\(b\)) \(\left(y+1\right)+\left(y+2\right)+\left(y+3\right)+......+\left(y+50\right)=1425\)

\(< =>50y+\left(1+49\right)+\left(2+48\right)+.....+\left(24+26\right)+25+50=1425\)

\(< =>50y+24\left(1+49\right)+75=1425\)

\(< =>50y+1275=1425\)

\(< =>50y=150\)

\(=>y=3\)

a thấy r :v

\(x=2\left(TM\right)\) mà ;v?

\(a\)) ĐK \(x>-1\)

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

<=>\(2x-1=x+1\)

<=>\(x=2\)\(\left(TM\right)\)

Vậy \(x=2\)

\(b\)) ĐK \(x\) ≤ 2

\(\sqrt{4-x^2}-x+2=0\)

<=> \(\sqrt{4-x^2}=x-2\)

<=>\(4-x^2=x^2-4x+4\)

<=>\(2x^2-4x=0\)

<=>\(2x\left(x-2\right)=0\)

<=>\(\left[{}\begin{matrix}2x=0\\x-2=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

Vậy \(S=\left\{0;2\right\}\)

Bài \(1\)

\(a\)) \(A=\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right).\sqrt{11}+3\sqrt{22}\)

\(A=\left(3\sqrt{11}-3\sqrt{2}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)

\(A=3.11-3\sqrt{22}-11+3\sqrt{22}\)

\(A=22\)

\(b\))\(B=\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\)

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

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

\(B=\left|\sqrt{3}+1\right|+\left|\sqrt{3}-1\right|\)

\(B=2\sqrt{3}\)

\(c\))\(C=\dfrac{5}{\sqrt{7}+\sqrt{2}}-\dfrac{7-\sqrt{7}}{\sqrt{7}-1}+6\sqrt{\dfrac{1}{2}}\)

\(C=\dfrac{5\left(\sqrt{7}-\sqrt{2}\right)}{5}-\dfrac{\sqrt{7}\left(\sqrt{7}-1\right)}{\left(\sqrt{7}-1\right)}+\sqrt{\dfrac{36}{2}}\)

\(C=\sqrt{7}-\sqrt{2}-\sqrt{7}+3\sqrt{2}\)

\(C=2\sqrt{2}\)

Bạn ơi chỗ \(x+y-4\)≥0 á

Biến đổi như vầy cũng được 

Ta có \(xy=4=>2\sqrt{xy}=4\)

Thay vào ta được \(x+y-2\sqrt{xy}\)≥0 <=>\(\left(x+y\right)^2\)≥0 ( Luôn đúng )
Vậy....

\(\dfrac{1}{x+3}+\dfrac{1}{y+3}=\dfrac{x+3+y+3}{\left(x+3\right)\left(y+3\right)}\)

\(=\dfrac{x+y+6}{3x+3y+13}\)(vì \(xy=4\))

=> \(\dfrac{x+y+6}{3x+3y+13}\)≤\(\dfrac{2}{5}\)

<=> \(5\left(x+y+6\right)\)≤\(2\left(3x+3y+13\right)\)

<=>\(6x+6y+26-5x-5y-30\)≥\(0\)

<=> \(x+y-4\)≥\(0\)

Áp dụng BĐT AM-GM \(\dfrac{a+b}{2}\)≥\(\sqrt{ab}\)

Ta có \(\dfrac{x+y}{2}\)≥\(\sqrt{xy}\)

<=>\(x+y\) ≥ 2\(\sqrt{xy}\)

=>2\(\sqrt{xy}-4\)≥\(0\)

<=> \(4-4\)≥0

<=>0≥0 ( Luôn đúng )

Vậy \(\dfrac{1}{x+3}+\dfrac{1}{y+3}\)≤\(\dfrac{2}{5}\)

à kh để ý :3