a)

ta có \(x^2+8x+7\)

\(=x^2+7x+x+7\)

\(=x\left(x+7\right)+\left(x+7\right)\)

\(=\left(x+7\right)\left(x+1\right)\)

b)

Ta có \(8x^2+30x+7\)

\(=8x^2+2x+28x+7\)

\(=2x\left(4x+1\right)+7\left(4x+1\right)\)

\(\left(4x+1\right)\left(2x+7\right)\)

A)(x+1)(x+7)

B)(x+1/4)(x+7/2)

giải rõ hơn dc không bạn ??

a) \(=3\left(x^2-10x+25\right)=3\left(x-5\right)^2\)

b) \(=x\left(x+y\right)+8\left(x+y\right)=\left(x+y\right)\left(x+8\right)\)

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

a) 

b) 

c) 

Mik ko viết 2 nhỏ đc nên thông cảm nha mn

a) \(8x^2+30x+7=0\)
\(\Rightarrow8x^2+2x+28x+7=0\)
\(\Rightarrow2x\left(4x+1\right)+7\left(4x+1\right)=0\)
\(\Rightarrow\left(2x+7\right)\left(4x+1\right)=0\)
\(\Rightarrow\)\(2x+7=0\)  hoặc  \(4x+1=0\)
\(\Rightarrow\)\(2x=-7\)          ;     \(4x=-1\)
\(\Rightarrow\)\(x=\frac{-7}{2}\)             ;     \(x=\frac{-1}{4}\)
Vậy \(x\in\left\{\frac{-7}{2};\frac{-1}{4}\right\}\)

b) \(x^3-11x^2+30x=0\)
\(\Rightarrow x\left(x^2-11x+30\right)=0\)
\(\Rightarrow x\left(x^2-6x-5x+30\right)=0\)
\(\Rightarrow x\left[x\left(x-6\right)-5\left(x-6\right)\right]=0\)
\(\Rightarrow x\left(x-5\right)\left(x-6\right)=0\)
\(\Rightarrow\)\(x=0\)  hoặc  \(x-5=0\)  hoặc  \(x-6=0\)
\(\Rightarrow\)\(x=0\)     ;      \(x=5\)               ;     \(x=6\)
Vậy \(x\in\left\{0;5;6\right\}\)

a)\(8x^2+30x+7=0\Leftrightarrow8x^2+2x+28x+7=0\Leftrightarrow2x\left(4x+1\right)+7\left(4x+1\right)=0\)

\(\Leftrightarrow\left(2x+7\right)\left(4x+1\right)=0\Leftrightarrow\orbr{\begin{cases}2x+7=0\\4x+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{7}{2}\\x=-\frac{1}{4}\end{cases}}\)

b)\(x^3-11x^2+30x=0\Leftrightarrow x\left(x^2-11x+30\right)=0\Leftrightarrow x\left(x^2-5x-6x+30\right)=0\)

\(\Leftrightarrow x\left[x\left(x-5\right)-6\left(x-5\right)\right]=0\Leftrightarrow x\left(x-6\right)\left(x-5\right)=0\)

<=>x=0 hoặc x-6=0 hoặc x-5=0 <=> x=0 hoặc x=6 hoặc x=5

( x² + 8x + 7 ) ( x² + 8x + 15 ) + 15

Đặt x² + 8x + 7 = y ta có:

   y ( y + 8 ) + 15 

= y² + 8y + 15 

= ( y + 3 ) ( y + 5 )

= ( x² + 8x + 10 ) ( x² + 8x + 12 ) 

= ( x² + 8x + 10 ) ( x + 2 ) ( x + 6 )

Đặt x² + 8x + 7 = y ta có:

   y ( y + 8 ) + 15 

= y² + 8y + 15 

= ( y + 3 ) ( y + 5 )

= ( x² + 8x + 10 ) ( x² + 8x + 12 ) 

= ( x² + 8x + 10 ) ( x + 2 ) ( x + 6 )

đặt t = x² + 8x + 7

có : t ( t + 8 ) + 15 

= t² + 8t + 15

= t² + 5t + 3t + 15

= ( t² + 5t ) + ( 3t + 15 )

= t ( t + 5 ) + 3 ( t + 5 )

= ( t + 3 ) ( t + 5 )

mà t = x² + 8x + 7

⇒ ( x² + 8x + 10 ) ( x² + 8x + 12 )

\(2x^4-8x^3-7x^3+28x^2+7x^2-28x-2x+8\\ =2x^3\left(x-4\right)-7x^2\left(x-4\right)+7x\left(x-4\right)-2\left(x-4\right)\\ =\left(x-4\right)\left(2x^3-7x^2+7x-2\right)\\ =\left(x-4\right)\left(2x^3-4x^2-3x^2+6x+x-2\right)\\ =\left(x-4\right)\left[2x^2\left(x-2\right)-3x\left(x-2\right)+\left(x-2\right)\right]\\ =\left(x-4\right)\left(x-2\right)\left(2x^2-2x-x+1\right)\\ =\left(x-4\right)\left(x-2\right)\left(2x-1\right)\left(x-1\right)\)

\(-\left(x+2y\right)^2\)

\(-\left(x-3\right)^2\)

\(\left(3-5x\right)^2\)

\(-x^2-4xy-4y^2=-\left(x+2y\right)^2\)

\(-x^2+6x-9=-\left(x-3\right)^2\)

\(25x^2-30x+9=\left(5x-3\right)^2\)

\(x^2-y^2+8x+6y+7\)

\(=\left(x-y\right)\left(x+y\right)+7\left(x+y\right)+x-y+7\)

\(=\left(x+y\right)\left(x-y+7\right)+\left(x-y+7\right)\)

\(=\left(x+y+1\right)\left(x-y+7\right)\)

Ta có x² - y² + 8x + 6y + 7

= x² + 8x + 16 - y² + 6y - 9

= \(x^2+4x+4x+16-y^2+3y+3y-9\)

= x(x + 4) + 4(x + 4) - y(y - 3) + 3(y - 3)

= (x + 4)² - (y - 3)²

= (x + 4 + y - 3)(x + 4 - y + 3) 

= (x + y + 1)(x - y + 7)