đk: \(\hept{\begin{cases}x\ge\frac{3}{2}\\y\ge\frac{3}{2}\end{cases}}\)

Xét y = 0 => PT vô nghiệm

Xét y khác 0:

Ta có: \(x^3+y^3-8xy\sqrt{2\left(x^2+y^2\right)}+7x^2y+7xy^2=0\)

\(\Leftrightarrow x^3+y^3+7xy\left(x+y\right)=8xy\sqrt{2\left(x^2+y^2\right)}\)

\(\Leftrightarrow\frac{\left(x^3+y^3\right)}{y^3}+\frac{7xy\left(x+y\right)}{y^3}=\frac{8xy\sqrt{2\left(x^2+y^2\right)}}{y^3}\)

\(\Leftrightarrow\left(\frac{x}{y}\right)^3+1+7\cdot\frac{x}{y}\cdot\left(1+\frac{x}{y}\right)=8\cdot\frac{x}{y}\cdot\sqrt{2+2\left(\frac{x}{y}\right)^2}\)

Đặt \(\frac{x}{y}=t>0\) khi đó: \(PT\Leftrightarrow t^3+1+7t\left(1+t\right)=8t\sqrt{2\left(1+t^2\right)}\)

\(=\left[8t\sqrt{2\left(1+t\right)^2}-8t\left(t+1\right)\right]+8t\left(t+1\right)\)

\(\Leftrightarrow t^3-t^2-t+1=8t\cdot\frac{2\left(1+t^2\right)-\left(t+1\right)^2}{\sqrt{2\left(1+t^2\right)}+t+1}\)

\(\Leftrightarrow t^2\left(t-1\right)-\left(t-1\right)=8t\cdot\frac{2+2t^2-t^2-2t-1}{\sqrt{2\left(1+t^2\right)}+t+1}\)

\(\Leftrightarrow\left(t-1\right)^2\left(t+1\right)=8t\cdot\frac{\left(t-1\right)^2}{\sqrt{2\left(1+t^2\right)}+t+1}\)

\(\Leftrightarrow\left(t-1\right)^2\left[t+1-\frac{1}{\sqrt{2\left(1+t^2\right)}+t+1}\right]=0\)

Mà \(t+1-\frac{1}{\sqrt{2\left(1+t^2\right)}+t+1}=\frac{t\left(\sqrt{2\left(1+t^2\right)}+t+1\right)+\sqrt{2\left(1+t^2\right)}+t}{\sqrt{2\left(1+t^2\right)}+t+1}>0\)

\(\Rightarrow t-1=0\Leftrightarrow t=1\Leftrightarrow\frac{x}{y}=1\Rightarrow x=y\)

Khi đó \(\sqrt{y}-\sqrt{2x-3}+2x=6\)

\(\Leftrightarrow\sqrt{x}-\sqrt{2x-3}=6-2x\)

\(\Leftrightarrow\frac{x-2x+3}{\sqrt{x}+\sqrt{2x-3}}=2\left(3-x\right)\)

\(\Leftrightarrow\frac{3-x}{\sqrt{x}+\sqrt{2x-3}}=2\left(3-x\right)\)

\(\Leftrightarrow\left(x-3\right)\left(2-\frac{1}{\sqrt{x}+\sqrt{2x-3}}\right)=0\)

Nếu \(2-\frac{1}{\sqrt{x}+\sqrt{2x-3}}=0\)

\(\Leftrightarrow\frac{1}{\sqrt{x}+\sqrt{2x-3}}=2\)

\(\Leftrightarrow\sqrt{x}+\sqrt{2x-3}=\frac{1}{2}\)

\(\Leftrightarrow\sqrt{x}=\frac{\frac{13}{2}-2x}{2}\) (CMT)

\(\Leftrightarrow4\sqrt{x}=13-4x\)

\(\Leftrightarrow16x=169-104x+16x^2\)

\(\Leftrightarrow16x^2-120x+169=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=y=\frac{15+2\sqrt{14}}{4}\\x=y=\frac{15-2\sqrt{14}}{4}\end{cases}}\)

Nếu \(x-3=0\Rightarrow x=y=3\)

Vậy ta có 3 cặp số (x;y) thỏa mãn: ...

Thử lại ta thấy cặp nghiệm vô tỉ:

\(x=y=\frac{15\pm2\sqrt{14}}{4}\) không thỏa mãn nên ta chỉ có 1 cặp nghiệm thỏa mãn:

\(x=y=3\)

\(\frac{\sqrt{\sqrt{5}+\sqrt{2}}}{\sqrt{3\sqrt{5}-3\sqrt{2}}}=\frac{\sqrt{\sqrt{5}+\sqrt{2}}}{\sqrt{3.\left(\sqrt{5}-\sqrt{2}\right)}}=\frac{\sqrt{\sqrt{5}+\sqrt{2}}}{\sqrt{3}.\sqrt{\sqrt{5}-\sqrt{2}}}\)

\(=\frac{(\sqrt{\sqrt{5}+\sqrt{2}})^2}{\sqrt{3}.\sqrt{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}}=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{3}.\sqrt{5-2}}\)

\(=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{3}.\sqrt{3}}=\frac{\sqrt{5}+\sqrt{2}}{3}\)

\(M=\frac{2\sqrt{a}}{\sqrt{a}+3}+\frac{\sqrt{a}+1}{\sqrt{a}-3}+\frac{3-11\sqrt{a}}{9-a}\)

