làm r mà bạn ei

Ta có:

\(x\sqrt{1-y^2}+y.\sqrt{1-x^2}\le\dfrac{1}{2}\left(x^2+1-y^2\right)+\dfrac{1}{2}\left(y^2+1-x^2\right)=1\)

Đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}x=\sqrt{1-y^2}\\y=\sqrt{1-x^2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^2=1-y^2\\y^2=1-x^2\end{matrix}\right.\)

\(\Rightarrow x^2+y^2=1\) (đpcm)

Lời giải:

$(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=2$

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

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

$\Leftrightarrow 2x+\sqrt{y^2+1}=2\sqrt{x^2+1}-y$

$\Rightarrow (2x+\sqrt{y^2+1})^2=(2\sqrt{x^2+1}-y)^2$
$\Leftrightarrow 4x^2+y^2+1+4x\sqrt{y^2+1}=4(x^2+1)+y^2-4y\sqrt{x^2+1}$

$\Leftrightarrow 4(x\sqrt{y^2+1})+y\sqrt{x^2+1})=3$

$\Leftrightarrow 4Q=3$

$\Leftrightarrow Q=\frac{3}{4}$ 

Áp dụng bất đẳng thức Bunhiacopxki ta có:

\(\left(x\cdot1+y\cdot1\right)^2\le\left(1^2+1^2\right)\left(x^2+y^2\right)=2\Rightarrow x+y\le\sqrt{2}\)

Áp dụng bất đẳng thức Bunhiacopxki ta có:

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

\(\Rightarrow x\sqrt{y+1}+y\sqrt{x+1}\ge\sqrt{2+\sqrt{2}}\)

Dấu = xảy ra khi \(x=y=\dfrac{1}{\sqrt{2}}\)

Đặt \(\left\{{}\begin{matrix}x+2=a\\y-1=b\end{matrix}\right.\)

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

\(\Rightarrow\left\{{}\begin{matrix}b+\sqrt{b^2+1}=\sqrt{a^2+1}-a\\a+\sqrt{a^2+1}=\sqrt{b^2+1}-b\end{matrix}\right.\)

\(\Rightarrow a+b+\sqrt{a^2+1}+\sqrt{b^2+1}=\sqrt{a^2+1}+\sqrt{b^2+1}-a-b\)

\(\Rightarrow a+b=0\)

\(\Rightarrow x+2+y-1=0\)

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

\(\sqrt{x^2+5x+4}\) hay \(\sqrt{x^2+4x+5}\) thế bạn

Lời giải:
ĐKĐB \(\Rightarrow (x+2-\sqrt{x^2+4x+5})(x+2+\sqrt{x^2+4x+5})(y-1+\sqrt{y^2-2y+2})=x+2-\sqrt{x^2+4x+5}\)

\(\Leftrightarrow -(y-1+\sqrt{y^2-2y+2})=x+2-\sqrt{x^2+4x+5}\)

\(\Leftrightarrow \sqrt{x^2+4x+5}-\sqrt{y^2-2y+2}=x+y+1(*)\)

ĐKĐB \(\Rightarrow (x+2+\sqrt{x^2+4x+5})(y-1+\sqrt{y^2-2y+2})(y-1-\sqrt{y^2-2y+2})=y-1-\sqrt{y^2-2y+2}\)

\(\Leftrightarrow -(x+2+\sqrt{x^2+4x+5})=y-1-\sqrt{y^2-2y+2}\)

\(\Leftrightarrow \sqrt{y^2-2y+2}-\sqrt{x^2+4x+5}=x+y+1(**)\)

Lấy $(*)+(**)\Rightarrow x+y+1=0$

$\Leftrightarrow x+y=-1$

Áp dụng bđt Cô-si vào 2 số dương có:

\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{2}{\sqrt{xy}}\Rightarrow\dfrac{1}{2}\ge\dfrac{2}{\sqrt{xy}}\Rightarrow\sqrt{xy}\ge4\)

\(\Rightarrow\sqrt{x}+\sqrt{y}\ge2\sqrt{\sqrt{xy}}=2\sqrt{4}=4\)

Dấu = xảy ra \(\Leftrightarrow x=y=4\)

`1/x+1/y>=2/(\sqrt{xy})`

`<=>1/2>=2/(\sqrt{xy})`

`<=>\sqrt{xy}>=4`

`=>\sqrt{x}+\sqrt{y}>=2.2=4`

Dấu "=" xảy ra khi `x=y=4`

\(\left\{{}\begin{matrix}x;y;z\ge0\\x+y+z=1\end{matrix}\right.\) \(\Rightarrow0\le x;y;z\le1\)

\(\Rightarrow\left\{{}\begin{matrix}x^2\le x\\y^2\le y\\z^2\le z\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x^2+x+1\le x^2+2x+1\\2y^2+y+1\le y^2+2y+1\\2z^2+z+1\le z^2+2z+1\end{matrix}\right.\)

\(\Rightarrow P\le\sqrt{\left(x+1\right)^2}+\sqrt{\left(y+1\right)^2}+\sqrt{\left(z+1\right)^2}=x+y+z+3=4\)

\(P_{max}=4\) khi \(\left(x;y;z\right)=\left(0;0;1\right)\) và các hoán vị