K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

15 tháng 9 2020

\(\left(x^2-4x+3\right)\left(x^2-6x+8\right)=8\) 

\(\left(x^2-3x-x+3\right)\left(x^2-4x-2x+8\right)=8\)  

\(\left[x\left(x-3\right)-1\left(x-3\right)\right]\left[x\left(x-4\right)-2\left(x-4\right)\right]=8\)

\(\left(x-1\right)\left(x-3\right)\left(x-2\right)\left(x-4\right)=8\) 

\(\left(x-1\right)\left(x-4\right)\left(x-2\right)\left(x-3\right)=8\) 

\(\left(x^2-5x+4\right)\left(x^2-5x+6\right)-8=0\)  

Đặt \(t=x^2-5x+4\) 

\(t\left(t+2\right)-8=0\) 

\(t^2+2t-8=0\) 

\(t^2+4t-2t-8=0\) 

\(t\left(t+4\right)-2\left(t+4\right)=0\) 

\(\left(t+4\right)\left(t-2\right)=0\) 

\(\orbr{\begin{cases}t+4=0\\t-2=0\end{cases}}\) 

\(\orbr{\begin{cases}t=-4\\t=2\end{cases}}\)  

\(\orbr{\begin{cases}x^2-5x+4=-4\\x^2-5x+4=2\end{cases}}\)  

\(\orbr{\begin{cases}x^2-5x+8=0\left(ptvn\right)\\x^2-5x+2=0\end{cases}}\) 

\(x^2-5x+2=0\) 

\(\orbr{\begin{cases}x=\frac{5+\sqrt{17}}{2}\\x=\frac{5-\sqrt{17}}{2}\end{cases}}\)

6 tháng 4 2018

VT>=0 suy ra 4x>=0

suy ra x>=0

..................................................................................................

8 tháng 4 2018

Do : VP ≥ 0

=> VT ≥ 0

=> 4x ≥ 0

=> x ≥ 0

nên Phương trình trên có dạng :

x + 2 + x + 9 + x + 2011 = 4x

<=> 3x + 2022 = 4x

<=> x = 2022 ( thỏa mãn )

KL....

10 tháng 2 2019

a, \(\Leftrightarrow\left(x+1+x-2\right)\left[\left(x+1\right)^2-\left(x+1\right)\left(x-2\right)+\left(x-2\right)^2\right]-\left(2x-1\right)^3=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x^2+2x+1-x^2+x+2+x^2-4x+4\right)-\left(2x-1\right)^3=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x^2-x+7-\left(2x-1\right)^2\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x^2-x+7-4x^2+4x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(-3x^2+3x+6\right)=0\)

\(\Leftrightarrow-3\left(2x-1\right)\left(x^2-x-2\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x+1\right)\left(x-2\right)=0\)

=>x=1/2 hoặc x=-1 hoặc x=2

Vậy pt có tập nghiệm là S={1/2;-1;2}

b, \(x^4=24x+32\Leftrightarrow x^4-24x-32=0\)

\(\Leftrightarrow x^4-2x^3-4x^2+2x^3-4x^2-8x+8x^2-16x-32=0\)

\(\Leftrightarrow x^2\left(x^2-2x-4\right)+2x\left(x^2-2x-4\right)+8\left(x^2-2x-4\right)=0\)

\(\Leftrightarrow\left(x^2-2x-4\right)\left(x^2+2x+8\right)=0\)

\(\Leftrightarrow x^2-2x-4=0\) (vì x^2+2x+8 > 0)

\(\Leftrightarrow\left(x-1\right)^2-5=0\Leftrightarrow\left(x-1\right)^2=5\Leftrightarrow x-1=\pm\sqrt{5}\Leftrightarrow x=1\pm\sqrt{5}\)

Vậy...

c, \(\left(x-6\right)^4+\left(x-8\right)^4=16\)

Đặt x-6=t => x-8=t-2

Ta có: \(t^4+\left(t-2\right)^4=16\Leftrightarrow t^4+t^4-8t^3+24t^2-32t+16=16\)

\(\Leftrightarrow2t^4-8t^3+24t^2-32t=0\Leftrightarrow t^4-4t^3+12t^2-16t=0\)

