hi army 

kbn vs mik nha

1.

a) \(x\left(1-x\right)+\left(x+1\right)\left(x-2\right)\)

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

\(-2\)

b) \(\left(x+3\right)^2-x^2=45\)

\(x^2+6x+9-x^2=45\)

\(6x+9=45\)

\(6x=36\)

\(x=6\)

Bài làm

a) x( 1 - x ) + ( x + 1 )( x - 2 )

= x - x² + x² - 2x + x - 2

= ( x - 2x + x ) + ( x² - 2x ) - 2

= -2 

b) ( x + 3 )² - x² = 45

=> ( x + 3 - x )( x + 3 + x ) = 45

=> 3. ( 2x + 3 ) = 45
=> 6x + 9 = 45

=> 6x = 45-9

=> 6x = 36

=> x = 6

Vậy x = 6

# Học tốt #

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

Vì n là số tự nhiên lẻ \(\Rightarrow n=2k+1\left(k\inℕ\right)\)

\(\Rightarrow n^4-1=\left(2k+1-1\right)\left(2k+1+1\right)\left(n^2+1\right)\)

                    \(=2k.\left(2k+2\right)\left(n^2+1\right)=4k\left(k+1\right)\left(n^2+1\right)\)

Vì \(k\)và \(k+1\)là 2 số tự nhiên liên tiếp \(\Rightarrow k\left(k+1\right)⋮2\)

\(\Rightarrow4k\left(k+1\right)⋮8\)\(\Rightarrow n^4-1⋮8\)

\(x^2-5x+6=0\)

\(x^2-2x-3x+6=0\)

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

\(x\left(x-2\right)-3\left(x-2\right)=0\)

\(\left(x-3\right)\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}}\)

\(x^2-5x+6=0\)

\(x^2-2x-3x+6=0\)

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

\(x\left(x-2\right)-3\left(x-2\right)=0\)

\(\left(x-2\right)\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\)            \(\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)