đk: \(x,y\ge-6\Rightarrow x+y\ge0\)

Theo bài ra, ta có: 

\(\left(x+y\right)^2=\left(\sqrt{x+6}+\sqrt{y+6}\right)^2\)

\(=x+y+12+2\sqrt{\left(x+6\right)\left(y+6\right)}\le x+y+12+x+6+y+6\)

Hay \(\left(x+y\right)^2\le2\left(x+y\right)+24\)

\(\Leftrightarrow\left(x+y+4\right)\left(x+y-6\right)\le0\)

\(\Leftrightarrow x+y-6\le0\Leftrightarrow x+y\le6\)

Dấu '=' xảy ra<=> x=y=3

=> GTNN của P là 6 <=> x=y=3

Đặt \(a=\sqrt{x+6};b=\sqrt{y+6}\Rightarrow a;b\ge0,a+b=a^2+b^2-12\)

và \(P=a^2+b^2-12=a+b\)

Ta có: \(a+b=\left(a+b\right)^2-2ab-12\Rightarrow a+b\le\left(a+b\right)^2-12\left(a;b\ge0\right)\)

Hay \(\left(a+b\right)^2-\left(a+b\right)-12\ge0\)

\(\Leftrightarrow\left(a+b+3\right)\left(a+b-4\right)\ge0\)

\(\Leftrightarrow a+b\ge4\Rightarrow P\ge4\)

Dấu '=' xảy ra <=> \(\hept{\begin{cases}a=4\\b=0\end{cases}}\) hoặc \(\hept{\begin{cases}a=0\\b=4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=10\\y=-6\end{cases}}\) hoặc \(\hept{\begin{cases}x=-6\\y=10\end{cases}}\)

ĐK: \(0\le x\le1\).

Ta có: 

với \(0\le x\le1\)thì \(0\le1-x\le1\Leftrightarrow\sqrt{1-x}\left(1-\sqrt{1-x}\right)\ge0\Leftrightarrow\sqrt{1-x}\ge1-x\)

do đó \(x+\sqrt{1-x}\ge x+1-x=1\Rightarrow\sqrt{x+\sqrt{1-x}}\ge1\).

Do đó để phương trình đã cho có nghiệm thì: 

\(\hept{\begin{cases}\sqrt{x}=0\\x+\sqrt{1-x}=1\end{cases}}\Leftrightarrow x=0\left(tm\right)\).

\(x^3-5x^2+14x-4=6\sqrt[3]{x^2-x+1}\)

\(\Leftrightarrow x^3-5x^2+11x-7=6\sqrt[3]{x^2-x+1}-3x-3\)

\(\Leftrightarrow x^3-5x^2+11x-7=3\frac{8x^2-8x+8-\left(x^3+3x^2+3x+1\right)}{4\sqrt[3]{\left(x^2-x+1\right)^2}+2\sqrt[3]{x^2-x+1}\left(x+1\right)+\left(x+1\right)^2}\)

\(\Leftrightarrow\left(x^3-5x^2+11x-7\right)\left(1+\frac{3}{4\sqrt[3]{\left(x^2-x+1\right)^2}+2\sqrt[3]{x^2-x+1}\left(x+1\right)+\left(x+1\right)^2}\right)=0\)

\(\Leftrightarrow x^3-5x^2+11x-7=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+7\right)=0\)

\(\Leftrightarrow x=1\).

cảm ơn anh đã giúp ạ

\(\Leftrightarrow\sqrt{x-2}=4-x\)

\(\Leftrightarrow x-2=16-8x+x^2\)

\(\Leftrightarrow x^2-9x+18=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=3\end{cases}}\)

\(x-4+\sqrt{x-2}=0\)

\(\Leftrightarrow\sqrt{x-2}=-x+4\)

bình phương 2 vế : \(\left|x-2\right|=\left(4-x\right)^2=x^2-8x+16\)

ĐK : \(\left(4-x\right)^2\ge0\Leftrightarrow x\le4\)

TH1 : \(x-2=x^2-8x+16\Leftrightarrow x^2-9x+18=0\)

\(\Delta=81-4.18=9>0\)

\(x_1=\frac{9-3}{2}=3\left(tm\right);x_2=\frac{9+3}{2}=6\left(ktm\right)\)

TH2 : \(-x+2=x^2-8x+16\Leftrightarrow x^2-7x+14=0\)

\(\Delta=49-4.14< 0\)phương trình vô nghiệm 

Vậy tập nghiệm của phương trình là S = { 3 } 

P đạt Max <=> x=5

=> P= \(5\sqrt{3}\)

\(A=\sqrt{x^2-6x+9}-\sqrt{x^2+6x+9}\)

\(A=\sqrt{x^2-6x+3^2}-\sqrt{x^2+6x+3^2}\)

\(A=\sqrt{\left(x-3\right)^2}-\sqrt{\left(x+3\right)^2}\)

b)\(\sqrt{\left(x-3\right)^2}-\sqrt{\left(x+3\right)^2}=1\)

\(TH1:x-3>=0\)

\(< =>x+3>=0\)

\(\left|x-3\right|-\left|x+3\right|=1\)

\(x-3-x-3=1\)

\(-6=1\)(loại)

\(TH2:x-3< =0\)

\(x+3>=0\)

\(< =>\left|x-3\right|-\left|x+3\right|=1\)

\(3-x-x-3\)

\(-2x=1\)

\(x=-\frac{1}{2}\left(TM\right)\)

\(TH3:x-3< =0\)

\(x+3< =0\)

\(< =>\left|x-3\right|-\left|x+3\right|=1\)

\(3-x+X+3=1\)

\(6=1\)(loại)

\(< =>x=\left\{\frac{1}{2}\right\}\)để \(A=1\)