Tìm MAX \(P=\frac{a+2b}{\sqrt{a^2+3b^2}+\sqrt{b^2+3a^2}+2b}\)

y+999=9999999999

y=9999999999-999

y=9999999000

\(x+100=900\)
\(x=900-100\)
\(x=800\)

=1128 so  nha bn k nha

Ta có: \(a+b\ge1\Rightarrow a\ge1-b\Rightarrow a^2\ge b^2-2b+1\)

Nên \(a^2+b^2\ge b^2+b^2-2b+1\)

\(\Rightarrow a^2+b^2\ge2b^2-2b+1=2\left(b^2-b+\frac{1}{4}\right)+\frac{1}{2}\)

\(\Rightarrow a^2+b^2\ge2\left(b-\frac{1}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\)

Đẳng thức xảy ra khi \(b=\frac{1}{2}\)

Vậy...............

Cho \(8x^3=11y^3=26z^3\) và \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)

Tính \(A=\sqrt{8x^2+11y^2+26z^2}\)

Tìm y biết \(y=\sqrt{6+\sqrt{6+\sqrt{6}+....}}\)

\(D=\frac{2\sqrt{x}}{x+3\sqrt{x}+3}\)

Tìm x để D nguyên.

2 + 1 x 3 = 2 + 3

                     = 5