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\(\sqrt{x-2016}+\sqrt{y-2017}+\sqrt{z-2018}+3024=\frac{1}{2}\left(x+y+z\right)\)
\(\Leftrightarrow2\left(\sqrt{x-2016}+\sqrt{y-2017}+\sqrt{z-2018}+3024\right)=x+y+z\)
\(\Leftrightarrow2\sqrt{x-2016}+2\sqrt{y-2017}+2\sqrt{z-2018}+6048=x+y+z\)
\(\Leftrightarrow x-2\sqrt{x-2016}+y-2\sqrt{y-2017}+z-2\sqrt{z-2018}+6048=0\)
\(\Leftrightarrow x-2016-2\sqrt{x-2016}+1+y-2017+2\sqrt{y-2017}+1+z-2018-2\sqrt{z-2018}+1=0\)
\(\Leftrightarrow\left(\sqrt{x-2016}-1\right)^2+\left(\sqrt{y-2017}-1\right)^2+\left(\sqrt{z-2018}-1\right)^2=0\)
\(ĐK:x\ge2016;y\ge2017;z\ge2018\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-2016}-1=0\\\sqrt{y-2017}-1=0\\\sqrt{z-2018}-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}\sqrt{x-2016}=1\\\sqrt{y-2017}=1\\\sqrt{z-2018}=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2017\\y=2018\\z=2019\end{cases}}}\)
Ta xét đẳng thức phụ : \(1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}=1^2+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}+2\left[\frac{1}{k-1}-\frac{1}{k\left(k-1\right)}+\frac{1}{k}\right]=\left(1+\frac{1}{k-1}-\frac{1}{k}\right)^2\)
\(\Rightarrow\sqrt{1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}}=\left|1+\frac{1}{k-1}-\frac{1}{k}\right|=1+\frac{1}{k-1}-\frac{1}{k}\)
Áp dụng được :
\(S=\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{2015^2}+\frac{1}{2016^2}}\)
\(=\left(1+\frac{1}{1}-\frac{1}{2}\right)+\left(1+\frac{1}{2}-\frac{1}{3}\right)+...+\left(1+\frac{1}{2015}-\frac{1}{2016}\right)=2015+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2015}-\frac{1}{2016}=2016-\frac{1}{2016}\)
\(\frac{2x-4}{2014}+\frac{2x-2}{2016}\) và \(\frac{2x-1}{2017}+\frac{2x-3}{2015}\)
VT = \(\frac{2x-4}{2014}+\frac{2x-2}{2016}\)
= \(\frac{2x-4}{2014}+1+\frac{2x-2}{2016}+1\)
= \(\frac{2x-2018}{2014}+\frac{2x-2018}{2016}\)
VP = \(\frac{2x-1}{2017}+\frac{2x-3}{2015}\)
= \(\frac{2x-1}{2017}+1+\frac{2x-3}{2015}+1\)
= \(\frac{2x-2018}{2017}+\frac{2x-2018}{2015}\)
Mà \(\frac{2x-2018}{2014}>\frac{2x-2018}{2015}\) và \(\frac{2x-2018}{2016}>\frac{2x-2018}{2017}\)
nên \(\frac{2x-4}{2014}+\frac{2x-2}{2016}\) > \(\frac{2x-1}{2017}+\frac{2x-3}{2015}\)
Chúc bn học tốt!!
tớ ko bt lm abc , tớ lm d thôi nha , thứ lỗi
\(\frac{5}{2x-3}-\frac{1}{x+2}=\frac{5}{x-6}-\frac{7}{2x-1}\)
\(\frac{3x+13}{2x^2+x-6}=\frac{5}{x-6}+\frac{7}{1-2x}\)
\(\frac{3x+13}{\left(x+2\right)\left(2x-3\right)}=\frac{3x+37}{\left(x-6\right)\left(2x-1\right)}\)
\(\frac{10-9x}{-4x^3+32x^2-51x+18}=0\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{10}{9}\end{cases}}\)
Đặt 2x2+x-2015=a; x2-5x-2016=b
phương trình tương đương a2+4b2=4ab
=> a2-4ab+4b2=0
=> (a-2b)2=0
=> a=2b
vậy 2x2+x-2015=2*(x2-5x-2016)
=> x=\(\frac{-2017}{11}\)
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
\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=\sqrt{1+2015^2+\frac{2015^2}{2016^2}}+\frac{2015}{2016}\)
\(\sqrt{x^2}-\sqrt{2x}+\sqrt{1}+\sqrt{x^2}-\sqrt{4x}+\sqrt{4}=\sqrt{1}+\sqrt{2015^2}+\sqrt{\frac{2015^2}{2016^2}}+\frac{2015}{2016}\)
\(\sqrt{x^2}-\sqrt{6x}+3=1+2015+\frac{2015}{2016}+\frac{2015}{2016}\)
\(x-\sqrt{6x}=1+\frac{2015}{1+2016+2016}-3\)
\(x-\sqrt{6x}=2-\frac{2015}{4033}\)
\(x-\sqrt{6x}=\frac{6051}{4033}\)