ĐK: x, y>=-2

\(pt\Leftrightarrow\sqrt{x+2}-\sqrt{y+2}+x^3-y^3=0\)

\(\Leftrightarrow\frac{x-y}{\sqrt{x+2}+\sqrt{y+2}}+\left(x-y\right)\left(x^2+xy+y^2\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(\frac{1}{\sqrt{x+2}+\sqrt{y+2}}+x^2+xy+y^2\right)=0\)

\(\Leftrightarrow x=y\)

Thay vào T=\(x^2+2x^2-2x^2+2x+10=x^2+2x+1+9=\left(x+1\right)^2+9\ge9\)

"=" xảy ra khi và chỉ khi x=y=-1 (thỏa mãn)

Vậy min T=9 khi x=y=-1

ý em mới hoc lớp 8 thui

Ta có \(\sqrt{x+2}-y^3=\sqrt{y+2}-x^3\)

\(\Leftrightarrow\sqrt{x+2}+x^3=\sqrt{y+2}+y^3\)

 Đặt \(f\left(x\right)=\sqrt{x+2}+x^3\). Ta chứng minh \(f\left(x\right)\) là hàm số đồng biến với \(x\ge-2\)

Giả sử \(f\left(a\right)>f\left(b\right)\) với \(a,b\ge-2\)

\(\Rightarrow\sqrt{a+2}+a^3>\sqrt{b+2}+b^3\)

\(\Leftrightarrow\sqrt{a+2}-\sqrt{b+2}+a^3-b^3>0\)

\(\Leftrightarrow\dfrac{a-b}{\sqrt{a+2}+\sqrt{b+2}}+\left(a-b\right)\left(a^2+ab+b^2\right)>0\)

\(\Leftrightarrow\left(a-b\right)\left(\dfrac{1}{\sqrt{a+2}+\sqrt{b+2}}+a^2-ab+b^2\right)>0\)     (*)

 Dễ thấy \(\dfrac{1}{\sqrt{a+2}+\sqrt{b+2}}+a^2+ab+b^2>0\) với mọi \(a,b\ge-2\)

 Do đó từ (*) suy ra \(a>b\).

 Vậy ta có \(f\left(a\right)>f\left(b\right)\Rightarrow a>b\). Do đó \(f\) là hàm số đồng biến.

 Theo trên, ta có \(f\left(x\right)=f\left(y\right)\Rightarrow x=y\)

 Thay vào biểu thức B, ta có \(B=x^2+2x+10\)

\(B=\left(x+1\right)^2+9\) \(\ge9\).

 Dấu "=" xảy ra \(\Leftrightarrow x=-1\) (nhận) \(\Rightarrow y=-1\)

 Vậy GTNN của B là 9, xảy ra khi \(\left(x;y\right)=\left(-1;-1\right)\)

\(\sqrt{x+2}-y^3=\sqrt{y+2}-x^3\)

\(\Leftrightarrow\left(\sqrt{x+2}-\sqrt{y+2}\right)+\left(x^3-y^3\right)=0\)

\(\Leftrightarrow\dfrac{x+2-y-2}{\sqrt{x+2}+\sqrt{y+2}}+\left(x-y\right)\left(x^2-xy+y^2\right)=0\)

\(\Leftrightarrow\left(\dfrac{1}{\sqrt{x+2}+\sqrt{y+2}}+x^2-xy+y^2\right)\left(x-y\right)=0\)

⇒ x = y. Thay vào A

\(\Rightarrow A=x^2+2x^2-2x^2+2x+10\)

\(=\left(x+1\right)^2+9\ge9\)

Suy ra Min A = 9 ⇔ x = y = - 1

\(A=x^2+2xy-2y^2+2y+10\)

\(\Leftrightarrow A=x^2+2xy+y^2-3y^2+2y-\dfrac{1}{3}+\dfrac{31}{3}\)

\(\Leftrightarrow A=\left(x^2+2xy+y^2\right)-\left(3y^2-2y+\dfrac{1}{3}\right)+\dfrac{31}{3}\)

\(\Leftrightarrow A=\left(x+y\right)^2-3\left(y^2-\dfrac{2}{3}y+\dfrac{1}{9}\right)+\dfrac{31}{3}\)

\(\Leftrightarrow A=\left(x+y\right)^2-3\left[y^2-2.y.\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2\right]+\dfrac{31}{3}\)

\(\Leftrightarrow A=\left(x+y\right)^2-3\left(y-\dfrac{1}{3}\right)^2+\dfrac{31}{3}\)

Vậy GTNN của \(A=\dfrac{31}{3}\) khi \(\left\{{}\begin{matrix}x+y=0\\y-\dfrac{1}{3}=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{1}{3}=0\\y=\dfrac{1}{3}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\)

Ta có: 

\(2\left(2x^2+xy+2y^2\right)=3\left(x^2+y^2\right)+\left(x+y\right)^2\ge\dfrac{3}{2}\left(x+y\right)^2+1\left(x+y\right)^2=\dfrac{5}{2}\left(x+y\right)^2\)

\(\Rightarrow\sqrt{2x^2+xy+2y^2}\ge\dfrac{\sqrt{5}}{2}\left(x+y\right)\)

Gợi ý. Dùng cái trên.

Mọi người giúp mình với a :))

Ta có : \(P=\sqrt{2x^2+xy+2y^2}+\sqrt{2y^2+yz+2z^2}+\sqrt{2z^2+xz+2x^2}\)

Xét : \(\sqrt{2x^2+xy+2y^2}=\sqrt{\dfrac{3}{4}.\left(x-y\right)^2+\dfrac{5}{4}.\left(x+y\right)^2}\)

\(\ge\sqrt{\dfrac{5}{4}.\left(x+y\right)^2}=\dfrac{\sqrt{5}}{2}.\left(x+y\right)\)

\(CMTT:\sqrt{2y^2+yz+2z^2}\ge\dfrac{\sqrt{5}}{2}.\left(y+z\right)\)

                \(\sqrt{2z^2+xz+2x^2}\ge\dfrac{\sqrt{5}}{2}.\left(x+z\right)\)

Do đó : \(P\ge\dfrac{\sqrt{5}}{2}.\left(x+y+y+z+z+x\right)=\dfrac{2\sqrt{5}.\left(x+y+z\right)}{2}\)

\(\Rightarrow P\ge\sqrt{5}.\left(x+y+z\right)\)

Ta có : BĐT : \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

\(\Rightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)

Mà : \(xy+yz+zx=3\)

\(\Rightarrow\left(x+y+z\right)^2\ge9\)

\(\Leftrightarrow x+y+z\ge3\)

\(\Rightarrow P_{min}=3\sqrt{5}\)

Dấu bằng xảy ra : \(\Leftrightarrow x=y=z=1\)