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14 tháng 8 2020
a) 16x^2 - (4x - 5)^2 = 15
<=> 16x^2 - 16x^2 + 40x - 25 = 15
<=> 40x = 40
<=> x = 1
b) (2x + 3)^2 - 4(x - 1)(x + 1) = 49
<=> 4x^2 + 12x + 9 - 4x^2 - 4x + 4x + 4 = 49
<=> 12x + 13 = 49
<=> 12x = 36
<=> x = 3
c) (2x + 1)(1 - 2x) + (1 - 2x)^2 = 18
<=> 1 - 4x^2 + 1 - 4x + 4x^2 = 18
<=> 2 - 4x = 18
<=> -4x = 16
<=> x = -4
d)2(x + 1)^2 - (x - 3)(x + 3) - (x - 4)^2 = 0
<=> 2x^2 + 4x + 2 - x^2 + 3^2 - x^2 + 8x - 16 = 0
<=> 12x - 5 = 0
<=> 12x = 5
<=> x = 5/12
e) (x - 5)^2 - x(x - 4) = 9
<=> x^2 - 10x + 25 - x^2 + 4x = 9
<=> -6x + 25 = 9
<=> -6x = 9 - 25
<=> -6x = -16
<=> x = -16/-6 = 8/3
f) (x - 5)^2 + (x - 4)(1 - x) = 0
<=> x^2 - 10x + 25 + x - x^2 - x - 4 + 4x = 0
<=> -5x + 21 = 0
<=> -5x = -21
<=> x = 21/5
21 tháng 8 2021
a, \(\left(2x+1\right)\left(1-2x\right)+\left(1-2x\right)^2=0\)
\(\Leftrightarrow\left(1-2x\right)\left(2x+1+1-2x\right)=0\Leftrightarrow x=\frac{1}{2}\)
b, \(2\left(x+1\right)^2-\left(x-3\right)\left(x+3\right)-\left(x-4\right)^2=0\)
\(\Leftrightarrow2\left(x^2+2x+1\right)-\left(x^2-9\right)-\left(x^2-8x+16\right)=0\)
\(\Leftrightarrow2x^2+4x+2-x^2+9-x^2+8x-16=0\Leftrightarrow12x-5=0\Leftrightarrow x=\frac{5}{12}\)
c, \(\left(x-5\right)^2-x\left(x-4\right)=9\Leftrightarrow x^2-10x+25-x^2+4x=9\)
\(\Leftrightarrow-6x+16=0\Leftrightarrow x=\frac{8}{3}\)
d, \(\left(x-5\right)^2+\left(x-4\right)\left(1-x\right)=0\)
\(\Leftrightarrow x^2-10x+25+x-x^2-4+4x=0\)
\(\Leftrightarrow-5x+21=0\Leftrightarrow x=\frac{21}{5}\)
NV
2
ML
10 tháng 7 2015
\(x^4+2x^3+2x^2+2x+1=0\)
\(\Leftrightarrow\left(x^4+2x^3+x^2\right)+\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x^2+x\right)^2+\left(x+1\right)^2=0\)
\(\Leftrightarrow x^2\left(x+1\right)^2+\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)^2\left(x^2+1\right)=0\)
\(\Leftrightarrow x+1=0\text{ (do }x^2+1>0\text{)}\)
\(\Leftrightarrow x=-1\)
11 tháng 12 2016
a) \(\left(x-2\right)\left(x^2+2x+7\right)+2\left(x^2-4\right)-5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+7+2x+4-5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+4x+6\right)=0\)
\(\Leftrightarrow x-2=0\) (Vì: \(x^2+4x+6>0\) )
\(\Leftrightarrow x=2\)
b) \(2x^3+x^2-6x=0\)
\(\Leftrightarrow x\left(2x^2+x-6\right)=0\)
\(\Leftrightarrow x\left[\left(2x^2+4x\right)-\left(3x+6\right)\right]=0\)
\(\Leftrightarrow x\left[2x\left(x+2\right)-3\left(x+2\right)\right]=0\)
\(\Leftrightarrow x\left(x+2\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x+2=0\\2x-3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-2\\x=\frac{3}{2}\end{array}\right.\)
c) \(4x^2+4xy+x^2-2x+1+y^2=0\)
\(\Leftrightarrow\left(4x^2+4xy+y^2\right)+\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2=0\)
\(\Leftrightarrow\begin{cases}2x+y=0\\x-1=0\end{cases}\)\(\Leftrightarrow\begin{cases}y=-2\\x=1\end{cases}\)
21 tháng 12 2016
\(x^4+2x^3+2x^2+2x+1=0\)
\(\Leftrightarrow\left(x^4+2x^3+x^2\right)+\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow x^2\left(x^2+2x+1\right)+\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow x^2\left(x+1\right)^2+\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)^2\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\left(x+1\right)^2=0\\x^2+1=0\left(loai\right)\end{array}\right.\)
\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
AT
0
DN
2
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
ML
12 tháng 8 2015
1)a)3(2x-1)(3x-1)-(2x-3)(9x-1)=0
<=>18x2-15x+1-18x2+29x-3=0
<=>14x-2=0
<=>14x=2
<=>x=1/7
b)4(x+1)2+(2x-1)2-8(x-1)(x+1)=11
<=>4x2+8x+4+4x2-4x+1-8x2+8=11
<=>4x+13=11
<=>4x=11-13
<=>4x=-2
<=>x=-1/2
c)Sai đề phải là dấu - chứ không phải +
(x-3)(x2+3x+9)-x(x-2)(x+2)=1
<=>x3-27-x3+4x=1
<=>4x=1+27
<=>4x=28
<=>x=7
2)a)(2x-3y)(2x+3y)-4(x-y)2-8xy
=4x2-9y2-4x2+8xy-4y2-8xy
=-13y2
b)(x-2)3-x(x+1)(x-1)+6x(x-3)
=x3-6x2+12x+8-x3+x+6x2-18x
=8-5x
c)(x-2)(x2-2x+4)(x+2)(x2+2x+4)
=(x-2)(x2+2x+4)(x+2)(x2-2x+4)
=(x3-8)(x3+8)
=x6-64