2) Viết nhầm thì phải, vế phải là 12 nhỉ

\(x\left(x-1\right)+y\left(y-1\right)=x^2+y^2-\left(x+y\right)\ge\dfrac{\left(x+y\right)^2}{2}-\left(x+y\right)\ge\dfrac{6^2}{2}-6=12\)

1) \(x\ge2y>0\Rightarrow x^3\ge8y^3\)

\(P=\dfrac{x^2+y^2}{xy}=\dfrac{x^2}{4xy}+\dfrac{x^2}{4xy}+\dfrac{x^2}{4xy}+\dfrac{x^2}{4xy}+\dfrac{4y^2}{4xy}\ge5\sqrt[5]{\dfrac{x^2}{4xy}.\dfrac{x^2}{4xy}.\dfrac{x^2}{4xy}.\dfrac{x^2}{4xy}.\dfrac{4y^2}{4xy}}=5\sqrt[5]{\dfrac{x^3}{256y^3}}\ge5\sqrt[5]{\dfrac{8y^3}{256y^3}}=5\sqrt[5]{\dfrac{1}{32}}=\dfrac{5}{2}\)

À, dấu bằng khi a=b=c=1/3 :v

Đề sai, ngược dấu rồi.

Ta chứng minh BĐT phụ sau: \(\dfrac{x}{x+1}\le\dfrac{9}{16}x+\dfrac{1}{16}\left(\forall x\in0;1\right)\)

Thật vậy: \(\dfrac{x}{x+1}\le\dfrac{9}{16}x+\dfrac{1}{16}\)

\(\Leftrightarrow0\le\dfrac{9x+1}{16}-\dfrac{x}{x+1}\)

\(\Leftrightarrow0\le\dfrac{\left(9x+1\right)\left(x+1\right)-16x}{16\left(x+1\right)}\)

\(\Leftrightarrow0\le9x^2-6x+1=\left(3x-1\right)^2\)(Luôn đúng \(\forall x\in0;1\))

Áp dụng vào bài, ta được:

\(\dfrac{a}{a+1}\le\dfrac{9}{16}a+\dfrac{1}{16}\)

\(\dfrac{b}{b+1}\le\dfrac{9}{16}b+\dfrac{1}{16}\)

\(\dfrac{c}{c+1}\le\dfrac{9}{16}c+\dfrac{1}{16}\)

Cộng vế theo vế ta được đpcm

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\(\left(\dfrac{3+\sqrt{5}}{2}\right)^2=\left(\dfrac{6+2\sqrt{5}}{4}\right)^2=\left(\dfrac{1+5+2\sqrt{5}}{4}\right)^2=\left[\dfrac{\left(1+\sqrt{5}\right)^2}{4}\right]^2\)

Towis ddaay thifsao nuwaxnhir

Chứng minh biểu thức đó <2

Với mọi \(n\in N^{\cdot}\), ta có

\(\dfrac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\)

\(\Leftrightarrow1< 2\left(n+1\right).\sqrt{n}\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\)

\(\Leftrightarrow0< n+1-2\sqrt{n+1}.\sqrt{n}+n\)

\(\Leftrightarrow0< \left(\sqrt{n+1}-\sqrt{n}\right)^2\)(Luôn đúng vì n thuộc N^*)

Do đó: \(\dfrac{1}{2}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{4\sqrt{3}}+...\dfrac{1}{2005\sqrt{2004}}< 2\left(\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{2004}}-\dfrac{1}{\sqrt{2005}}\right)\)

\(=2\left(1-\dfrac{1}{\sqrt{2005}}\right)< 2\)

ĐK: \(x,y>0\)

\(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}=4-\sqrt{x}-\sqrt{y}\)

\(\Leftrightarrow\dfrac{1}{\sqrt{x}}+\sqrt{x}-2+\dfrac{1}{\sqrt{y}}+\sqrt{y}-2=0\)

\(\Leftrightarrow\left(\dfrac{1}{\sqrt{x}}-\sqrt{x}\right)^2+\left(\dfrac{1}{\sqrt{y}}-\sqrt{y}\right)^2=0\)
\(\Rightarrow\)\(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x}}=\sqrt{x}\\\dfrac{1}{\sqrt{y}}=\sqrt{y}\end{matrix}\right.\)\(\)\(\)
\(\Rightarrow\)\(\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

\(\sqrt{117,5^2-26,5^2-1440}\)

\(=\sqrt{\left(117,5+26,5\right)\left(117,5-26,5\right)-144.10}\)

\(=\sqrt{144.91-144.10}\)

\(=\sqrt{144\left(91-10\right)}=\sqrt{144.81}=\sqrt{144}.\sqrt{81}=12.9=108\)

Ta có: \(\dfrac{\sqrt{6+2\sqrt{5}}}{1+\sqrt{5}}=\dfrac{\sqrt{1+5+2\sqrt{5}}}{1+\sqrt{5}}=\dfrac{\sqrt{\left(1+\sqrt{5}\right)^2}}{1+\sqrt{5}}=\dfrac{1+\sqrt{5}}{1+\sqrt{5}}=1\)