a) 

b)

c)

e)

Những câu còn lại mk hổng bt làm đâu

Từ pt thứ nhất: \(\Leftrightarrow x+1+\sqrt{\left(x+1\right)^2+1}=\left(-y\right)+\sqrt{\left(-y\right)^2+1}\)

Xét hàm \(f\left(t\right)=t+\sqrt{t^2+1}\Rightarrow f'\left(t\right)=1+\dfrac{t}{\sqrt{t^2+1}}=\dfrac{t+\sqrt{t^2+1}}{\sqrt{t^2+1}}\)

\(f'\left(t\right)>\dfrac{t+\sqrt{t^2}}{\sqrt{t^2+1}}=\dfrac{t+\left|t\right|}{\sqrt{t^2+1}}\ge0\Rightarrow f'\left(t\right)>0\) ; \(\forall t\)

\(\Rightarrow f\left(t\right)\) đồng biến trên R

\(\Rightarrow x+1=-y\Rightarrow y=-x-1\)

Thế xuống pt dưới:

\(x^3-\left(3x^2-2x-8\right)\sqrt{2x^2+x-1}=0\)

Bạn coi lại đề, pt vô tỉ này ko giải được

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:

\(x=y=3\)

\(A=\sqrt{\frac{x}{2y^2z^2+xyz}}+\sqrt{\frac{y}{2x^2z^2+xyz}}+\sqrt{\frac{z}{2x^2y^2+xyz}}\)

\(A=\sqrt{\frac{x^2}{2xyz.yz+xz.xy}}+\sqrt{\frac{y^2}{2xyz.xz+xy.yz}}+\sqrt{\frac{z^2}{2xyz.xy+xz.yz}}\)

\(A=\sqrt{\frac{x^2}{yz\left(xy+yz+xz\right)+xz.xy}}+\sqrt{\frac{y^2}{xz\left(xy+yz+xz\right)+xy.yz}}+\sqrt{\frac{z^2}{xy\left(xy+yz+xz\right)+xz.yz}}\)

\(A=\sqrt{\frac{x^2}{\left(yz+xy\right)\left(yz+xz\right)}}+\sqrt{\frac{y^2}{\left(xz+xy\right)\left(xz+yz\right)}}+\sqrt{\frac{z^2}{\left(xy+yz\right)\left(xy+xz\right)}}\)

Áp dụng bđt \(\sqrt{ab}\le\frac{a+b}{2}\) ta có:

\(2A\le\frac{x}{yz+xy}+\frac{x}{yz+xz}+\frac{y}{xz+xy}+\frac{y}{xz+yz}+\frac{z}{xy+yz}+\frac{z}{xy+xz}\)

\(=\frac{x+z}{yz+xy}+\frac{x+y}{yz+xz}+\frac{y+z}{xz+xy}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)

Mà: \(xy+yz+xz=2xyz\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)

\(\Rightarrow2A\le2\Rightarrow A\le1."="\Leftrightarrow a=b=c=\frac{3}{2}\)

\(\sqrt{2x+yz}=\sqrt{x\left(x+y+z\right)+yz}=\sqrt{\left(x+y\right)\left(x+z\right)}\le\dfrac{1}{2}\left(x+y+x+z\right)=\dfrac{1}{2}\left(2x+y+z\right)\)

Tương tự: \(\sqrt{2y+xz}\le\dfrac{1}{2}\left(x+2y+z\right)\) ; \(\sqrt{2z+xy}\le\dfrac{1}{2}\left(x+y+2z\right)\)

Cộng vế:

\(P\le\dfrac{1}{2}\left(4x+4y+4z\right)=4\)

\(P_{max}=4\) khi \(x=y=z=\dfrac{2}{3}\)

P = \(1.\sqrt{2x+yz}+1.\sqrt{2y+xz}+1.\sqrt{2z+xy}\)

\(\le\sqrt{\left(1^2+1^2+1^2\right)\left(2x+yz+2y+xz+2z+xy\right)}\)

\(=\sqrt{3.\left(4+xy+yz+zx\right)}\)

Đã biết x² + y² + z² \(\ge\)xy + yz + zx

=> xy + yz + zx \(\le\dfrac{\left(x+y+z\right)^2}{3}\)

Khi đó \(P\le\sqrt{3\left(4+xy+yz+zx\right)}\le\sqrt{3\left[4+\dfrac{\left(x+y+z\right)^2}{3}\right]}\)

= 4 

Dấu "=" xảy ra <=> x = 2/3