so sánh \(\sqrt{2015}+\sqrt{2017}va2\sqrt{2016}\)
tổng quát: chứng minh \(\sqrt{x-1}+\sqrt{x+1}<2\sqrt{x},x>1\)
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b: \(\sqrt{2017}-\sqrt{2016}=\dfrac{1}{\sqrt{2016}+\sqrt{2017}}\)
\(\sqrt{2016}-\sqrt{2015}=\dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)
mà \(\sqrt{2016}+\sqrt{2017}< \sqrt{2016}+\sqrt{2015}\)
nên \(\sqrt{2017}-\sqrt{2016}>\sqrt{2016}-\sqrt{2015}\)
a.
\(x=9-\dfrac{1}{\sqrt{\dfrac{9-4\sqrt{5}}{4}}}+\dfrac{1}{\sqrt{\dfrac{9+4\sqrt{5}}{4}}}\\ x=9-\dfrac{1}{\dfrac{\sqrt{5}-2}{2}}+\dfrac{1}{\dfrac{\sqrt{5}+2}{2}}\\ x=9-\left(\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}\right)=9-8=1\\ \Rightarrow f\left(x\right)=f\left(1\right)=\left(1-1+1\right)^{2016}=1\)
c.
\(=\sin x\cdot\cos x+\dfrac{\sin^2x}{1+\dfrac{\cos x}{\sin x}}+\dfrac{\cos^2x}{1+\dfrac{\sin x}{\cos x}}\\ =\sin x\cdot\cos x+\dfrac{\sin^2x}{\dfrac{\sin x+\cos x}{\sin x}}+\dfrac{\cos^2x}{\dfrac{\sin x+\cos x}{\cos x}}\\ =\sin x\cdot\cos x+\dfrac{\sin^3x}{\sin x+\cos x}+\dfrac{\cos^3x}{\sin x+\cos x}\\ =\sin x\cdot\cos x+\dfrac{\left(\sin x+\cos x\right)\left(\sin^2x-\sin x\cdot\cos x+\cos^2x\right)}{\sin x+\cos x}\\ =\sin x\cdot\cos x-\sin x\cdot\cos x+\sin^2x+\cos^2x\\ =1\)
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
Ta có :
\(\left(\sqrt{2015}+\sqrt{2017}\right)^2=2015+2\sqrt{2015.2017}+2017=8064+2\sqrt{2015.2017}\)
\(\left(2\sqrt{2016}\right)^2=8064\)
Vì \(\left(\sqrt{2015}+\sqrt{2017}\right)^2>\left(2\sqrt{2016}\right)^2\) nên \(\sqrt{2015}+\sqrt{2017}>2\sqrt{2016}\)
Vậy...
Chúc bạn học tốt ~
1.
ĐKXĐ: $x\geq 1; y\geq 2; z\geq 3$
PT \(\Leftrightarrow x+y+z+8-2\sqrt{x-1}-4\sqrt{y-2}-6\sqrt{z-3}=0\)
\(\Leftrightarrow [(x-1)-2\sqrt{x-1}+1]+[(y-2)-4\sqrt{y-2}+4]+[(z-3)-6\sqrt{z-3}+9]=0\)
\(\Leftrightarrow (\sqrt{x-1}-1)^2+(\sqrt{y-2}-2)^2+(\sqrt{z-3}-3)^2=0\)
\(\Rightarrow \sqrt{x-1}-1=\sqrt{y-2}-2=\sqrt{z-3}-3=0\)
\(\Leftrightarrow \left\{\begin{matrix} x=2\\ y=6\\ z=12\end{matrix}\right.\)
2.
ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow \sqrt{x+1}=1-\sqrt{x}$
$\Rightarrow x+1=(1-\sqrt{x})^2=x+1-2\sqrt{x}$
$\Leftrightarrow 2\sqrt{x}=0$
$\Leftrightarrow x=0$
Thử lại thấy thỏa mãn
Vậy $x=0$
Ta có: \(\left(\sqrt{2015}+\sqrt{2018}\right)^2=4033+2\sqrt{2015.2018}\)
\(\left(\sqrt{2016}+\sqrt{2017}\right)^2=4033+2\sqrt{2016.2017}\)
\(2015.2018=2015.2017+2015=2017\left(2015+1\right)-2017+2015=2017.2016-2\)\(\Rightarrow2015.2018< 2016.2017\)
\(\Rightarrow4033+2\sqrt{2015.2018}< 4033+2\sqrt{2016.2017}\)
\(\Rightarrow\sqrt{2015}+\sqrt{2018}< \sqrt{2016}+\sqrt{2017}\left(đpcm\right)\)
\(\sqrt{2017}-\sqrt{2016}=\dfrac{1}{\sqrt{2017}+\sqrt{2016}}\)
\(\sqrt{2016}-\sqrt{2015}=\dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)
2017>2015
=>căn 2017>căn 2015
=>\(\sqrt{2017}+\sqrt{2016}>\sqrt{2016}+\sqrt{2015}\)
=>\(\dfrac{1}{\sqrt{2017}+\sqrt{2016}}< \dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)
=>\(\sqrt{2017}-\sqrt{2016}< \sqrt{2016}-\sqrt{2015}\)
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 = \(-\sqrt{2017}+\sqrt{2016}=\sqrt{2016}-\sqrt{2017}\)