\(A=2\sqrt{x+1}+\sqrt{x-2}-x+2016\)

\(2A=4\sqrt{x+1}+2\sqrt{x-2}-2x+4032\)

\(=-\left(x+1-4\sqrt{x+1}+4\right)-\left(x-2-2\sqrt{x-2}+1\right)+4036\)

\(=4036-\left(\sqrt{x+1}-2\right)^2-\left(\sqrt{x-2}-1\right)^2\le4036\)

\(\Rightarrow A\le2018\)

đánh thừa 1 con 2016

Ta có:

\(\frac{1}{n\sqrt{n+4}+\left(n+4\right)\sqrt{n}}=\frac{1}{\sqrt{n\left(n+4\right)}\left(\sqrt{n}+\sqrt{n+4}\right)}\)

\(=\frac{\sqrt{n+4}-\sqrt{n}}{4\sqrt{n\left(n+4\right)}}=\frac{1}{4}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+4}}\right)\)

Áp dụng vào bài toán ta được

\(\frac{1}{1\sqrt{5}+5\sqrt{1}}+\frac{1}{5\sqrt{9}+9\sqrt{5}}+...+\frac{1}{2009\sqrt{2013}+2013\sqrt{2009}}\)

\(=\frac{1}{4}.\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{9}}+...+\frac{1}{\sqrt{2009}}-\frac{1}{\sqrt{2013}}\right)\)

\(=\frac{1}{4}.\left(1-\frac{1}{\sqrt{2013}}\right)\)

Xem lại đề nhé

\(\sqrt{281108+291108\sqrt{1+x}}-\sqrt{281108+291108\sqrt{1+x}}=1\)=1

điều này vô lí

vì hiệu cùa hai số bằng nhau phải bằng 0

=> phương trình vô nghiệm

\(64=\left(a^2+a+2\right)\left(b+1\right)^2\left(c^2+3c\right)\)

\(=\left(a^2+a+1+1\right)\left(b^2+b+b+1\right)\left(c^2+c+c+c\right)\)

\(\ge4.\sqrt[4]{a^3}.4.\sqrt[4]{b^4}.4.\sqrt[4]{c^5}\)

\(=64\sqrt[4]{a^3b^4c^5}\)

\(\Rightarrow\sqrt[4]{a^3b^4c^5}\le1\)

\(\Leftrightarrow a^3b^4c^5\le1\)

minh khong biet 

\(\sqrt{x^2-5x+6}+\sqrt{x+1}=\sqrt{x-2}+\sqrt{x^2-2x-3}\)

ĐK:  x^2-5x+6>=0<=> x<=2 hoặc x>=3

       x^2-2x-3>=0<=> x<=-1 hoặc x>=3

<=>\(\sqrt{x^2-5x+6}+\sqrt{x+1}=\sqrt{x-2}+\sqrt{x^2-2x-3}\)

<=>\(\sqrt{\left(x-3\right)\left(x-2\right)}+\sqrt{x+1}=\sqrt{x-2}+\sqrt{\left(x-3\right)\left(x+1\right)}\)

<=> \(\sqrt{x-2}\left(\sqrt{x-3}-1\right)+\sqrt{x+1}\left(1-\sqrt{x-3}\right)=0\)

<=> \(\sqrt{x-2}\left(\sqrt{x-3}-1\right)-\sqrt{x+1}\left(\sqrt{x-3}-1\right)=0\)

<=> \(\left(\sqrt{x-3}-1\right)\left(\sqrt{x-2}-\sqrt{x+1}\right)=0\)

<=>\(\orbr{\orbr{\begin{cases}\sqrt{x-3}-1=0\\\sqrt{x-2}-\sqrt{x+1}=0\end{cases}}}\)

<=>\(\orbr{\begin{cases}\sqrt{x-3}=1\\\sqrt{x-2}=\sqrt{x+1}\end{cases}}\)

<=> \(\orbr{\begin{cases}x=4\left(nhan\right)\\0x=3\left(vôly\right)=>loai\end{cases}}\)

S={4} 

ngungu

\(\frac{a^4}{a\left(b+c\right)}+\frac{b^4}{b\left(a+c\right)}+\frac{c^4}{c\left(a+b\right)}\)

ap dung bdt cauchy -schwaz dang engel ta co 

\(\frac{a^4}{a\left(b+c\right)}+\frac{b^4}{b\left(a+c\right)}+\frac{c^4}{c\left(a+b\right)}\ge\frac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ac\right)}\)\(\)

ma \(a^2+b^2+c^2\ge ab+bc+ac\)

\(\Rightarrow VT\ge\frac{1}{2\left(a^2+b^2+c^2\right)}=\frac{1}{2}\)

dau =xay ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)

Áp dụng BĐT Mincopski ta có:

\(VT=\sqrt{a^2+\left(1-b\right)^2}+\sqrt{b^2+\left(1-c\right)^2}+\sqrt{c^2+\left(1-b\right)^2}\)

\(\ge\sqrt{\left(a+b+c\right)^2+\left(3-a-b-c\right)^2}\)

Đặt \(a+b+c=x>0\) thì ta có:

\(\ge\sqrt{x^2+\left(3-x\right)^2}=\sqrt{2x^2-6x+9}\)

\(=\sqrt{2\left(x-\frac{3}{2}\right)^2+\frac{9}{2}}\ge\sqrt{\frac{9}{2}}=\frac{3\sqrt{2}}{2}\)