\(ab\le\frac{a^2+b^2}{2}\le\frac{16}{2}=8\)

Ta có: \(N^2=\left(a\sqrt{9b\left(a+8b\right)}+b\sqrt{9a\left(b+8a\right)}\right)^2\)

\(\le\left(a^2+b^2\right)\left[9b\left(a+8b\right)+9a\left(b+8a\right)\right]\)

\(\le16\left(18ab+72\left(a^2+b^2\right)\right)\le16\left(18.8+72.16\right)\)

\(=20736\)

=> \(N\le144\)

Dấu "=" xảy ra <=> a = b = \(\sqrt{8}\)

Vậy max N = 144 tại a = b = \(\sqrt{8}\)

Ta có \(2ab\le a^2+b^2\)

Áp dụng BĐT Bunhia:

\(M^2\le\left(a^2+b^2\right)\left(b\left(a+8b\right)+a\left(b+8a\right)\right)\)

\(\Rightarrow M^2\le\left(a^2+b^2\right)\left(2ab+8b^2+8a^2\right)\)

\(\Rightarrow M^2\le\left(a^2+b^2\right)\left(8a^2+8b^2+a^2+b^2\right)\)

\(\Rightarrow M^2\le9\left(a^2+b^2\right)^2\Rightarrow M\le3\left(a^2+b^2\right)=48\)

\(\Rightarrow M_{max}=48\) khi \(a=b=2\sqrt{2}\)

Ta có \(2=a^2+b^2\ge2ab\)

\(\Leftrightarrow ab\le1\)

\(M\le\sqrt{\left(a^2+b^2\right)\left(36ab+45b^2+36ab+45a^2\right)}\)

\(=\sqrt{2\left(72ab+90\right)}\)\(\le\sqrt{2\left(72+90\right)}=\sqrt{324}=18\)

GTLN là 18 đạt được khi a = b = 1

a: Ta có: \(x^2=3-2\sqrt{2}\)

nên \(x=\sqrt{2}-1\)

Thay \(x=\sqrt{2}-1\) vào A, ta được:

\(A=\dfrac{\left(\sqrt{2}+1\right)^2}{\sqrt{2}-1}=\dfrac{3+2\sqrt{2}}{\sqrt{2}-1}=7+5\sqrt{2}\)

Áp dụng BĐT AM - GM ta có:

\(4\sqrt{ab}=2\sqrt{a.4b}\le a+4b\)

\(4\sqrt{bc}=2\sqrt{b.4c}\le b+4c\)

\(4\sqrt[3]{abc}=\sqrt[3]{a.4b.16c}\le\frac{a+4b+16c}{3}\)

Cộng theo vế 3 BĐT ta được:

\(8a+3b+4\left(\sqrt{ab}+\sqrt{bc}+\sqrt[3]{abc}\right)\le\frac{28}{3}\left(a+b+c\right)\)

\(\Rightarrow P\le\frac{28\left(a+b+c\right)}{3+3\left(a+b+c\right)^2}=\frac{14}{3}-\frac{14\left(a+b+c-1\right)^2}{3\left[\left(a+b+c\right)^2+1\right]}\le\frac{14}{3}\)

\(\Rightarrow Max_P=\frac{14}{3}\)

Đẳng thức xảy ra \(\Leftrightarrow a+b+c=1\)và \(a=4b=16c\)

\(\Leftrightarrow a=\frac{16}{21};b=\frac{4}{21};c=\frac{1}{21}\)