\(2015\sqrt{2015x-2014}+\sqrt{2016x-2015}=2016\)

ĐK:\(x\ge\frac{2015}{2016}\)

\(\Leftrightarrow2015\left(\sqrt{2015x-2014}-1\right)+\sqrt{2016x-2015}-1=0\)

\(\Leftrightarrow2015\frac{2015x-2014-1}{\sqrt{2015x-2014}+1}+\frac{2016x-2015-1}{\sqrt{2016x-2015}+1}=0\)

\(\Leftrightarrow2015\frac{2015x-2015}{\sqrt{2015x-2014}+1}+\frac{2016x-2016}{\sqrt{2016x-2015}+1}=0\)

\(\Leftrightarrow2015\frac{2015\left(x-1\right)}{\sqrt{2015x-2014}+1}+\frac{2016\left(x-1\right)}{\sqrt{2016x-2015}+1}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{2015^2}{\sqrt{2015x-2014}+1}+\frac{2016}{\sqrt{2016x-2015}+1}\right)=0\)

Dễ thấy: \(\frac{2015^2}{\sqrt{2015x-2014}+1}+\frac{2016}{\sqrt{2016x-2015}+1}>0\)

\(\Rightarrow x-1=0\Rightarrow x=1\)

tui ko bít bạn học lớp mí

lớp999999

P(x) = x²⁰¹⁶ - 2015x²⁰¹⁵ - 2015x²⁰¹⁴ - ... - 2015x² - 2015x 

<=> P(x) = x²⁰¹⁶ - 2016x²⁰¹⁵ + x²⁰¹⁵ - 2016x²⁰¹⁴ + x²⁰¹⁴ - ... - 2016x² + x² - 2016x + x 

<=> P(2016) = 2016²⁰¹⁶ - 2016.2016²⁰¹⁵ + 2016²⁰¹⁵ - 2016.2016²⁰¹⁴ + 2016²⁰¹⁴ -...- 2016.2016² + 2016² - 2016.2016 + 2016 

<=> P(2016)=2016²⁰¹⁶ - 2016²⁰¹⁶ + 2016²⁰¹⁵ - 2016²⁰¹⁵ + 2016²⁰¹⁴ - ... - 2016³ + 2016² - 2016² + 2016

<=> P(2016) = 2016

Vậy P(2016) = 2016

Ta có:

P(2016) = 2016²⁰¹⁶ - 2015 . 2016²⁰¹⁵ - 2015 . 2016²⁰¹⁴ -.....- 2015 . 2016² - 2015 . 2016 - 1

P(2016) = 2016²⁰¹⁶ - ( 2016 - 1 ) . 2016²⁰¹⁵ - ( 2016 -1 ) . 2016²⁰¹⁴ - ..... - ( 2016 - 1 ) . 2016² - ( 2016 - 1 ) . 2016 - 1

P(2016)= 2016²⁰¹⁶ - 2016²⁰¹⁶ + 2016²⁰¹⁵ - 2016²⁰¹⁵ + 2016²⁰¹⁴  - ..... - 2016³ + 2016² - 2016² + 2016 - 1

P(2016) = 2016 - 1

P(2016) = 2015.

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

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

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

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

Ta có: \(\sqrt{2015}-\sqrt{2014}=\dfrac{2015-2014}{\sqrt{2015}+\sqrt{2014}}>\dfrac{2016-2015}{\sqrt{2016}+\sqrt{2015}}=\sqrt{2016}-\sqrt{2015}\)

Ta có:  √2015−√2014=2015−2014√2015+√2014>2016−2015√2016+√2015=√2016−√2015

x=2015

=> x+1=2016

=> A=x²⁰¹⁶-(x+1).x²⁰¹⁵+(x+1).x²⁰¹⁴-(x+1).x²⁰¹³+...+(x+1)x²-(x+1)x+2016

=x²⁰¹⁶-x²⁰¹⁶-x²⁰¹⁵+x²⁰¹⁵+x²⁰¹⁴-x²⁰¹⁴-x²⁰¹³+...+x³+x²-x²-x+2016

=-x+2016

=-2015+2016

=1

Vậy A=1.