ĐK: \(\hept{\begin{cases}a\ge0\\b\ge0\\a+b\ge2011\end{cases}}\)

pt => \(a+b-2011=a+b+2011-2\sqrt{2011}\left(\sqrt{a}+\sqrt{b}\right)+2\left(\sqrt{ab}\right)\)

<=> \(2.2011-2\sqrt{2011}.\sqrt{a}-2\sqrt{2011}\sqrt{b}+2\sqrt{ab}=0\)

<=> \(\sqrt{2011}\left(\sqrt{2011}-\sqrt{a}\right)-\sqrt{b}\left(\sqrt{2011}-\sqrt{a}\right)=0\)

<=> \(\left(\sqrt{2011}-\sqrt{a}\right)\left(\sqrt{2011}-\sqrt{b}\right)=0\)

<=> a = 2011 và b = 2011 ( thỏa mãn đk )

Thử lại với phương trình ta thấy thỏa mãn

Vậy a= b = 2011.

a) 

b) 

https://hoc24.vn/cau-hoi/c-voi-a-b-c-la-cac-so-duong-thoa-man-dieu-kien-a-b-c-2-tim-max-q-sqrt2abcsqrt2bcasqrt2cab.8298826302

Bạn có thể tham khảo ở đây. Đừng quên like giúp mik nha bạn. Thx

We put \(n^2-14n+38=k^2\)

\(\Rightarrow n^2-14n+49-11=k^2\)

\(\Rightarrow\left(n-7\right)^2-11=k^2\)

\(\Rightarrow\left(n-7\right)^2-k^2=11\)

\(\Rightarrow\left(n-7-k\right)\left(n-7+k\right)=11=1.11=11.1=\left(-1\right).\left(-11\right)\)

\(=\left(-11\right).\left(-1\right)\)

Prints:

Case by case, we have \(n\in\left\{13;1\right\}\)

\(\left(a-\sqrt{2011}\right)\left(b+\sqrt{2011}\right)=14\)

\(\Leftrightarrow ab+\sqrt{2011}\left(a-b\right)=2025\)

Có: a,b nguyên => a-b nguyên

=> VP=VT <=> \(\sqrt{2011}\left(a-b\right)\)nguyên

=> a-b=0 <=> a=b

=> pt <=> a^2=2025

Làm nốt nhé. 

\(b+c\le\sqrt{2\left(b^2+c^2\right)}\Rightarrow\dfrac{a^2}{b+c}\ge\dfrac{a^2}{\sqrt{2\left(b^2+c^2\right)}}=\dfrac{1}{\sqrt{2}}.\dfrac{a^2}{\sqrt{b^2+c^2}}\)

Sau đó làm tiếp như bài đó là được

Đặt vế trái BĐT cần chứng minh là P

Ta có:

\(P=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{a^2}{\sqrt{2\left(b^2+c^2\right)}}+\dfrac{b^2}{\sqrt{2\left(a^2+c^2\right)}}+\dfrac{c^2}{\sqrt{2\left(a^2+b^2\right)}}\)

Đặt \(\left(\sqrt{b^2+c^2};\sqrt{c^2+a^2};\sqrt{a^2+b^2}\right)=\left(x;y;z\right)\Rightarrow x+y+z=\sqrt{2011}\)

Đồng thời: \(\left\{{}\begin{matrix}y^2+z^2-x^2=2a^2\\z^2+x^2-y^2=2b^2\\x^2+y^2-z^2=2c^2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^2=\dfrac{y^2+z^2-x^2}{2}\\b^2=\dfrac{z^2+x^2-y^2}{2}\\c^2=\dfrac{x^2+y^2-z^2}{2}\end{matrix}\right.\)

\(\Rightarrow P\ge\dfrac{1}{2\sqrt{2}}\left(\dfrac{y^2+z^2-x^2}{x}+\dfrac{z^2+x^2-y^2}{y}+\dfrac{x^2+y^2-z^2}{z}\right)\)

\(\Rightarrow P\ge\dfrac{1}{2\sqrt{2}}\left(\dfrac{y^2+z^2}{x}+\dfrac{z^2+x^2}{y}+\dfrac{x^2+y^2}{z}-\left(x+y+z\right)\right)\)

\(\Rightarrow P\ge\dfrac{1}{2\sqrt{2}}\left(\dfrac{\left(y+z\right)^2}{2x}+\dfrac{\left(z+x\right)^2}{2y}+\dfrac{\left(x+y\right)^2}{2z}-\left(x+y+z\right)\right)\)

