a) \(\left(\sqrt{11}+\sqrt{14}\right)^2=25+\sqrt{154}\)

\(\left(2\sqrt{12}\right)^2=24+\sqrt{144}\)

Vậy \(2\sqrt{12}< \sqrt{11}+\sqrt{14}\)

b) \(\left(\sqrt{a+1}+\sqrt{a+3}\right)^2=2a+4+\sqrt{\left(a+1\right)\left(a+3\right)}\)

\(\left(2\sqrt{a+2}\right)^2=2a+4+\sqrt{\left(a+2\right)\left(a+2\right)}\)

Vậy \(\sqrt{a+1}+\sqrt{a+3}< 2\sqrt{a+2}\)

\(\Leftrightarrow\hept{\begin{cases}\left(x-y\right)^2=1\\2x^2+3y^2-4xy+1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1+y\\x=-1+y\end{cases}}\)

Thay vô rùi giải nhá

Vô nghiệm đúng không bạn :3333

\(DK:x\notin\left(0;2\right)\)

Dat \(\hept{\begin{cases}\sqrt{2x^2+1}=a\\\sqrt{x^2-2x}=b\end{cases}\left(a,b\ge0\right)}\)

\(\Rightarrow\hept{\begin{cases}\sqrt{x^2-x+2}=b^2+x+2\\\sqrt{2x^2+x+3}=a^2+x+2\end{cases}}\)

PT tro thanh

\(a+b^2+x+2=a^2+x+2+b\)

\(\Leftrightarrow a^2-b^2+b-a=0\)

\(\Leftrightarrow\left(a+b\right)\left(a-b\right)-\left(a-b\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(a+b-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=b\left(1\right)\\a+b=1\left(2\right)\end{cases}}\)

PT(1)\(\Leftrightarrow\sqrt{2x^2+1}=\sqrt{x^2-2x}\)

\(\Leftrightarrow2x^2+1=x^2-2x\)

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

\(\Leftrightarrow x=-1\left(n\right)\)

PT(2)\(\Leftrightarrow\sqrt{2x^2+1}+\sqrt{x^2-2x}=1\)

\(\Leftrightarrow3x^2-2x+2\sqrt{\left(2x^2+1\right)\left(x^2-2x\right)}=0\)

\(\Leftrightarrow2\sqrt{2x^4-4x^3+x^2-2x}=2x-3x^2\)

\(\Leftrightarrow8x^4-16x^3+4x^2-8x=4x^2-12x^3+9x^4\)

\(\Leftrightarrow x^4+4x^3+8x=0\)

\(\Leftrightarrow x\left(x^3+4x^2+8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^3+4x^2+8=0\end{cases}}\)

Cái PT \(x^3+4x^2+8=0\)có nghiệm nên mỉnh gọi là alpha nhé

Vay nghiem cua PT la \(x_1=-1;x_2=0;x_3=\alpha\)

Cau o duoi lam 

\(DK:x\notin\left(0;2\right)\)

\(\Leftrightarrow3x^2-x+3+2\sqrt{\left(2x^2+1\right)\left(x^2-x+2\right)}=3x^2-x+3+2\sqrt{\left(x^2-2x\right)\left(2x^2+x+3\right)}\)

\(\Leftrightarrow2x^4-2x^3+5x^2-x+2=2x^4-3x^3+x^2-6x\)

\(\Leftrightarrow x^3+4x^2+5x+2=0\)

\(\Leftrightarrow\left(x^3+1\right)+\left(4x^2+5x+1\right)=0\)

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

\(\Leftrightarrow\left(x+1\right)\left(x^2+3x+2\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)

Vay nghiem cua PT la \(x=-1;x=-2\)

Dat \(P=\sqrt{a^4+1}+\sqrt{b^4+1}\)

Ta co:\(\sqrt{\left(a^4+1\right)\left(1+16\right)}\ge a^2+4\)

\(\sqrt{\left(b^4+1\right)\left(1+16\right)}\ge b^2+4\)

\(\Rightarrow\sqrt{17}\left(\sqrt{a^4+1}+\sqrt{b^4}+1\right)\ge a^2+b^2+8\ge\frac{1}{2}+8=\frac{17}{2}\)

\(\Leftrightarrow\sqrt{a^4+1}+\sqrt{b^4+1}\ge\frac{17}{2\sqrt{17}}\)

Dau '=' ra khi \(a=b=\frac{1}{2}\)

Vay \(P_{min}=\frac{17}{2\sqrt{17}}\)khi \(a=b=\frac{1}{2}\)

Cảm ơn bạn nhưng lúc chiều vừa được cô giảng rồi (-_-)

\(\sqrt{a}>0\) nên A < 0 \(\Leftrightarrow a-1< 0\)

\(\Leftrightarrow0< a< 1\)

\(A< 0\Leftrightarrow\frac{a-1}{\sqrt{a}}< 0\)

Mà \(\sqrt{a}>0\Rightarrow a-1< 0\)

\(\Rightarrow a< 1\)

\(\Rightarrow\)Để \(\frac{a-1}{\sqrt{a}}< 0\Leftrightarrow0< a< 1\)

\(pt\Leftrightarrow\sqrt{x-y+2010}-\sqrt{2010}=\sqrt{x}-\sqrt{y}\)

\(\Leftrightarrow\frac{x-y}{\sqrt{x-y+2010}+\sqrt{2010}}=\frac{x-y}{\sqrt{x}+\sqrt{y}}\)

\(\Leftrightarrow\left(x-y\right)\left(\frac{1}{\sqrt{x-y+2010}+\sqrt{2010}}-\frac{1}{\sqrt{x}+\sqrt{y}}\right)=0\)

MK giải đc đến đây bạn làm nốt hộ mk nhá :)