Check lại đề, sai đề rồi nhé, thay a=-2, b=1, c=0 không thoả mãn nhé.

Áp dụng BĐT Bunhiacopxki ta có: 

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

\(=3\left(2a+2b+2c\right)=3.2\left(a+b+c\right)=6.2021=12126\)

\(\Rightarrow\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\le\sqrt{12126}\)

Dấu ''='' xảy ra khi \(a=b=c=\dfrac{2021}{3}\)

Ủa bị lỗi hả:v? 

Đặt \(A=\dfrac{1}{\sqrt[3]{a+7b}}+\dfrac{1}{\sqrt[3]{b+7c}}+\dfrac{1}{\sqrt[3]{c+7a}}\)

\(A=\dfrac{\sqrt[3]{64}}{\sqrt[3]{8.8\left(a+7b\right)}}+\dfrac{\sqrt[3]{64}}{\sqrt[3]{8.8\left(b+7c\right)}}+\dfrac{\sqrt[3]{64}}{\sqrt[3]{8.8\left(c+7a\right)}}\)

\(\ge\dfrac{4}{\dfrac{8+8+a+7b}{3}}+\dfrac{4}{\dfrac{8+8+b+7c}{3}}+\dfrac{4}{\dfrac{8+8+c+7a}{3}}\ge\dfrac{\left(2+2+2\right)^2}{\dfrac{8+8+a+7b+8+8+b+7c+8+8+c+7a}{3}}\)

\(=\dfrac{36.3}{8\left(a+b+c\right)+48}=\dfrac{3}{2}\)

Vậy \(A_{min}=\dfrac{3}{2}\Leftrightarrow a=b=c=1\)

\(x^2+y^2-2x+y=9\)

\(\Leftrightarrow4x^2+4y^2-8x+4y=36\)

\(\Leftrightarrow4x^2-8x+4+4y^2+4y+1=41\)

\(\Leftrightarrow\left(2x-2\right)^2+\left(2y+1\right)^2=41\)

Vì \(\left(2x-2\right)^2\ge0\) với \(\forall x\)

\(\Rightarrow\left(2y+1\right)^2\le41\)

Mà \(\left(2y+1\right)^2\) là số lẻ \(\Rightarrow\left(2y+1\right)^2\in\left\{1;9;25\right\}\)

\(\Rightarrow y\in\left\{0;1;2;-1;-2;-3\right\}\)

Tìm được y rồi thì thay vào tìm x nhé.

2 cái kìa còn lại làm tương tự rồi sau đó cộng lại với nhau sẽ ra 1 số tự nhiên nhé, dễ nên lười đánh nốt lắm :v

ĐKXĐ: \(x\ge0\)

Ta có:  \(P-4=\dfrac{x+2\sqrt{x}+5}{\sqrt{x}+1}-4=\dfrac{x+2\sqrt{x}+5-4\sqrt{x}-4}{\sqrt{x}+1}=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\ge0\)\(\Leftrightarrow P-4\ge0\Leftrightarrow P\ge4\)

 Vậy \(P_{min}=4\Leftrightarrow x=1\)

b) ĐKXĐ: \(x\ge0;y\ge0;x\ne y\)

Ta có: \(\left(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right)\left(\dfrac{\sqrt{x}+\sqrt{y}}{x-y}\right)^2\)

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

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

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

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

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

=> Đpcm.

Bài 2a: Thấy bạn khoanh 2a nên mình làm 2a thôi nhé.

ĐKXĐ: \(x\ge-1\)

\(\sqrt{9x+9}-\sqrt{x+1}=4\)

\(\Leftrightarrow\sqrt{9\left(x+1\right)}-\sqrt{x+1}=4\)

\(\Leftrightarrow3\sqrt{x+1}-\sqrt{x+1}=4\)

\(\Leftrightarrow2\sqrt{x+1}=4\)

\(\Leftrightarrow4\left(x+1\right)=16\Leftrightarrow x+1=4\Leftrightarrow x=3\) (Tm ĐKXĐ)

Vậy...