\(\left(1+x^2\right)\left(1+y^2\right)+4xy+2\left(x+y\right)\left(1+xy\right)\)

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

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

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

\(=\left(x+y+1+xy\right)^2\) là SCP

(1+x²)(1+y²)+4xy+2(x+y)(1+xy)

 = 1+y²+x²+x²y²+2xy+2xy+2(x+y)(1+xy)

 =(x²+2xy+y²)+(x²y²+2xy+1)+2(x+y)(1+xy)

 =(x+y)²+(xy+1)²+2(x+y)(1+xy)

 =(x+y+xy+1)²

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

\(x^2+2x-x^2+4x-4=8\)

\(6x-4=8\)

\(x=2\)

Lời giải:

a.

Tại $x=5$ thì $B=\frac{5+3}{5-2}=\frac{8}{3}$

b.

\(A=\frac{x^2-x+1}{(x-2)(x+2)}+\frac{2(x+2)}{(x-2)(x+2)}-\frac{x-2}{(x-2)(x+2)}=\frac{x^2-x+1+2(x+2)-(x-2)}{(x-2)(x+2)}\)

\(=\frac{x^2+7}{(x-2)(x+2)}\)

c.

\(P=A:B(x+2)=\frac{x^2+7}{(x-2)(x+2)}:\frac{x+3}{x-2}.(x+2)=\frac{x^2+7}{x+3}\)

Áp dụng BĐT Cô-si:

$x^2+1\geq 2|x|\geq 2x$

$\Rightarrow x^2+7\geq 2x+6=2(x+3)$

$\Rightarrow P\geq \frac{2(x+3)}{x+3}=2$
Vậy $P_{\min}=2$. Giá trị này đạt tại $x^2=1\Leftrightarrow x=\pm 1$ (tm)

Áp dụng ddL pytago vào Δ abc( góc a =90 )

BC² =AB² +AC²

⇒BC²=16+9=25

⇒BC=5

Xét Δabc vông tại a có:

AH=\(\dfrac{1}{2}\) BC=\(\dfrac{5}{2}\) =2.5 (CM)

Vậy AH=2.5cm