\(ĐKXĐ:x\ge3\)

\(2\sqrt{9x-27}-\frac{1}{5}\sqrt{25x-75}-\frac{1}{7}\sqrt{49x-147}=0\)

\(2\sqrt{9\left(x-3\right)}-\frac{1}{5}\sqrt{25\left(x-3\right)}-\frac{1}{7}\sqrt{49\left(x-3\right)}=0\)

\(6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=0\)

\(4\sqrt{x-3}=0\)

\(x-3=0\)

\(x=3\left(TM\right)\)

\(\sqrt{x-3}-2\sqrt{x^2-9}=0\)ĐK : \(x\le-3;x\ge3\)

\(\Leftrightarrow\sqrt{x-3}\left(1-2\sqrt{x+3}\right)=0\)

TH1 : \(\sqrt{x-3}=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)(tm)

TH2 : \(1-2\sqrt{x+3}=0\Leftrightarrow2\sqrt{x+3}=1\)

\(\Leftrightarrow4\left(x+3\right)=1\Leftrightarrow x+3=\frac{1}{4}\Leftrightarrow x=-\frac{11}{4}\)(ktm) 

\(ĐKXĐ:x\ge3\)

\(\sqrt{x-3}-2\sqrt{x^2-9}=0\)

\(\sqrt{x-3}-2\sqrt{\left(x-3\right)}\sqrt{\left(x+3\right)}=0\)

\(\sqrt{x-3}\left(1-2\sqrt{x+3}\right)=0\)

\(\orbr{\begin{cases}\sqrt{x-3}=0\\1-2\sqrt{x+3}=0\end{cases}\orbr{\begin{cases}x=3\\\sqrt{x+3}=\frac{1}{2}\end{cases}\orbr{\begin{cases}x=3\\x+3=\frac{1}{4}\end{cases}\orbr{\begin{cases}x=3\left(TM\right)\\x=-\frac{11}{4}\left(KTM\right)\end{cases}}}}}\)

7sin²a+5cos²a=13/2

2sin²a+(5sin²a+5cos²a)=13/2

2sin²a+5=13/2(do sin²a+cos²a=1, công thức này bạn có thể xem chứng minh trên youtube nha, cái này ai cũng phải biết)

2sin²a=3/2

sin²a=3/4

sin(a)=\(\frac{\sqrt{3}}{2}\)

suy ra a=60⁰

\(A=\frac{\frac{1-\sqrt{5}}{1+\sqrt{5}}-\frac{1+\sqrt{5}}{1-\sqrt{5}}}{\sqrt{5}}=\frac{\frac{6-2\sqrt{5}-6-2\sqrt{5}}{1-5}}{\sqrt{5}}\)

\(=\frac{\frac{4\sqrt{5}}{4}}{\sqrt{5}}=\frac{\sqrt{5}}{\sqrt{5}}=1\)

Ta có: \(2a+b^2=2a\left(a+b+c\right)+b^2=b^2+2a^2+2ab+2ac\)

\(\ge4ab+2ac+a^2\)

\(\Rightarrow\frac{a}{2a+b^2}\le\frac{a}{4ab+2ac+a^2}=\frac{1}{4b+2c+a}\)

\(\le\frac{1}{49}.\frac{49}{4b+2c+a}=\frac{1}{49}.\frac{\left(4+2+1\right)^2}{4b+2c+a}\)

\(\le\frac{1}{49}\left(\frac{16}{4b}+\frac{4}{2c}+\frac{1}{a}\right)=\frac{1}{49}\left(\frac{4}{b}+\frac{2}{c}+\frac{1}{a}\right)\)

CMTT: \(\frac{b}{2b+c^2}\le\frac{1}{49}\left(\frac{4}{c}+\frac{2}{a}+\frac{1}{b}\right);\frac{c}{2c+a^2}\le\frac{1}{49}\left(\frac{4}{a}+\frac{2}{b}+\frac{1}{c}\right)\)

\(\Rightarrow\frac{a}{2a+b^2}+\frac{b}{2b+c^2}+\frac{c}{2c+a^2}\le\frac{1}{7}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)( đpcm )

Ta có \(\sqrt{x}\ge0\forall x\ne0;x\ne1\Rightarrow-5\sqrt{x}\le0\Rightarrow2-5\sqrt{x}\le2\)

\(\Rightarrow A\le\frac{2}{3}\forall x\ge0;x\ne1\)

Vậy GTLN của A  là 2/3 <=> x=0