i dont no

5x² - 5x - 3x² + 3x

= 5x( x - 1 ) - 3x( x - 1 )

= ( x - 1 )( 5x - 3x )

4x² - 8xy + 4y² - 16z²

= ( 2x - 2y )² - (4z)²

= ( 2x - 2y - 4z )( 2x - 2y + 4z )

3x² + 7x + 2

= 3x² + 6x + x + 2

= 3x( x + 2 ) + ( x + 2 )

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

HỌC TỐT !

\(x^2-25=6x-9\)

\(\Rightarrow x^2-25-\left(6x-9\right)=0\)

\(\Rightarrow x^2-25-6x+9=0\)

\(\Rightarrow\left(x^2-6x+9\right)-25=0\)

\(\Rightarrow\left(x^2-2x.3+3^2\right)-5^2=0\)

\(\Rightarrow\left(x-3\right)^2-5^2=0\)

\(\Rightarrow\left(x-3-5\right)\left(x-3+5\right)=0\)

\(\Rightarrow\left(x-8\right)\left(x+2\right)=0\)

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

\(y=\frac{x^2+2}{x-1}\left(x\ne1\right)\)

Để \(y\in Z\Rightarrow\frac{x^2+2}{x-1}\in Z\Rightarrow x^2+2⋮\left(x-1\right)\)

\(\Rightarrow\left(x-1\right)\left(x+1\right)+3⋮\left(x-1\right)\)

\(\Rightarrow3⋮\left(x-1\right)\Rightarrow\left(x-1\right)\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

\(\Rightarrow x\in\left\{-2;0;2;4\right\}\) (thỏa mãn x khác 1)

Từ đó thay lần lượt x vào \(y=\frac{x^2+2}{x-1}\) ,tìm được 

\(\left(x,y\right)\in\left\{\left(-2;-2\right),\left(0;-2\right),\left(2;6\right),\left(4;6\right)\right\}\)

\(7x^2+50x+7=\left(7x^2+49x\right)+\left(x+7\right)=7x\left(x+7\right)+\left(x+7\right)=\left(x+7\right)\left(7x+1\right)\)

7x² + 50x + 7

= 7x² + 49x + x + 7

= ( 7x² + 49x ) + ( x + 7 )

= 7x( x + 7 ) + ( x + 7 )

= ( 7x + 1 )( x + 7 )

hok tốt

\(ax^3+bx-24=\left(x+1\right)Q\left(x\right)\)(1)

\(ax^3+bx-24=\left(x+3\right)P\left(x\right)\) (2) (P(x),Q(x) là các thương)

Thay x = -1 vào (1) và x = -3 vào (2), ta có: 

\(\hept{\begin{cases}a.\left(-1\right)^3+b.\left(-1\right)-24=0\\a.\left(-3\right)^3+b.\left(-3\right)-24=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}-a-b=24\\-27a-3b=24\end{cases}}\Rightarrow\hept{\begin{cases}-3a-3b=72\\-27a-3b=24\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}-3a-3b-\left(-27a-3b\right)=72-24\\-a-b=24\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}24a=48\\a+b=-24\end{cases}\Rightarrow}\hept{\begin{cases}a=2\\b=-26\end{cases}}\)

\(2x^2-x-1=0\)     

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

\(\Rightarrow2x\left(x-1\right)+\left(x-1\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}x-1=0\\2x+1=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=-\frac{1}{2}\end{cases}}\)

Đặt \(\left(a-1\right)^2=t\)

Ta có: \(\left(a-1\right)^4-11\left(a-1\right)^2+30\)

\(=t^2-11t+30\)

\(=t\left(t-5\right)-6\left(t-5\right)=\left(t-5\right)\left(t-6\right)\)

\(=\left[\left(a-1\right)^2-5\right]\left[\left(a-1\right)^2-6\right]\)

\(=\left(a^2-2a-4\right)\left(a^2-2a-5\right)\)

Đặt \(a^2-2a=k\)

Ta có: \(3\left(a-1\right)^4-18\left(a^2-2a\right)-3\)

\(=3\left(a^2-2a+1\right)^2-18\left(a^2-2a\right)-3\)

\(=3\left(k+1\right)^2-18k-3\)

\(=3k^2+6k+3-18k-3\)

\(=3k^2-12k=3k\left(k-4\right)\)

\(=3\left(a^2-2a\right)\left(a^2-2a-4\right)\)(Ở đây bạn ghi thêm điều kiện nhé)

Khi đó: \(N=\frac{\left(a^2-2a-4\right)\left(a^2-2a-5\right)}{3\left(a^2-2a\right)\left(a^2-2a-4\right)}=\frac{a^2-2a-5}{3\left(a^2-2a\right)}\)