đặt a+b=t => ab=t²-4

Ta có: \(a^2+b^2=4\Leftrightarrow2ab=\left(a+b\right)^2-4\)

\(\Rightarrow2M=\frac{\left(a+b\right)^2-4}{a+b+2}=a+b-2\)

Ta có: \(a+b\le\sqrt{2\left(a^2+b^2\right)}=2\sqrt{2}\Rightarrow M\le\sqrt{2}-1\)

Dấu "="\(\Leftrightarrow a=b=\sqrt{2}\)

Vậy \(M_{max}=\sqrt{2}-1\Leftrightarrow a=b=\sqrt{2}\)

\(A-\frac{1}{A}=\frac{x+9}{6\sqrt{x}}-\frac{6\sqrt{x}}{x+9}=\frac{\left(x+9\right)^2-36x}{6\sqrt{x}\left(x+9\right)}=\frac{\left(x-9\right)^2}{6\sqrt{x}\left(x+9\right)}\ge0\)

\(\Rightarrow A\ge\frac{1}{A}\)

Dấu "=" xảy ra khi x-9 = 0 hay x= 9.

\(a=\left(4+\sqrt{15}\right).\left(\sqrt{10}-\sqrt{6}\right).\sqrt{4-\sqrt{15}}\)

=\(\left(4+\sqrt{15}\right).\left(\sqrt{5}\sqrt{2}-\sqrt{3}\sqrt{2}\right).\sqrt{4-\sqrt{15}}\)

=\(\left(4+\sqrt{15}\right).\left(\sqrt{5}-\sqrt{3}\right).\sqrt{2}.\sqrt{4-\sqrt{15}}\)

=\(\left(4+\sqrt{15}\right).\left(\sqrt{5}-\sqrt{3}\right).\sqrt{2\left(4-\sqrt{15}\right)}\)

=\(\left(4+\sqrt{15}\right).\left(\sqrt{5}-\sqrt{3}\right).\sqrt{8+2\sqrt{15}}\)

=\(\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{5-2\sqrt{5}\sqrt{3}+3}\)

\(=\left(4+\sqrt{15}\right).\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)

\(=\left(4+\sqrt{15}\right).\left(\sqrt{5}-\sqrt{3}\right).\left(\sqrt{5}-\sqrt{3}\right)\)

=\(\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

=\(2.\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)\)

=\(2.\left(16-15\right)=2\)

ĐK: \(x\ge0\)

\(pt\Leftrightarrow\left(\sqrt{x}+4\right)\left(\sqrt{x}-7\right)=0\)

\(\Leftrightarrow\sqrt{x}=7\text{ (do }\sqrt{x}+4>0\text{)}\)

\(\Leftrightarrow x=49\)

Kết luận: x = 49.

\(\Leftrightarrow\left(\frac{3}{n}-\frac{4}{15}\right)\sqrt{x-1}=\frac{17}{n}-\frac{23}{15}\)

\(\Leftrightarrow\sqrt{x-1}=\frac{\frac{17}{n}-\frac{23}{15}}{\frac{3}{n}-\frac{4}{15}}=\frac{23n-255}{4n-45}\)

+Nếu \(\frac{23x-255}{4x-45}

mão có đuôi là o thì bỏ o thành mã