\(\frac{1}{\sqrt{a+2b}}+\frac{1}{\sqrt{b+2a}}\ge\frac{2}{a+2b+1}+\frac{2}{b+2a+1}\)

\(\ge2.\frac{4}{3a+3b+2}=\frac{8}{\frac{3.2}{3}+2}=2\)

Dấu = xảy ra khi \(a=b=\frac{1}{3}\)

A = 1²-2²+3²-4²+...-2014²+2015²

A= (1-2)(1+2)+(3-4)(3+4)+...+(2013-2014)(2013+2014)+2015²

A= (-1)(1+2+3+4+...+2014)+2015²

A= -2029105 + 4060225 = 2031120

(* Nếu bạn không biết tính 1+2+3+...+2014 hãy bấm trên CASIO nút xích ma)

\(A=1^2-2^2+...+2013^2-2014^2+2015^2\)

\(=1^2+\left(3^2-2^2\right)+\left(5^2-4^2\right)+...+\left(2015^2-2014^2\right)\)

\(=1+5+9+...+4029\)

\(=\frac{1008.\left(4029+1\right)}{2}=2031120\)

Luôn có \(\left(x-y\right)^2+\left(x-z\right)^2+\left(y-x\right)^2\ge0\Leftrightarrow2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)\ge0\)

\(\Leftrightarrow\left(x^2+y^2+z^2\right)\ge xy+yz+xz\ge-1\)

\(P_{min}=-1\)dấu "=" sảy ra khi (x,y,z) là hoán vị của 3 phần tử (0,0,-1)

Ta có:

\(xy+yz+zx=-1\)

\(\Leftrightarrow2\left(xy+yz+zx\right)=-2\)

\(\Leftrightarrow2\left(xy+yz+zx\right)+x^2+y^2+z^2=-2+x^2+y^2+z^2\)

\(\Leftrightarrow P=x^2+y^2+z^2=\left(x+y+z\right)^2+2\ge2\)

Dấu = xảy ra khi \(\hept{\begin{cases}x+y+z=0\\xy+yz+zx=-1\end{cases}}\)

Chỉ ra 1 bộ số thỏa mãn cái đấy nhé là: \(\hept{\begin{cases}x=0\\y=1\\z=-1\end{cases}}\)

\(\sqrt[33]{8589934592}\)= 2

tk mình nha

mình nè

\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+ac+bc\right)=\frac{9}{4}\)\(\Rightarrow2\left(ab+ac+bc\right)=\frac{9}{4}-\left(a^2+b^2+c^2\right)\)

mà ta có \(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\)

\(\Leftrightarrow2\left(a^2+b^2+c^2\right)-2\left(ab+ac+bc\right)\ge0\)\(\Leftrightarrow2\left(a^2+b^2+c^2\right)-\frac{9}{4}+\left(a^2+b^2+c^2\right)\ge0\)

\(3\left(a^2+b^2+c^2\right)\ge\frac{9}{4}\Leftrightarrow\left(a^2+b^2+c^2\right)\ge\frac{3}{4}\)có \(\left(a^2+b^2+c^2\right)\)đạt min là 3/4 khi và chỉ khi a=b=c=1/2

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

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

\(=4+2.\sqrt{4-3}=4+2=6\)

\(\Rightarrow\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\sqrt{6}\)

Cách khác:

Ta có: \(A=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)

\(\Rightarrow\sqrt{2}A=\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\)

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

\(\Rightarrow A=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)

chia cả 2 vế cho x ta được
(x+2+\(\frac{1}{x}\))\(\sqrt{x^2+\frac{3}{x}}\)=x²+2x+\(\frac{5}{x}\)
<=>(x+\(\frac{1}{x}\)+2)\(\sqrt{x^2+\frac{3}{x}}\)=x²+\(\frac{3}{x}\)+2(x+\(\frac{1}{x}\))
đặt x+\(\frac{1}{x}\)=b
\(\sqrt{x^2+\frac{3}{x}}\)=a
ptr trở thành
a(b+2)=a²+2b
=>(a-b)(a-2)=0
=>a,b thay vào tính