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

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

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

\(\Rightarrow S=2x\sqrt{\frac{y+1}{x+1}}+2y\sqrt{\frac{x+1}{y+1}}+\left(x+y\right)\sqrt{\left(x+1\right)\left(y+1\right)}\)

\(=\frac{2x}{x+1}\sqrt{\left(x+1\right)\left(y+1\right)}+\frac{2y}{y+1}\sqrt{\left(x+1\right)\left(y+1\right)}+\sqrt{2}\left(x+y\right)\)

\(=\sqrt{2}\left(\frac{2x}{x+1}+\frac{2y}{y+1}+x+y\right)=\sqrt{2}\left(5-\frac{2}{x+1}-\frac{2}{y+1}+x+y\right)\)

\(=\sqrt{2}\left[5-\frac{2\left(x+y+2\right)}{\left(x+1\right)\left(y+1\right)}+x+y\right]=\sqrt{2}\left[5-\left(x+y+2\right)+x+y\right]\)

\(=3\sqrt{2}\)

Khai  triển nó ra,ta có:

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

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

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

Ta có:\(P=\Sigma x\sqrt{\frac{\left(y+x\right)\left(y+z\right)\left(z+x\right)\left(z+y\right)}{\left(x+y\right)\left(x+z\right)}}\)

\(\Sigma x\cdot\left(y+z\right)\)

Rút gọn dc như vậy rồi chị làm nốt ạ

zZz Cool Kid zZz ghê zữ nhen:D

a) ĐK: \(x\ge-2\) (chắc vậy:D)

Chú ý x = -2 không phải là một nghiệm. Xét x > -2

PT \(\Leftrightarrow2\left(x^2-10x-12\right)+20x+40=5\sqrt{2x^3+16}\)

\(\Leftrightarrow2\left(x^2-10x-12\right)+5\left[\left(4x+8\right)-\sqrt{2x^3+16}\right]=0\)

\(\Leftrightarrow2\left(x^2-10x-12\right)-\frac{10\left(x+2\right)\left(x^2-10x-12\right)}{4x+8+\sqrt{2x^3+16}}=0\)

\(\Leftrightarrow\left(x^2-10x-12\right)\left[2-\frac{10\left(x+2\right)}{4x+8+\sqrt{2x^3+16}}\right]=0\)

Xét cái ngoặc to: \(=\frac{2\left[\sqrt{2x^3+16}-\left(x+2\right)\right]}{4x+8+\sqrt{2x^3+16}}\)

\(=\frac{\frac{2\left(x+2\right)\left(2x^2-5x+6\right)}{\sqrt{2x^3+16}+x+2}}{4x+8+\sqrt{2x^3+16}}>0\forall x>-2\)

Do đó\(x^2-10x-12=0\Rightarrow...\)

b) Nghiệm quá xấu -> Chịu.

P/s: Em ko chắc đâu á, nhất là chỗ xét x> -2 ấy, ko biết có được ko? Với cả xử lý cái ngoặc to em ko chắc là mình nhầm chỗ nào đâu đấy nhá!

a/ \(5\sqrt{2x^3+16}=2\left(x^2+8\right)\)

\(\Leftrightarrow25\left(2x^3+16\right)=4\left(x^2+8\right)\)

\(\Leftrightarrow2x^4-50x^3+32x^2-72=0\)

\(\Leftrightarrow\left(x^2-10x-12\right)\left(2x^2-5x+6\right)=0\)

Dễ thấy \(2x^2-5x+6>0\)

\(\Rightarrow x^2-10x-12=0\)

\(\(b)\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\left(a,b\ge0;a,b\ne1\right)\)\)

\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{a}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab+1}\right)}\)\)

\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)

\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)

\(\(=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{ab}-1\right)}\left(a,b\ge0.a,b\ne1\right)\)\)

_Minh ngụy_

\(\(c)\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)\)( tự ghi điều kiện )

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

\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(x\sqrt{x}+x\sqrt{y}-2x\sqrt{y}-2y\sqrt{x}+y\sqrt{x}+y\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)

\(\(=\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)\)( phá ngoặc và tính )

\(\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\sqrt{xy}\)\)

_Minh ngụy_

\(\(d)\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\left(x\ge0\right)\)\)

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

\(\(=\frac{|\sqrt{x}-1|}{|\sqrt{x}+1|}\)\)

\(\(=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)\)( vì \(\(x\ge0\)\))

_Minh ngụy_

giúp mình với ^^