\(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{2015}+\sqrt{2016}}=.\)

\(\frac{2-1}{1+\sqrt{2}}+\frac{3-2}{\sqrt{2}+\sqrt{3}}+\frac{4-3}{\sqrt{3}+\sqrt{4}}+...+\frac{2016-2015}{\sqrt{2015}+\sqrt{2016}}=.\)

\(\frac{\left(\sqrt{2}\right)^2-1}{1+\sqrt{2}}+\frac{\left(\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2}{\sqrt{2}+\sqrt{3}}+\frac{\left(\sqrt{4}\right)^2-\left(\sqrt{3}\right)^2}{\sqrt{3}+\sqrt{4}}+...+\frac{\left(\sqrt{2016}\right)^2-\left(\sqrt{2015}\right)^2}{\sqrt{2015}+\sqrt{2016}}=.\)

\(\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}{1+\sqrt{2}}+\frac{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}}+\frac{\left(\sqrt{4}+\sqrt{3}\right)\left(\sqrt{4}-\sqrt{3}\right)}{\sqrt{3}+\sqrt{4}}+...=.\)

\(=-1+\sqrt{2}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2016}-\sqrt{2015}\)

\(=\sqrt{2016}-1\). đpcm

\(\frac{3}{2}\sqrt{4x-8}-9\sqrt{\frac{x-2}{81}}=6\)

đkxđ x>=2,x>0

\(\frac{3}{2}\sqrt{4\left(x-2\right)}-9\sqrt{\frac{x-2}{81}}=6\)

đặt t=x-2

\(\frac{3}{2}\sqrt{4t}-9\sqrt{\frac{t}{81}}=6\)

\(\frac{3}{2}.2\sqrt{t}-9\frac{\sqrt{t}}{9}=6\)

\(3\sqrt{t}-\sqrt{t}=6\)

\(2\sqrt{t}=6\)

\(\sqrt{t}=3=>t=9\)

thế t vào x-2 ta được 

x-2=9<=> x=11 (thỏa)

S={11}

=>|x-1|+|x-2|=2016

TH1: x<1

Pt sẽ là 1-x+2-x=2016

=>-2x+3=2016

=>-2x=2013

=>x=-2013/2(nhận)

TH2: 1<=x<2

Pt sẽ là x-1+2-x=2016

=>1=2016(loại)

TH3: x>=2

Pt sẽ là 2x-3=2016

=>2x=2019

=>x=2019/2(nhận)

Ta có:

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

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

Thế vô bài toán được

\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\)

\(=1-\frac{1}{\sqrt{2016}}\)

Bài này dài lắm, mình học qua rùi cũng bỏ xó luôn ....... Ko biết còn quyển vở ko để xem lại

giúp đi

Giải tổng quát nha : 

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

\(x=\dfrac{1}{2}\cdot\sqrt{\left(\sqrt{2}-1\right)^2}=\dfrac{\sqrt{2}-1}{2}\)

\(A=\left[4\cdot\left(\dfrac{\sqrt{2}-1}{2}\right)^4+4\cdot\left(\dfrac{\sqrt{2}-1}{2}\right)^3-5\cdot\left(\dfrac{\sqrt{2}-1}{2}\right)^2+5\cdot\dfrac{\sqrt{2}-1}{2}-2\right]^{2015}+2016\)

=-1,13+2016=2014,87

Ta có

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

\(=\frac{1}{2}.\frac{\left(\sqrt{2}-1\right)}{1}=\frac{\sqrt{2}-1}{2}\)

Ta lại có

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

\(\Rightarrow x\left(x+1\right)=\frac{\sqrt{2}-1}{2}.\frac{\sqrt{2}+1}{2}=\frac{1}{4}\)

Ta lại có

\(4x^4+4x^3-5x^2+5x-2=4x^3\left(x+1\right)-5x^2+5x-2\)

\(=x^2-5x^2+5x-2=-4x^2\left(x+1\right)+9x-2\)

\(=-1+9x-2=-3+\frac{\sqrt{2}-1}{2}=\frac{\sqrt{2}-7}{2}\)

Giải tới đây thì mình nghĩ là bạn sai đề rồi. Bạn xem lại đề nhé

ờ sai đề thật , mk cũng làm đc rùi nhưng vẫn cảm ơn nha