Ta có: \(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n}+\sqrt{n+1}\right)}\)

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

Thế vô bài toán ta được

\(A=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{2012}}-\dfrac{1}{\sqrt{2013}}=1-\dfrac{1}{\sqrt{2013}}\)

Ta có: \(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n}+\sqrt{n+1}\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n.\left(n+1\right)}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)

Sau đó thế vô bài toán và làm tiếp như bác ctv là ta hoàn thành bài toán!

@Akai Haruma,@Ace Legona

Điều kiện: \(x\ge2012;y\ge2013;z\ge2014\)

Áp dụng bất đẳng thức Cauchy, ta có:

\(\left\{{}\begin{matrix}\dfrac{\sqrt{x-2012}-1}{x-2012}=\dfrac{\sqrt{4\left(x-2012\right)}-2}{2\left(x-2012\right)}\le\dfrac{\dfrac{4+x-2012}{2}-2}{2\left(x-2012\right)}=\dfrac{1}{4}\\\dfrac{\sqrt{y-2013}-1}{y-2013}=\dfrac{\sqrt{4\left(y-2013\right)}-2}{2\left(y-2013\right)}\le\dfrac{\dfrac{4+y-2013}{2}-2}{2\left(y-2013\right)}=\dfrac{1}{4}\\\dfrac{\sqrt{z-2014}-1}{z-2014}=\dfrac{\sqrt{4\left(z-2014\right)}-2}{2\left(z-2014\right)}\le\dfrac{\dfrac{4+z-2014}{2}-2}{2\left(z-2014\right)}=\dfrac{1}{4}\end{matrix}\right.\)

Cộng vế theo vế, ta được:

\(\dfrac{\sqrt{x-2012}-1}{x-2012}+\dfrac{\sqrt{y-2013}-1}{y-2013}+\dfrac{\sqrt{z-2014}-1}{z-2014}\le\dfrac{3}{4}\)

Đẳng thức xảy ra khi \(x=2016;y=2017;z=2018\)

Vậy....

ĐKXĐ : \(\left\{{}\begin{matrix}x\ge2011\\y\ge2012\\z\ge2013\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}a=\sqrt{x-2011}\ge0\\b=\sqrt{y-2012}\ge0\\c=\sqrt{z-2013}\ge0\end{matrix}\right.\) ta có :

\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{a^2}-\frac{1}{a}+\frac{1}{4}+\frac{1}{b^2}-\frac{1}{b}+\frac{1}{4}+\frac{1}{c^2}-\frac{1}{c}+\frac{1}{4}=0\)

\(\Leftrightarrow\left(\frac{1}{a}-\frac{1}{2}\right)^2+\left(\frac{1}{b}-\frac{1}{2}\right)^2+\left(\frac{1}{c}-\frac{1}{2}\right)^2=0\)

\(\Leftrightarrow a=b=c=2\Leftrightarrow\left\{{}\begin{matrix}x=2015\\y=2016\\z=2017\end{matrix}\right.\)

\(M=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2012}+\sqrt{2013}}\)

\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{2013}-\sqrt{2012}\)

\(=\sqrt{2013}-1\)

2 = 1.2 => \(\dfrac{1}{2}\) = \(\dfrac{1}{1.2}\) = 1 - \(\dfrac{1}{2}\)

TT \(\dfrac{1}{6}=\dfrac{1}{2}-\dfrac{1}{3}\)

.................

=> VT = 1 - \(\dfrac{1}{x+1}\) = \(\dfrac{\sqrt{2012-x}+2012}{\sqrt{2012-x}+2013}\)

Đặt \(\sqrt{2012-x}+2012=y\)

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

=> \(\dfrac{x}{x+1}\) = \(\dfrac{y}{y+1}\)

=> x = y

<=> x = \(\sqrt{2012-x}+2012\)

<=> 2012 - x + \(\sqrt{2012-x}\) = 0

<=> \(\sqrt{2012-x}=0\)

<=> x = 2012

Câu a :

Áp dụng BĐT \(\dfrac{1}{\sqrt{ab}}>\dfrac{2}{a+b}\left(a\ne b;a,b>0\right)\) ta có :

\(\dfrac{1}{\sqrt{1.1998}}>\dfrac{2}{1+1998}=\dfrac{2}{1999}\)

\(\dfrac{1}{\sqrt{2.1997}}>\dfrac{2}{2+1997}=\dfrac{2}{19999}\)

.......................................................

\(\dfrac{1}{\sqrt{1998.1}}>\dfrac{2}{1998+1}=\dfrac{2}{1999}\)

Cộng tất cả vế với nhau ta được : \(P>2.\dfrac{1998}{1999}\)

\(\Rightarrowđpcm\)

Câu a, b sao tính chất cái cuối khác những cái còn lại thế. Vậy sao biết tới đâu thì nó dừng.

Bài 1: 

a: \(\Leftrightarrow2-3\sqrt{x}+5\sqrt{x}=8\)

=>2 căn x=6

=>căn x=3

=>x=9

b: \(\Leftrightarrow\dfrac{1}{\sqrt{x}}\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{6}\right)=\dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{1}{\sqrt{x}}=\dfrac{2}{3}:\dfrac{2}{3}=1\)

=>x=1