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

\(\Leftrightarrow x^2+6x+9-x^2+1=6x+10=2\)

\(\Leftrightarrow6x=2-10=-8\Leftrightarrow x=-\frac{8}{6}=-\frac{4}{3}\)

Vậy ......

a.\(\left(x+2\right)\left(2-x\right)-\left(2x-1\right)\left(x+3\right)=4-x^2-\left(2x^2+5x-3\right)=7-5x-x^2\)

b.\(\left(x+1\right)^2-2\left(x^2-1\right)+\left(x-1\right)^2=x^2+2x+1-2x^2+2+x^2-2x+1=4\)

c.\(\left(x+y\right)^3-\left(x-y\right)^3-6x^2y=x^3+3x^2y+3xy^2+y^3-\left(x^3-3x^2y+3xy^2-y^3\right)-6x^2y\)

\(=2y^3\)

2.\(\left(x^2-16y^2\right)-3x+12y=\left(x-4y\right)\left(x+4y\right)-3\left(x-4y\right)=\left(x-4y\right)\left(x+4y-3\right)\)

4.\(x^3+6x^2+12x+8=\left(x+2\right)^3\)

bạn ơi  bạn viết sai đề có đề nào viết  thế này đâu?????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????

Uchiha Itachi

a. \(A=4x-x^2+3=7-\left(x^2-4x\right)+4=7-\left(x-2\right)^2\le7\)

b.\(B=x-x^2=\frac{1}{4}-\left(x^2-x+\frac{1}{4}\right)=\frac{1}{4}-\left(x-\frac{1}{2}\right)^2\le\frac{1}{4}\)

c.\(C=2x-2x^2-5=-\frac{9}{2}-2\left(x^2-x+\frac{1}{4}\right)=-\frac{9}{2}-2\left(x-\frac{1}{2}\right)^2\le-\frac{9}{2}\)

a) \(x^2-2x-4y^4-4y^2=\left(x^2-2x+1\right)-\left(4y^4+4y^2+1\right)\)

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

b) \(x^3-4x^2+12x-27=x^3-3x^2-x^2+3x+9x-27\)

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

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

c) \(x^3+2x^2+2x+1=x^3+x^2+x^2+x+x+1\)

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

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

d) \(\left(ab-1\right)^2+\left(a+b\right)^2=a^2b^2-2ab+1+a^2+2ab+b^2\)

\(=a^2b^2+a^2+b^2+1\)

\(=\left(a^2+1\right)\left(b^2+1\right)\)

\(1,\)

\(\left(x^2-9y^2\right)\left(4x+12y\right)\)

\(=\left(x-3y\right)\left(x+3y\right)-4\left(x+3y\right)\)

\(=\left(x+3y\right)\left(x-3y-4\right)\)

\(3,\)

\(-x^2+2xy-y^2+25\)

\(=-\left(x^2-2xy+y^2\right)+25\)

\(=25-\left(x-y\right)^2\)

\(=5^2-\left(x-y\right)^2\)

\(=\left(5-x+y\right)\left(5+x-y\right)\)

\(\left(x+2\right)^2+\left|x-5\right|-x^2-14=0\)

\(\Leftrightarrow x^2+4x+4+\left|x-5\right|-x^2-14=0\)

\(\Leftrightarrow4x+4-14+\left|x-5\right|=0\)

\(\Leftrightarrow\left|x-5\right|=10-4x\)

\(\Leftrightarrow\orbr{\begin{cases}x-5=10-4x\left(ĐK:x\ge5\right)\\x-5=4x-10\left(ĐK:x< 5\right)\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5x=15\\3x=5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\left(KhôngTM\right)\\x=\frac{5}{3}\left(TM\right)\end{cases}}\)

a, \(\left(x+2\right)\left(2-x\right)-\left(2x-1\right)\left(x+3\right)\)

\(=4-x^2-\left(2x^2+5x-3\right)=4-x^2-2x^2-5x+3=-3x^2-5x+7\)

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

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

c, \(\left(x+y\right)^3-\left(x-y\right)^2-6x^2y\)

\(=x^3+3x^2y+3xy^2+y^3-x^2+2xy-y^2-6x^2y\)

\(=x^3-3x^2y+3xy^2+y^3-x^2+2xy-y^2\)