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28 tháng 7 2022
a: Đặt a=2017
\(A=\sqrt{1+\left(\dfrac{1}{a}+\dfrac{1}{a+2}\right)^2}\)
\(=\sqrt{1+\left(\dfrac{2a+2}{a\left(a+2\right)}\right)^2}\)
\(=\sqrt{1+\dfrac{4a^2+8a+4}{a^2\cdot\left(a+2\right)^2}}=\sqrt{\dfrac{\left(a^2+a\right)^2+4a^2+8a+4}{a^2\left(a+2\right)^2}}\)
\(=\sqrt{\dfrac{\left(a^2+a\right)^2+4\left(a+1\right)^2}{a^2\left(a+2\right)^2}}\)
\(=\dfrac{\sqrt{\left(a^2+a\right)^2+4\left(a+1\right)^2}}{a\left(a+2\right)}\)
\(=\dfrac{\sqrt{\left(2017^2+2017\right)^2+4\cdot2018^2}}{2017\cdot2019}\)
b: Đặt 2017=a
\(B=\sqrt{a^2+a^2\cdot\left(a+1\right)^2+\left(a+1\right)^2}\)
\(=\sqrt{2a^2+2a+1+\left(a^2+a\right)^2}\)
\(=\sqrt{\left(a^2+a+1\right)^2}=a^2+a+1\)
\(=2017^2+2017+1=4070307\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 10 2019
a/ ĐKXĐ: \(x\ge2\)
\(\Leftrightarrow2\sqrt{\left(x-2\right)\left(x+2\right)}-6\sqrt{x-2}+\sqrt{x+2}-3=0\)
\(\Leftrightarrow2\sqrt{x-2}\left(\sqrt{x+2}-3\right)+\sqrt{x+2}-3=0\)
\(\Leftrightarrow\left(2\sqrt{x-2}+1\right)\left(\sqrt{x+2}-3\right)=0\)
\(\Leftrightarrow\sqrt{x+2}-3=0\Rightarrow x=11\)
b/ ĐKXĐ: ....
Đặt \(\left\{{}\begin{matrix}\sqrt{x-2016}=a>0\\\sqrt{y-2017}=b>0\\\sqrt{z-2018}=a>0\end{matrix}\right.\)
\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{4}-\frac{a-1}{a^2}+\frac{1}{4}-\frac{b-1}{b^2}+\frac{1}{4}-\frac{c-1}{c^2}=0\)
\(\Leftrightarrow\frac{\left(a-2\right)^2}{a^2}+\frac{\left(b-2\right)^2}{b^2}+\frac{\left(c-2\right)^2}{c^2}=0\)
\(\Leftrightarrow a=b=c=2\Rightarrow\left\{{}\begin{matrix}x=2020\\y=2021\\z=2022\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 10 2019
a/ ĐK: \(x\ge0\)
\(\Leftrightarrow\sqrt{3+x}=x^2-3\)
Đặt \(\sqrt{3+x}=a>0\Rightarrow3=a^2-x\) pt trở thành:
\(a=x^2-\left(a^2-x\right)\)
\(\Leftrightarrow x^2-a^2+x-a=0\)
\(\Leftrightarrow\left(x-a\right)\left(x+a+1\right)=0\)
\(\Leftrightarrow x=a\) (do \(x\ge0;a>0\))
\(\Leftrightarrow\sqrt{3+x}=x\Leftrightarrow x^2-x-3=0\)
d/ ĐKXĐ: ...
\(\sqrt{6x^2+1}=\sqrt{2x-3}+x^2\)
\(\Leftrightarrow\sqrt{2x-3}-1+x^2+1-\sqrt{6x^2+1}\)
\(\Leftrightarrow\frac{2\left(x-2\right)}{\sqrt{2x-3}+1}+\frac{x^4+2x^2+1-6x^2-1}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}=0\)
\(\Leftrightarrow\frac{2\left(x-2\right)}{\sqrt{2x-3}+1}+\frac{x^2\left(x+2\right)\left(x-2\right)}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{2}{\sqrt{2x-3}+1}+\frac{x^2\left(x+2\right)}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}\right)=0\)
\(\Leftrightarrow x=2\) (phần trong ngoặc luôn dương với mọi \(x\ge\frac{3}{2}\))
1 tháng 8 2018
a) Ta có: \(\left(\sqrt{2017}+\sqrt{2019}\right)^2=2017+2019+2\sqrt{2017.2019}\)
\(=4036+2\sqrt{\left(2018-1\right).\left(2018+1\right)}\)
\(=4036+2\sqrt{2018^2-1}< 4036+2\sqrt{2018^2}=2018.4=\left(2\sqrt{2018}\right)^2\)
Vậy x < y
AS
0
DN
2
TD
16 tháng 6 2018
Đat 2017,5=t Ta có
\(\sqrt{\dfrac{\left(t+0,5\right)^2+\left(t-0,5\right)^2\cdot\left(t+0,5\right)^2+\left(t-0,5\right)^2}{\left(t+0,5\right)^2}}+\dfrac{t-0,5}{t+0,5}\\ =\sqrt{\dfrac{t^2+t+0,25+t^4-0,5t^2+0,0625+t^2-t+0,25}{\left(t+0,5\right)^2}}+\dfrac{t-0,5}{t+0,5}\\ =\dfrac{\sqrt{t^4+1,5t^2+0,5625}}{t+0,5}+\dfrac{t-0,5}{t+0,5}\\ =\dfrac{t^2+0,75+t-0,5}{t+0,5}\\ =\dfrac{\left(t+0,5\right)^2}{t+0,5}\\ =t+0,5\)thay t=2017,5 vào suy ra A=2017,5+0,5=2018
HH
16 tháng 6 2018
Giải:
\(\sqrt{1+2017^2+\dfrac{2017^2}{2018^2}}+\dfrac{2017}{2018}\)
\(=\sqrt{\dfrac{1}{1^2}+\dfrac{1}{\left(\dfrac{1}{2017}\right)^2}+\dfrac{1}{\left(-\dfrac{2018}{2017}\right)^2}}+\dfrac{2017}{2018}\)
\(=\sqrt{\left(\dfrac{1}{1}+\dfrac{1}{\dfrac{1}{2017}}+\dfrac{1}{-\dfrac{2018}{2017}}\right)^2}+\dfrac{2017}{2018}\) (\(\left\{{}\begin{matrix}1>0\\2017^2>0\\\dfrac{2017^2}{2018^2}>0\end{matrix}\right.\Leftrightarrow1+2017^2+\dfrac{2017^2}{2018^2}>0\ne0\))
\(=1+2017+-\dfrac{2017}{2018}+\dfrac{2017}{2018}\)
\(=2018\)
Vậy ...