\(x^2+3x+x+3=\left(x+1\right)\left(x+3\right)\)

\(=x^2+x+3x+3=x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(x+3\right)\)

\(x^2+x+3x+3=x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(x+3\right)\)

\(=x^2+8x-x-8=x\left(x+8\right)-\left(x+8\right)=\left(x+8\right)\left(x-1\right)\)

\(x^2+8x-x-8=8\left(x-1\right)+x\left(x-1\right)=\left(8+x\right)\left(x-1\right)\)

   x²+7x-8
= x²-x+8x-8
= (x²-x)+(8x-8)
= x(x-1)+8(x-1)
= (x-1)(x+8)

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

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

\(=100^2=10000\)

\(\Leftrightarrow x^2+3x+2-x^2+6x-9=11\)

=>x=2

\(\left(x+1\right)\left(x+2\right)-\left(x-3\right)^2=11\)

\(\Leftrightarrow x^2+3x+2-x^2+6x-9-11=0\)

\(\Leftrightarrow9x-18=0\)

<=> 9. (x - 2) = 0

<=> x = 2

\(\Leftrightarrow x^2+3x+2-x^2+6x-9=11\)

hay x=2

\(3x+2\left(5-x\right)=0\Leftrightarrow3x+10-2x=0\Leftrightarrow x=-10\)

\(=x^2+2\cdot6\cdot x+6^2=\left(x+6\right)^2\\ =\left(94+6\right)^2=100^2=10000\)

a. x\(^2\) + 12x + 36 = x\(^2\) + 2 . x . 6 + 6\(^2\) = ( x + 6 )\(^2\)

thay x = 94 vào biểu thức, ta có:

( 94 + 6 )\(^2\) = 100\(^2\) = 10000

vậy x + 12x + 36 = 10000 tại x = 94.

\(3x+2\left(5-x\right)=0\\ \Rightarrow3x+\left(10-2x\right)=0\\ \Rightarrow3x+10-2x=0\\ \Rightarrow\left(3x-2x\right)+10=0\\ \Rightarrow x+10=0\\ \Rightarrow x=-10\)