Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
B = \(\dfrac{1}{\sqrt{x}+\sqrt{x+1}}+\dfrac{1}{\sqrt{x+1}+\sqrt{x+2}}+...+\dfrac{1}{\sqrt{x+2015}+\sqrt{x+2016}}\)
B = \(\dfrac{\sqrt{x}-\sqrt{x+1}}{x-x-1}+\dfrac{\sqrt{x+1}-\sqrt{x+2}}{x+1-x-2}+...+\dfrac{\sqrt{x+2015}-\sqrt{x+2016}}{x+2015-x-2016}\)
B = \(\dfrac{\sqrt{x}-\sqrt{x+1}}{-1}+\dfrac{\sqrt{x+1}-\sqrt{x+2}}{-1}+...+\dfrac{\sqrt{x+2015}-\sqrt{x+2016}}{-1}\)
B = \(-\sqrt{x}+\sqrt{x+1}-\sqrt{x+1}+\sqrt{x+2}-...-\sqrt{2015}+\sqrt{2016}\)
B = \(-\sqrt{x}+\sqrt{2016}\)
Khi x = 2017
B = \(-\sqrt{2017}+\sqrt{2016}=\sqrt{2016}-\sqrt{2017}\)
Áp dụng bđt \(\frac{\sqrt{a}+\sqrt{b}}{2}< \sqrt{\frac{a+b}{2}}\) với a > 0; b > 0; a khác b ta có:
\(\frac{\sqrt{2016}+\sqrt{2014}}{2}< \sqrt{\frac{2016+2014}{2}}\)
\(\Rightarrow\frac{\sqrt{2016}+\sqrt{2014}}{2}< \sqrt{\frac{4030}{2}}\)
\(\Rightarrow\sqrt{2016}+\sqrt{2014}< \sqrt{2015}.2\)
\(\Rightarrow\sqrt{2016}-\sqrt{2015}< \sqrt{2015}-\sqrt{2014}\)
Xét: \(\sqrt{1+n^2+\frac{n^2}{\left(n+1\right)^2}}=\sqrt{\frac{\left(n+1\right)^2+n^2\left(n+1\right)^2+n^2}{\left(n+1\right)^2}}\) (với \(n\inℕ\))
\(=\sqrt{\frac{n^2+2n+1+n^4+2n^3+n^2+n^2}{\left(n+1\right)^2}}\)
\(=\sqrt{\frac{n^4+n^2+1+2n^3+2n^2+2n}{\left(n+1\right)^2}}\)
\(=\sqrt{\frac{\left(n^2+n+1\right)^2}{\left(n+1\right)^2}}=\frac{n^2+n+1}{n+1}=n+\frac{1}{n+1}\)
Áp dụng vào ta tính được: \(\sqrt{1+2015^2+\frac{2015^2}{2016^2}}+\frac{2015}{2016}=2015+\frac{1}{2016}+\frac{2015}{2016}\)
\(=2015+1=2016\)
Khi đó: \(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=2016\)
\(\Leftrightarrow\left|x-1\right|+\left|x-2\right|=2016\)
Đến đây xét tiếp các TH nhé, ez rồi:))
chẳng biết đúng ko,mới lớp 5
\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=\sqrt{1+2015^2+\frac{2015^2}{2016^2}}+\frac{2015}{2016}\)
\(\sqrt{x^2}-\sqrt{2x}+\sqrt{1}+\sqrt{x^2}-\sqrt{4x}+\sqrt{4}=\sqrt{1}+\sqrt{2015^2}+\sqrt{\frac{2015^2}{2016^2}}+\frac{2015}{2016}\)
\(\sqrt{x^2}-\sqrt{6x}+3=1+2015+\frac{2015}{2016}+\frac{2015}{2016}\)
\(x-\sqrt{6x}=1+\frac{2015}{1+2016+2016}-3\)
\(x-\sqrt{6x}=2-\frac{2015}{4033}\)
\(x-\sqrt{6x}=\frac{6051}{4033}\)
\(B=B_1+B_2+...+B_{2016}\)
\(B_1=\dfrac{\sqrt{x+1}-\sqrt{x}}{\left(\sqrt{x}+\sqrt{x+1}\right)\left(\sqrt{x+1}-\sqrt{x}\right)}=\dfrac{\sqrt{x+1}-\sqrt{x}}{x+1-x}\)
\(B_1=\sqrt{x+1}-\sqrt{x}\)
\(B_2=\sqrt{x+2}-\sqrt{x+1}\)
\(B_3=\sqrt{x+3}-\sqrt{x+2}\)
...
\(B_{2015}=\sqrt{x+2015}-\sqrt{x+2014}\)
\(B_{2016}=\sqrt{x+2016}-\sqrt{x+2015}\)
\(B=\sqrt{x+2016}-\sqrt{x}\)
\(B\left(2017\right)=\sqrt{2017+2016}-\sqrt{2017}\)
\(\sqrt{x-2016}+\sqrt{y-2017}+\sqrt{z-2018}+3024=\frac{1}{2}\left(x+y+z\right)\)
\(\Leftrightarrow2\left(\sqrt{x-2016}+\sqrt{y-2017}+\sqrt{z-2018}+3024\right)=x+y+z\)
\(\Leftrightarrow2\sqrt{x-2016}+2\sqrt{y-2017}+2\sqrt{z-2018}+6048=x+y+z\)
\(\Leftrightarrow x-2\sqrt{x-2016}+y-2\sqrt{y-2017}+z-2\sqrt{z-2018}+6048=0\)
\(\Leftrightarrow x-2016-2\sqrt{x-2016}+1+y-2017+2\sqrt{y-2017}+1+z-2018-2\sqrt{z-2018}+1=0\)
\(\Leftrightarrow\left(\sqrt{x-2016}-1\right)^2+\left(\sqrt{y-2017}-1\right)^2+\left(\sqrt{z-2018}-1\right)^2=0\)
\(ĐK:x\ge2016;y\ge2017;z\ge2018\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-2016}-1=0\\\sqrt{y-2017}-1=0\\\sqrt{z-2018}-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}\sqrt{x-2016}=1\\\sqrt{y-2017}=1\\\sqrt{z-2018}=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2017\\y=2018\\z=2019\end{cases}}}\)
sprt là gì
bằng nhau. vì
= sqrt(2017-2016) =sqrt (1)
=sqrt(2016-2015) =sqrt (2)
từ (1) (2) => 2 cái đó bằng nhau.
đây là cách trình bày nháp. khi bạn viết ra bài thì ghi đề ra nha. CHÚC HỌC TỐT!