=> \(\orbr{\begin{cases}2x-3=0\\x^2-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}2x=3\\x^2=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{2}\\x=\pm1\end{cases}}\)
Vậy \(x\in\left\{\frac{3}{2};\pm1\right\}\)