(x+1) * (x² +x+1) * (x-1) * (x²-x+1)   = 7

[(x+1) * (x² +x+1) ]*[(x-1) * (x²-x+1)]= 7  [Áp dụng hằng đẳng thức a³+b³=(a+b)*(a²+ab+b²)]

(x³+1³) * (x³-1³)                               = 7

x³ * x³ - x³ * 1³ + x³ * 1³ - 1³ *1³     =7

(x³)² - 1                                            = 7

(x³)²                                                  =7+1

(x³)²                                                  =8

suy ra x = 3 căn 2

a: \(\Leftrightarrow\left(x-1\right)^2=81\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-8\end{matrix}\right.\)

Mình đang cần gắp 

bạn còn

c) Ta có : \(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)

\(\Rightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\)\(\left(\frac{x+6}{2003}+1\right)\)

\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)

\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)

\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)

Mà : \(\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\ne0\)

Nên x + 2009 = 0 => x = -2009

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

\(\Rightarrow\left[{}\begin{matrix}x-1=0\Rightarrow x=1\\2x-4=0\Rightarrow x=2\end{matrix}\right.\)

b) \(\left(x^2+5\right)\left(x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2+5=0\Rightarrow x=-\sqrt{5}\\x-5=0\Rightarrow x=5\end{matrix}\right.\)

mà \(x\in Z\Rightarrow x=5\)

c) \(\left(x^2+5\right)\left(x^2-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2+5=0\Rightarrow x=-\sqrt{5}\\x^2-2=0\Rightarrow x=\sqrt{2}\end{matrix}\right.\)

mà \(x\in Z\Rightarrow x\in\varnothing\)

\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{100}\right).200x=4036\)

\(\Leftrightarrow\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{99}{100}.200x=4036\)

\(\Leftrightarrow\frac{1.2.3...99}{2.3.4....100}.200x=4036\)

\(\Leftrightarrow\frac{1}{100}.200x=4036\)

\(\Leftrightarrow\frac{1}{100}.200x=4036\)

\(\Leftrightarrow2x=4036\)

\(\Leftrightarrow x=4036:2=2018\)

\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times...\times\left(1-\frac{1}{100}\right)\times200\times x=4036\)

=> \(\frac{1}{2}\times\frac{2}{3}\times...\times\frac{99}{100}\times200\times x=4036\)

=> \(\frac{1\times2\times...\times99}{2\times3\times...\times100}\times200\times x=4036\)

\(\Rightarrow\frac{1}{100}\times200\times x=4036\)

\(\Rightarrow2\times x=4036\)

=> x = 2018 

\(2006.\left|x-1\right|+\left(x-1\right)^2=2005.\left|1-x\right|\) (để thỏa mản là chúng bằng nhau thì ta cần tích của chúng bằng 0)

Ta tính vế phải:

\(2005.\left|x-1\right|=0\)

\(\left|x-1\right|=0\)

Ta có: x - 1 = 0

=> x = 0 + 1 = 1

Mà vế trái bằng vế phải nên x = 1