\(\Rightarrow P\ge\dfrac{1}{2\sqrt{2}}\left(\dfrac{\left(y+z+z+x+x+y\right)^2}{2x+2y+2z}-\left(x+y+z\right)\right)=\dfrac{1}{2\sqrt{2}}\left(x+y+z\right)=\dfrac{1}{2}\sqrt{\dfrac{2011}{2}}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\sqrt{\dfrac{2011}{2}}\)

\(VT\ge\dfrac{1}{2}\sqrt{\dfrac{2011}{2}}\)

Đặt \(\left(x,y,z\right)=\left(\sqrt{a^2+b^2},\sqrt{b^2+c^2},\sqrt{c^2+a^2}\right)\).

Ta có \(x+y+z=\sqrt{2011}\).

BĐT cần cm trở thành: 

\(\dfrac{y^2+z^2-x^2}{2\sqrt{2}x}+\dfrac{z^2+x^2-y^2}{2\sqrt{2}y}+\dfrac{x^2+y^2-z^2}{2\sqrt{2}z}\ge\dfrac{1}{2}\sqrt{\dfrac{2011}{2}}\)

\(\Leftrightarrow\dfrac{y^2+z^2-x^2}{x}+\dfrac{z^2+x^2-y^2}{y}+\dfrac{x^2+y^2-z^2}{z}\ge x+y+z\)

\(\Leftrightarrow\left(\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{y}\right)+\left(\dfrac{y^2}{x}+\dfrac{z^2}{y}+\dfrac{x^2}{z}\right)\ge2\left(x+y+z\right)\).

Theo bđt AM - GM:

\(\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{y}=\left(\dfrac{x^2}{y}+y\right)+\left(\dfrac{y^2}{z}+z\right)+\left(\dfrac{z^2}{x}+x\right)-x-y-z\ge2x+2y+2z-x-y-z=x+y+z\).

Tương tự, \(\dfrac{y^2}{x}+\dfrac{z^2}{y}+\dfrac{x^2}{z}\ge x+y+z\).

Dễ có điều phải chứng minh.

\(P\sqrt{2}\ge\dfrac{a^2}{\sqrt{b^2+c^2}}+\dfrac{b^2}{\sqrt{c^2+a^2}}+\dfrac{c^2}{\sqrt{a^2+b^2}}\)

Đặt \(\left(\sqrt{b^2+c^2};\sqrt{c^2+a^2};\sqrt{a^2+b^2}\right)=\left(x;y;z\right)\)

\(\Rightarrow\left\{{}\begin{matrix}x+y+z=\sqrt{2011}\\a^2=\dfrac{y^2+z^2-x^2}{2}\\b^2=\dfrac{z^2+x^2-y^2}{2}\\c^2=\dfrac{x^2+y^2-z^2}{2}\end{matrix}\right.\)

\(\Rightarrow P2\sqrt{2}\ge\dfrac{y^2+z^2-x^2}{x}+\dfrac{z^2+x^2-y^2}{y}+\dfrac{x^2+y^2-z^2}{z}\)

\(P4\sqrt{2}\ge\dfrac{\left(y+z\right)^2}{2x}+\dfrac{\left(z+x\right)^2}{2y}+\dfrac{\left(x+y\right)^2}{2z}-\left(x+y+z\right)\)

\(P2\sqrt{2}\ge\dfrac{4\left(x+y+z\right)^2}{2\left(x+y+z\right)}-\left(x+y+z\right)=x+y+z=\sqrt{2011}\)

\(\Rightarrow P\ge\dfrac{\sqrt{2011}}{2\sqrt{2}}\)

Đề sai

ta có:

\(x\left(\sqrt{2011}+\sqrt{2010}\right)+y\left(\sqrt{2011}-\sqrt{2010}\right)=x\sqrt{2011}+x\sqrt{2010}+y\sqrt{2011}-y\sqrt{2010}\)

 pt tương đương với:

\(\left(x+y\right)\sqrt{2011}+\left(x-y\right)\sqrt{2010}=\sqrt{2011^3}+\sqrt{2010^3}\)

vì x,y là số hữu tỉ nên

\(\hept{\begin{cases}\sqrt{2011}\left(x+y\right)=\sqrt{2011^3}\\\sqrt{2010}\left(x-y\right)=\sqrt{2010^3}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x+y=2011\\x-y=2010\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{4021}{2}\\y=\frac{1}{2}\end{cases}}\)   

tích trước trả lời sau

là sao bạn???