\(\Leftrightarrow t^4-2t^3-2t^3+4t^2+8t^2-16t=0\)

\(\Leftrightarrow t^3\left(t-2\right)-2t^2\left(t-2\right)+8t\left(t-2\right)=0\)

\(\Leftrightarrow\left(t-2\right)\left(t^3-2t^2+8t\right)=0\Leftrightarrow\left(t-2\right)t\left(t^2-2t+8\right)=0\)

Mà t^2-2t+8=(t-1)^2+7 > 0

\(\Rightarrow\orbr{\begin{cases}t-2=0\\t=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x-6-2=0\\x-6=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=8\\x=6\end{cases}}}\)

Vậy...

22 tháng 8 2016

\(\left(2-x\right)\left(2x-1\right)+\left(4x^2-4x+1\right)=0\)

\(\Leftrightarrow\left(2-x\right)\left(2x-1\right)+\left(2x-1\right)^2=0\)

\(\Leftrightarrow\left(2x-1\right)\left(2-x+2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-1=0\\x+1=0\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{2}\\x=-1\end{array}\right.\)

Vậy phương trình có tập nghiệm \(\left\{-1;\frac{1}{2}\right\}\)

22 tháng 8 2016

(2-x)(2x-1)+(4x^2-4x+1)=0

 Ta có:  (2x-1)(2-x)+(2x-1)^2=0

                (2x-1)(2-x+2x-1)=0

Sau đó bn tự lam nha tại vì mk làm bằng phone

3 tháng 2 2021

1) Ta có: \(\left(x^2-1\right)^2-x\left(x^2-1\right)-2x^2=0\)

\(\Leftrightarrow\left[\left(x^2-1\right)^2+x\left(x^2-1\right)\right]-\left[2x\left(x^2-1\right)+2x^2\right]=0\)

\(\Leftrightarrow\left(x^2-1\right)\left(x^2+x-1\right)-2x\left(x^2+x-1\right)=0\)

\(\Leftrightarrow\left(x^2-2x-1\right)\left(x^2+x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-2x-1=0\\x^2+x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^2=2\\\left(x+\frac{1}{2}\right)^2=\frac{5}{4}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=\pm\sqrt{2}\\x+\frac{1}{2}=\pm\frac{\sqrt{5}}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\pm\sqrt{2}\\x=-\frac{1\pm\sqrt{5}}{2}\end{cases}}\)

3 tháng 2 2021

2) Ta có: \(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2=0\)

\(\Leftrightarrow\left[\left(x^2+4x+8\right)^2+x\left(x^2+4x+8\right)\right]+\left[2x\left(x^2+4x+8\right)+2x^2\right]=0\)

\(\Leftrightarrow\left(x^2+4x+8\right)\left(x^2+5x+8\right)+2x\left(x^2+5x+8\right)=0\)

\(\Leftrightarrow\left(x^2+6x+8\right)\left(x^2+5x+8\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x+4\right)\left(x^2+5x+8\right)=0\)

Vì \(x^2+5x+8=\left(x^2+5x+\frac{25}{4}\right)+\frac{7}{4}=\left(x+\frac{5}{2}\right)^2+\frac{7}{4}>0\)

\(\Rightarrow\orbr{\begin{cases}x+2=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-2\\x=-4\end{cases}}\)

Vậy x = -2 hoặc x = -4

23 tháng 3 2019

\(\frac{4x}{x^2+4x+3}-1=6\left(\frac{1}{x+3}-\frac{1}{2x+2}\right)\) \(ĐK:x\ne-1;x\ne-3\)

\(\Leftrightarrow\frac{4x}{x^2+4x+3}-\frac{x^2+4x+3}{x^2+4x+3}=6\left[\frac{2\left(x+1\right)}{2\left(x+3\right)\left(x+1\right)}-\frac{x+3}{2\left(x+1\right)\left(x+3\right)}\right]\)

\(\Leftrightarrow\frac{4x-x^2-4x-3}{x^2+4x+3}=6\left[\frac{2\left(x+1\right)-x-3}{2\left(x+3\right)\left(x+1\right)}\right]\)

