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\(\dfrac{1}{7}\sqrt{51}với\dfrac{1}{9}\sqrt{150}\)
<=> \(\dfrac{\sqrt{51}}{7}với\dfrac{\sqrt{150}}{9}\)
<=> \(9\sqrt{51}với7\sqrt{150}\)
<=> \(\sqrt{4131}với\sqrt{7350}\)
=> \(\sqrt{4131}< \sqrt{7350}\)
=> \(\dfrac{1}{7}\sqrt{51}< \dfrac{1}{9}\sqrt{150}\)
\(\dfrac{\sqrt{x-2015}-1}{x-2015}+\dfrac{\sqrt{y-2016}-1}{y-2016}=\dfrac{1}{2}\)
Điều kiện \(\left\{{}\begin{matrix}x>2015\\y>2016\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x-2015}}-\dfrac{1}{x-2015}+\dfrac{1}{\sqrt{y-2016}}-\dfrac{1}{y-2016}=\dfrac{1}{2}\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x-2015}}=a>0\\\dfrac{1}{\sqrt{y-2016}}=b>0\end{matrix}\right.\) thì ta có:
\(a-a^2+b-b^2=\dfrac{1}{2}\)
\(\Leftrightarrow\left(2a^2-2a+\dfrac{1}{2}\right)+\left(2b^2-2b+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow\left(\sqrt{2}a-\dfrac{1}{\sqrt{2}}\right)^2+\left(\sqrt{2}b-\dfrac{1}{\sqrt{2}}\right)^2=0\)
\(\Leftrightarrow a=b=\dfrac{1}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x-2015}}=\dfrac{1}{4}\\\dfrac{1}{\sqrt{y-2016}}=\dfrac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2019\\y=2020\end{matrix}\right.\)
Lời giải:
a)
Ta có: \(\frac{1}{7}\sqrt{51}< \frac{1}{7}\sqrt{64}=\frac{8}{7}\)
\(\frac{1}{9}\sqrt{150}> \frac{1}{9}\sqrt{144}=\frac{12}{9}=\frac{4}{3}=\frac{8}{6}> \frac{8}{7}\)
Do đó: \(\frac{1}{7}\sqrt{51}< \frac{1}{9}\sqrt{150}\)
b)
\(\sqrt{2017}-\sqrt{2016}=\frac{2017-2016}{\sqrt{2017}+\sqrt{2016}}=\frac{1}{\sqrt{2017}+\sqrt{2016}}< \frac{1}{\sqrt{2016}+\sqrt{2015}}\)
\(\sqrt{2016}-\sqrt{2015}=\frac{2016-2015}{\sqrt{2016}+\sqrt{2015}}=\frac{1}{\sqrt{2016}+\sqrt{2015}}\)
Do đó:
\(\sqrt{2017}-\sqrt{2016}< \sqrt{2016}-\sqrt{2015}\)
\(\text{a) }\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}}\\ =\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2\left(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}\right)-2\left(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}\right)}\\ =\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2\cdot\dfrac{x+y+z}{xyz}}\\ =\left|\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right|\)
\(\text{b) }\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+...+\sqrt{1+\dfrac{1}{2017^2}+\dfrac{1}{2018^2}}\\ =1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+...+1+\dfrac{1}{2017}-\dfrac{1}{2018}\\ =2016+\dfrac{1}{2}-\dfrac{1}{2018}\\ =\dfrac{2034698}{1009}\)
Lời giải:
Trong TH này ta thêm điều kiện $x$ là số nguyên dương.
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x(x+1)}=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{(x+1)-x}{x(x+1)}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\)
\(=1-\frac{1}{x+1}=\frac{x}{x+1}\)
Vậy \(\frac{x}{x+1}=\frac{\sqrt{2017-x}+2016}{\sqrt{2016-x}+2017}\)
\(\Rightarrow x\sqrt{2016-x}+2017x=(x+1)\sqrt{2017-x}+2016(x+1)\)
\(\Leftrightarrow x\sqrt{2016-x}=(x+1)\sqrt{2017-x}+2016-x\)
\(\Leftrightarrow x(\sqrt{2017-x}-\sqrt{2016-x})+\sqrt{2017-x}+2016-x=0\)
\(\Leftrightarrow \frac{x}{\sqrt{2017-x}+\sqrt{2016-x}}+\sqrt{2017-x}+(2016-x)=0\)
Hiển nhiên ta thấy:
\(\frac{x}{\sqrt{2017-x}+\sqrt{2016-x}}>0\)
\(\sqrt{2017-x}\geq 0\)
\(2016-x\geq 0\)
Do đó pt trên vô nghiệm
Tức là không tìm đc $x$ thỏa mãn.
.... giúp với nha . Thank trước !
Bài 1:
\(A=\dfrac{2}{\sqrt{2017}+\sqrt{2015}}\)
\(B=\dfrac{2}{\sqrt{2019}+\sqrt{2017}}\)
mà \(\sqrt{2015}< \sqrt{2019}\)
nên A>B
=>|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)
3) Gợi ý: Thay 1=xy+yz+xz
\(x\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}=x\sqrt{\dfrac{\left(y^2+xy+yz+xz\right)\left(z^2+xy+yz+xz\right)}{x^2+xy+yz+xz}}=x\sqrt{\dfrac{\left(y+z\right)\left(x+y\right)\left(x+z\right)\left(y+z\right)}{\left(x+z\right)\left(x+y\right)}}=x\sqrt{\left(y+z\right)^2}=x\left(y+z\right)\)
Tương tự rồi cộng vào
Đặt \(a=\sqrt{x-2015};b=\sqrt{y-2016};c=\sqrt{z-2017}\left(a,b,c>0\right)\)
Khi đó phương trình trở thành:
\(\dfrac{a-1}{a^2}+\dfrac{b-1}{b^2}+\dfrac{c-1}{c^2}=\dfrac{3}{4}\\ \Leftrightarrow\left(\dfrac{1}{4}-\dfrac{1}{a}+\dfrac{1}{a^2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{b}+\dfrac{1}{b^2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{c}+\dfrac{1}{c^2}\right)=0\\ \Leftrightarrow\left(\dfrac{1}{2}-\dfrac{1}{a}\right)^2+\left(\dfrac{1}{2}-\dfrac{1}{b}\right)^2+\left(\dfrac{1}{2}-\dfrac{1}{c}\right)^2=0\\ \Leftrightarrow a=b=c=2\\ \Leftrightarrow x=2019;y=2020;z=2021\)
Tick plz