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\(2015\sqrt{2015x-2014}+\sqrt{2016x-2015}=2016\)
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: \(\frac{2015^2}{\sqrt{2015x-2014}+1}+\frac{2016}{\sqrt{2016x-2015}+1}>0\)
\(\Rightarrow x-1=0\Rightarrow x=1\)
\(\sqrt{2016}+\sqrt{2015}>\sqrt{2015}+\sqrt{2014}\)
=> \(\frac{1}{\sqrt{2016}+\sqrt{2015}}<\frac{1}{\sqrt{2015}+\sqrt{2014}}\)
=> \(\frac{\sqrt{2016}-\sqrt{2015}}{\left(\sqrt{2016}-\sqrt{2015}\right)\left(\sqrt{2016}+\sqrt{2015}\right)}<\frac{\sqrt{2015}-\sqrt{2014}}{\left(\sqrt{2015}-\sqrt{2014}\right)\left(\sqrt{2015}+\sqrt{2014}\right)}\)
=> \(\sqrt{2016}-\sqrt{2015}<\sqrt{2015}-\sqrt{2014}\)
A = \(\frac{2016-2015}{\sqrt{2016}+\sqrt{2015}}=\frac{1}{\sqrt{2016}+\sqrt{2015}}\); B = \(\frac{2015-2014}{\sqrt{2015}+\sqrt{2014}}=\frac{1}{\sqrt{2015}+\sqrt{2014}}\)
Mà \(\sqrt{2016}+\sqrt{2015}>\sqrt{2015}+\sqrt{2014}\) ( Vì \(\sqrt{2016}>\sqrt{2014}\))
Nên \(\frac{1}{\sqrt{2016}+\sqrt{2015}}<\frac{1}{\sqrt{2015}+\sqrt{2014}}\) => A < B
Lời giải:
Ta có:
\(A-B=(\sqrt{2016}-\sqrt{2014})+(\sqrt{2017}-\sqrt{2015})+(\sqrt{2018}-\sqrt{2022})\)
\(=\frac{2}{\sqrt{2016}+\sqrt{2014}}+\frac{2}{\sqrt{2017}+\sqrt{2015}}-\frac{4}{\sqrt{2018}+\sqrt{2022}}\)
Dễ thấy:
\(0< \sqrt{2016}+\sqrt{2014}< \sqrt{2018}+\sqrt{2022}; 0< \sqrt{2017}+\sqrt{2015}< \sqrt{2018}+\sqrt{2022}\)
\(\Rightarrow \frac{1}{\sqrt{2016}+\sqrt{2014}}>\frac{1}{\sqrt{2018}+\sqrt{2022}};\frac{1}{\sqrt{2017}+\sqrt{2015}}>\frac{1}{\sqrt{2018}+\sqrt{2022}}\)
\(\Rightarrow A-B=2\left(\frac{1}{\sqrt{2016}+\sqrt{2014}}-\frac{1}{\sqrt{2018}+\sqrt{2022}}+\frac{1}{\sqrt{2017}+\sqrt{2015}}-\frac{1}{\sqrt{2018}+\sqrt{2022}}\right)>0\)
\(\Rightarrow A>B\)
giúp vs tth Trần Thanh Phương Nguyễn Văn Đạt Nguyễn Việt Lâm Akai Haruma
\(2015\sqrt{2015x-2014} + \sqrt{2016x-2015} = 2016\)
\(pt\Leftrightarrow 2015\sqrt{2015x-2014}-2015+\sqrt{2016x-2015}-1=0\)
\(\Leftrightarrow 2015(\sqrt{2015x-2014}-1)+(\sqrt{2016x-2015}-1)=0\)
\(\Leftrightarrow \frac{2015^2(x-1)}{\sqrt{2015x-2014}+1}+\frac{2016(x-1)}{\sqrt{2016-2015}+1}=0\)
\(\Leftrightarrow (x-1)(\frac{2015^2}{\sqrt{2015x-2014}+1}+\frac{2016}{\sqrt{2016x-2015}+1})=0\)
Dễ thấy: \(\frac{2015^2}{\sqrt{2015x-2014}+1}+\frac{2016}{\sqrt{2016x-2015}+1}=0\) vô nghiệm nên
\(x-1=0\Rightarrow x=1\)
dệ mà m :v bình phương đi :v