\(\Leftrightarrow\frac{-x^2-3}{x^2+4x+3}=6\left[\frac{2x+2-x-3}{2\left(x^2+4x+3\right)}\right]\)

\(\Leftrightarrow\frac{-x^2-3}{x^2+4x+3}=\frac{6\left(x-1\right)}{2\left(x^2+4x+3\right)}\)

\(\Leftrightarrow\frac{-x^2-3}{x^2+4x+3}=\frac{3\left(x-1\right)}{x^2+4x+3}\)

\(\Leftrightarrow-x^2-3=3x-3\)

\(\Leftrightarrow-x^2-3x=0\)

\(\Leftrightarrow-x\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-3\left(loại\right)\end{cases}}\) 

Vậy x = 0 

23 tháng 3 2019

\(ĐK:x\ne\frac{-1}{2};x\ne\frac{-3}{2}\)

\(\frac{3}{2x+1}=\frac{6}{2x+3}+\frac{8}{4x^2+8x+3}\)

\(\Leftrightarrow\frac{3}{2x+1}-\frac{6}{2x+3}=\frac{8}{4x^2+8x+3}\)

\(\Leftrightarrow\frac{3\left(2x+3\right)-6\left(2x+1\right)}{\left(2x+1\right)\left(2x+3\right)}=\frac{8}{4x^2+8x+3}\)

\(\Leftrightarrow\frac{6x+9-12x-6}{4x^2+8x+3}=\frac{8}{4x^2+8x+3}\)

\(\Leftrightarrow-6x+3=8\)

\(\Leftrightarrow x=-\frac{5}{6}\)

Vậy ... 

25 tháng 2 2019

\(a,\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)  ĐKXĐ : \(x\ne0;x\ne\frac{3}{2}\)

\(\Leftrightarrow\frac{x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{5\left(2x-3\right)}{x\left(2x-3\right)}\)

\(\Leftrightarrow x-3=10x-15\)

\(\Leftrightarrow x-10x=3-15\)

\(\Leftrightarrow-9x=-12\)

\(\Leftrightarrow x=\frac{-12}{-9}=\frac{4}{3}\)(TMĐKXĐ)

KL :....

25 tháng 2 2019

\(b,\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\)   ĐKXĐ : \(x\ne0;2\)

\(\Leftrightarrow\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)

\(\Leftrightarrow x^2+2x-x+2=2\)

\(\Leftrightarrow x^2+x=2-2\)

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

KL ::

18 tháng 10 2020

a) ( 3 - x )( x2 + 2x - 7 ) + ( x - 3 )( x2 + x - 5 )

= ( 3 - x )( x2 + 2x - 7 ) - ( 3 - x )( x2 + x - 5 )

= ( 3 - x )( x2 + 2x - 7 - x2 - x + 5 )

= ( 3 - x )( x - 2 )

b) ( x - 5 )2 + 3( 5 - x )

= ( x - 5 )2 - 3( x - 5 )

= ( x - 5 )( x - 5 - 3 ) = ( x - 5 )( x - 8 )

c) 2x( x - 1 )2 - ( 1 - x )3

= 2x( 1 - x )2 - ( 1 - x )3

= ( 1 - x )2( 2x - 1 + x ) = ( 1 - x )2( 3x - 1 )

d) x2 + 8x + 16 = ( x + 4 )2

e) x2 - 4xy + 4y2 = ( x - 2y )2

g) 4x2 - 25y2 = ( 2x )2 - ( 5y )2 = ( 2x - 5y )( 2x + 5y )

h) 25( x + 1 )2 - 4( x - 3 )2

= 52( x + 1 )2 - 22( x - 3 )2

= ( 5x + 5 )2 - ( 2x - 6 )2

= ( 5x + 5 - 2x + 6 )( 5x + 5 + 2x - 6 )

= ( 3x + 11 )( 7x - 1 )

i) x3 + 27 = ( x + 3 )( x2 - 3x + 9 )

k) 8x3 - 125 = ( 2x )3 - 53 = ( 2x - 5 )( 4x2 + 10x + 25 )

l) x3 + 6x2 + 12x + 8 = ( x + 2 )3

m) -x3 + 9x2 - 27x + 27 = -( x3 - 9x2 + 27x - 27 ) = -( x - 3 )3