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ML
10 tháng 7 2015
\(x^4+2x^3+2x^2+2x+1=0\)
\(\Leftrightarrow\left(x^4+2x^3+x^2\right)+\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x^2+x\right)^2+\left(x+1\right)^2=0\)
\(\Leftrightarrow x^2\left(x+1\right)^2+\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)^2\left(x^2+1\right)=0\)
\(\Leftrightarrow x+1=0\text{ (do }x^2+1>0\text{)}\)
\(\Leftrightarrow x=-1\)
21 tháng 12 2016
\(x^4+2x^3+2x^2+2x+1=0\)
\(\Leftrightarrow\left(x^4+2x^3+x^2\right)+\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow x^2\left(x^2+2x+1\right)+\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow x^2\left(x+1\right)^2+\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)^2\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\left(x+1\right)^2=0\\x^2+1=0\left(loai\right)\end{array}\right.\)
\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
24 tháng 6 2018
\(x^4+2x^3+2x^2+2x+1=0\)
\(\Leftrightarrow\) \(x^4+x^3+x^3+2x^2+2x+1=0\)
\(\Leftrightarrow\) \(x^3\left(x+1\right)+2x\left(x+1\right)+\left(x^3+1\right)=0\)
\(\Leftrightarrow\) \(x^3\left(x+1\right)+2x\left(x+1\right)+\left(x+1\right)\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\) \(\left(x+1\right)\left(x^3+2x+x^2-x+1\right)=0\)
\(\Leftrightarrow\) \(\left(x+1\right)\left[x^2\left(x+1\right)+\left(x+1\right)\right]=0\)
\(\Leftrightarrow\) \(\left(x+1\right)\left(x+1\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\) \(\left(x+1\right)^2\left(x^2+1\right)=0\)
\(\Leftrightarrow\)\(\hept{\begin{cases}\left(x+1\right)^2=0\\x^2+1=0\end{cases}}\)
\(\Leftrightarrow\)\(\hept{\begin{cases}x=-1\\x^2=-1\rightarrow kotm\end{cases}}\)
Vậy.....................................................
YT
23 tháng 6 2018
\(x^4+x^3+x^3+x^2+x^2+x+x+1=0\)
\(x^3(x+1)+x^2(x+1)+x(x+1)=0\)
\((x+1)(x^3+x^2+x+1)=0\)
\((x+1)[x^2(x+1)+(x+1)]=0\)
\((x+1)^2(x^2+1)=0\)
\(\orbr{\begin{cases}x+1=0\\x^2+1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\sqrt{-1}\left(loai\right)\end{cases}}\)
vay \(x=-1\)
NẾU CÓ SAI BN THÔNG CẢM
10 tháng 7 2015
\(x^4+2x^3+2x^2+2x+1=0\)
\(\Rightarrow x^4+x^3+x^3+x^2+x^2+x+x+1=0\)
\(\Leftrightarrow x^3\left(x+1\right)+x^2\left(x+1\right)+x\left(x+1\right)+x+1\)
\(\Leftrightarrow\left(x+1\right)\left(x^3+x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[x^2\left(x+1\right)+x+1\right]=0\Leftrightarrow\left(x+1\right)\left(x^2+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\left(x^2+1\right)=0\)
Vì \(x^2\ge0\Rightarrow x^2+1>0\)
=> x + 1 = 0
=> x = - 1
VẬy x = -1
LH
0
ZT
1
NN
25 tháng 10 2020
b) \(ĐKXĐ:x\ne0\)
\(\left(5x^4-3x^3\right):2x^3=\frac{1}{2}\)
\(\Leftrightarrow x^3.\left(5x-2\right):2x^3=\frac{1}{2}\)
\(\Leftrightarrow\frac{5x-2}{2}=\frac{1}{2}\)\(\Leftrightarrow5x-2=1\)
\(\Leftrightarrow5x=3\)\(\Leftrightarrow x=\frac{3}{5}\)( thỏa mãn ĐKXĐ )
Vậy \(x=\frac{3}{5}\)
c) \(ĐKXĐ:x\ne2\)
\(\frac{x^4-2x^2-8}{x-2}=0\)\(\Rightarrow x^4-2x^2-8=0\)
\(\Leftrightarrow\left(x^4-4x^2\right)+\left(2x^2-8\right)=0\)
\(\Leftrightarrow x^2.\left(x^2-4\right)+2\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x^2+2\right)=0\)
Vì \(x^2\ge0\forall x\)\(\Rightarrow x^2+2\ge2\)
\(\Rightarrow\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
So sánh với ĐKXĐ ta thấy: \(x=-2\)thỏa mãn
Vậy \(x=-2\)
ZT
1
B
25 tháng 10 2020
a) \(x^2=2x+1\)
\(\Leftrightarrow x^2-2x-1=0\)
\(\Leftrightarrow x^2-2x+1-2=0\)
\(\Leftrightarrow\left(x-1\right)^2-2=0\)
\(\Leftrightarrow\left(x-1-\sqrt{2}\right)\left(x-1+\sqrt{2}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1+\sqrt{2}\\x=1-\sqrt{2}\end{cases}}\)
b) ĐKXĐ : x khác 0
\(\frac{5x^4-3x^3}{2x^3}=\frac{1}{2}\)
\(\Leftrightarrow\frac{x^3\left(5x-3\right)}{2x^3}=\frac{1}{2}\)
\(\Leftrightarrow\frac{5x-3}{2}=\frac{1}{2}\)
\(\Leftrightarrow5x-3=1\Leftrightarrow x=\frac{4}{5}\)( thỏa mãn ĐKXĐ )
c) ĐKXĐ : x khác 2
\(\frac{x^4-2x^2-8}{x-2}=0\)
\(\Leftrightarrow x^4-2x^2-8=0\)
\(\Leftrightarrow x^4-4x^2+2x^2-8=0\)
\(\Leftrightarrow x^2\left(x^2-4\right)+2\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-2\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\left(tm\right)\\x=2\left(ktm\right)\end{cases}}\)
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