\(1,=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\\ 2,=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\\ 3,=2y\left(y^2+4y+4\right)=2y\left(y+2\right)^2\\ 4,=2\left(x^2+2x+1-y^2\right)=2\left[\left(x+1\right)^2-y^2\right]\\ =2\left(x+y+1\right)\left(x-y+1\right)\\ 5,=16-\left(x-y\right)^2=\left(4-x+y\right)\left(4+x-y\right)\)

2) \(=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\)

3) \(=2y\left(y^2+4y+4\right)=2y\left(y+2\right)^2\)

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

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

5) \(=16-\left(x^2-2xy+y^2\right)=16-\left(x-y\right)^2\)

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

trang ?

 BẠN PHẢI NÓI TRANG CHO MÌNH LÀM CHỨ KHÔNG CHO TRANG THÌ SAO LÀM ĐC

Trang mấy???

1.

Ta có: $4x^2+4x+3=(4x^2+4x+1)+2=(2x+1)^2+2\geq 0+2=2$

$\Rightarrow A=\frac{6}{4x^2+4x+3}\leq \frac{6}{2}=3$

Vậy $A_{\max}=3$. Giá trị này đạt tại $2x+1=0\Leftrightarrow x=\frac{-1}{2}$

2.

$6+4x+x^2=(x^2+4x+4)+2=(x+2)^2+2\geq 0+2=2$

$\Rightarrow \frac{4}{6+4x+x^2}\leq \frac{4}{2}=2$

$\Rightarrow \frac{-4}{6+4x+x^2}\geq -2$

$\Rightarrow B\geq -2$

Vậy $B_{\min}=-2$. Giá trị này đạt tại $x+2=0\Leftrightarrow x=-2$

4.

$D=x+\frac{1}{x}=\frac{15x}{16}+(\frac{x}{16}+\frac{1}{x})$

Do $x\geq 4$ nên $\frac{15x}{16}\geq \frac{15.4}{16}=\frac{15}{4}$

$\frac{x}{16}+\frac{1}{x}\geq 2\sqrt{\frac{1}{16}}=\frac{1}{2}$

$\Rightarrow D=\frac{15x}{16}+(\frac{x}{16}+\frac{1}{x})\geq \frac{15}{4}+\frac{1}{2}=\frac{17}{4}$

Vậy $D_{\min}=\frac{17}{4}$. Giá trị này đạt được khi $x=4$