ĐK : \(\hept{\begin{cases}a\ge0\\a\ne9\end{cases}}\)

\(=\frac{2\sqrt{a}}{\sqrt{a}+3}+\frac{\sqrt{a}+1}{\sqrt{a}-3}+\frac{11\sqrt{a}-3}{a-9}\)

\(=\frac{2\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}+\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}+\frac{11\sqrt{a}-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\)

\(=\frac{2a-6\sqrt{a}}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}+\frac{a+4\sqrt{a}+3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}+\frac{11\sqrt{a}-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\)

\(=\frac{2a-6\sqrt{a}+a+4\sqrt{a}+3+11\sqrt{a}-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\)

\(=\frac{3a+9\sqrt{a}}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\)

\(=\frac{3\sqrt{a}\left(\sqrt{a}+3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}=\frac{3\sqrt{a}}{\sqrt{a}-3}\)

b) Thiếu đề

pt \(\Leftrightarrow M=\frac{2\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}+\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}-\frac{3-11\sqrt{a}}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)

\(\Leftrightarrow M=\frac{2a-6\sqrt{a}+a+3+4\sqrt{a}-3+11\sqrt{a}}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)

\(\Leftrightarrow M=\frac{3a-9\sqrt{a}}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)

\(\Leftrightarrow M=\frac{3\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)

\(\Leftrightarrow M=\frac{3\sqrt{a}}{\sqrt{a}+3}\)

Iem cần gấp

Ta có công thức tổng quát: \(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{n+1}-\sqrt{n}\)(*)

Áp dụng (*), ta được: \(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}=\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{3}-\sqrt{2}\right)+\left(\sqrt{4}-\sqrt{3}\right)+...+\left(\sqrt{100}-\sqrt{99}\right)=\sqrt{100}-\sqrt{1}=9\left(đpcm\right)\)

Trục căn thức ở mẫu : 

\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)

\(=\frac{\sqrt{1}-\sqrt{2}}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{1}-\sqrt{2}\right)}+\frac{\sqrt{2}-\sqrt{3}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}+\frac{\sqrt{3}-\sqrt{4}}{\left(\sqrt{3}+\sqrt{4}\right)\left(\sqrt{3}-\sqrt{4}\right)}+...+\frac{\sqrt{99}-\sqrt{100}}{\left(\sqrt{99}+\sqrt{100}\right)\left(\sqrt{99}-\sqrt{100}\right)}\)

\(=\frac{\sqrt{1}-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+\frac{\sqrt{3}-\sqrt{4}}{3-4}+...+\frac{\sqrt{99}-\sqrt{100}}{99-100}\)

\(=\frac{\sqrt{1}-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1}+\frac{\sqrt{3}-\sqrt{4}}{-1}+...+\frac{\sqrt{99}-\sqrt{100}}{-1}\)

\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{100}-\sqrt{99}\)

\(=\sqrt{100}-\sqrt{1}\)

\(=10-1=9\)

=> đpcm

\(\sqrt{x^2+5x+20}=4\)

ĐK : x ∈ R

Bình phương hai vế

<=> \(x^2+5x+20=16\)

<=> \(x^2+5x+20-16=0\)

<=> \(x^2+5x+4=0\)(1)

Ta có : a - b + c = 1 - 5 + 4 = 0

=> (1) có hai nghiệm \(\hept{\begin{cases}x_1=-1\\x_2=-\frac{c}{a}=-\frac{4}{1}=-4\end{cases}}\)

\(\sqrt{x^2+5x+20}=4\)

\(\Rightarrow\left(\sqrt{x^2+5x+20}\right)^2=16\)

\(\Leftrightarrow x^2+5x+20=16\)

\(\Leftrightarrow x^2+5x+4=0\)

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

\(\Leftrightarrow x\left(x+1\right)+4\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-4\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{-1;-4\right\}\)

\(ĐKXĐ:y\ge0\)

\(x^2-4x+y-6\sqrt{y}+13=0\)

\(\Leftrightarrow x^2-4x+4+y-6\sqrt{y}+9=0\)

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

Vì \(\left(x-2\right)^2\ge0\forall x\); \(\left(\sqrt{y}-3\right)^2\ge0\forall y\ge0\)

\(\Rightarrow\left(x-2\right)^2+\left(\sqrt{y}-3\right)^2\ge0\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x-2=0\\\sqrt{y}-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\\sqrt{y}=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=9\left(t/m\right)\end{cases}}\)( t/m là thỏa mãn )

Vậy \(x=2\)và \(y=9\)

\(x^2-4x+y-6\sqrt{y}+13=0\)

<=> \(\left(x^2-4x+4\right)+\left(y-6\sqrt{y}+9\right)=0\)

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

Ta có : \(\hept{\begin{cases}\left(x-2\right)^2\ge0\forall x\\\left(\sqrt{y}-3\right)^2\ge0\forall y\ge0\end{cases}}\Leftrightarrow\left(x-2\right)^2+\left(\sqrt{y}-3\right)^2\ge0\forall x,\left(y\ge0\right)\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-2=0\\\sqrt{y}-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=9\left(tm\right)\end{cases}}\)

\(A=x^3+y^3-2xy=\left(x+y\right)\left(x^2-xy+y^2\right)-2xy=2x^2-4xy-2y^2\)

\(=2\left(x-y\right)^2\ge0.\text{ Dấu bằng: }x=y=1\)