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a: Ta có: \(A=2018^2-2017^2=2018+2017\)
\(B=2017^2-2016^2=2017+2016\)
mà 2018>2016
nên A>B
Câu 1:
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)..\left(1-\frac{1}{2016}\right)\left(1-\frac{1}{2017}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2015}{2016}.\frac{2016}{2017}\)
\(A=\frac{1}{2017}\)
Vậy ..............................
Phần giống nhau là gạch ý!
Câu 2
\(S=2^{2010}-2^{2009}-2^{2008}-...-2-1\)
\(\Rightarrow S=2^{2010}-\left(2^{2009}+2^{2008}+...+2+1\right)\)
Đặt \(Q=2^{2009}+2^{2008}+...+2+1\)
\(\Rightarrow2Q=2^{2010}+2^{2009}+...+4+2\)
\(\Rightarrow2Q-Q=\left(2^{2010}+2^{2009}+...+4+2\right)-\left(2^{2009}+2^{2008}+...+2+1\right)\)
\(\Rightarrow Q=2^{2010}-1\)
\(\Rightarrow S=2^{2010}-\left(2^{2010}-1\right)\)
\(\Rightarrow S=2^{2010}-2^{2010}+1\)
\(\Rightarrow S=1\)
Vậy .........................
b) \(S=2^{2010}-2^{2009}-2^{2008}-...-2-1\)
\(\Rightarrow2S=2^{2011}-2^{2010}-2^{2009}-...-2^2-2\)
\(\Rightarrow2S-S=\left(2^{2011}-2^{2010}-2^{2009}-...-2^2-2\right)-\left(2^{2010}-2^{2009}-2^{2008}-...-2-1\right)\)
\(\Rightarrow S=2^{2011}-2^{2010}-2^{2009}-...-2^2-2-2^{2010}+2^{2009}+2^{2008}+...+2+1\)
\(\Rightarrow S=2^{2011}-2^{2010}-2^{2010}+1\)
\(\Rightarrow S=2^{2011}-2.2^{2010}+1\)
\(\Rightarrow S=2^{2011}-2^{2011}+1\)
\(\Rightarrow S=0+1\)
\(\Rightarrow S=1.\)
Vậy \(S=1.\)
Chúc bạn học tốt!
\(5^x+5^{x-1}+5^{x-2}=155\)
\(\Rightarrow5^x:1+5^x:5+5^x:25=155\)
\(\Rightarrow5^x:\left(1+5+25\right)=155\)
\(\Rightarrow5^x:31=155\)
\(\Rightarrow5^x=4805\)
2)
\(x^3=x\)
\(\Rightarrow x^3-x=0\)
\(\Rightarrow x^2\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2=0\Rightarrow x=0\\x-1=0\Rightarrow x=1\end{matrix}\right.\)
a)\(\frac{2016}{2017}< 1;\frac{2015}{2016}< 1\)
b)\(\frac{2017}{2016}>1;\frac{2016}{2015}>1\)
=> \(\frac{2016}{2017}\)và
\(\frac{2016}{2017}< 1;\frac{2016}{2015}< 1\)
\(\frac{2017}{2016}>1;\frac{2016}{2015}>1\)
=> \(\frac{2016}{2017}\)và \(\frac{2015}{2016}\)< \(\frac{2017}{2016}\)và \(\frac{2016}{2015}\)
Ta thấy : \(\left|x-1\right|\ge0\)
\(\left|y+2007\right|\ge0\)
\(\Rightarrow B=\left|x-1\right|=2\left|y+2007\right|-2010\ge-2010\)
\(MaxB=-2010\Leftrightarrow\hept{\begin{cases}x-1=0\\y+2007=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2007\end{cases}}}\)
a)có ng` lm r`
b)Áp dụng Bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)ta có:
\(C-10\ge\left|x-2+2009-x\right|=2007\)
\(\Rightarrow C\ge2017\)
Dấu = khi x=2 hoặc x=2009
Vậy MinC=2017 khi x=2 hoặc x=2009
c)Xét từng trường hợp và ta có:
MinD=-1 khi \(x\ge1\)
d)\(\left|x-1\right|+\left|x-5\right|+\left|x-7\right|\)
\(\ge\left|x-1+0+7-x\right|=6\)
\(\Rightarrow E\ge6\)
Dấu = khi \(\hept{\begin{cases}x-1\ge0\\x-5=0\\x-7\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge1\\x=5\\x\le7\end{cases}}\Leftrightarrow x=5\)
Vậy MinE=6 khi x=5
a, \(\dfrac{2009}{2010}\) và \(\dfrac{2010}{2011}\)
Ta có:
\(2009.2011=4040099\)
\(2010.2010=4040100\)
Vì \(2009.2011< 2010.2010\)
nên \(\dfrac{2009}{2010}< \dfrac{2010}{2011}\)
b, \(\dfrac{2008}{2008.2009}\) và \(\dfrac{2009}{2009.2010}\)
Ta có:
\(\dfrac{2008}{2008.2009}=\dfrac{1}{2009};\dfrac{2009}{2009.2010}=\dfrac{1}{2010}\)
Vì \(\dfrac{1}{2009}>\dfrac{1}{2010}\) nên \(\dfrac{2008}{2008.2009}>\dfrac{2009}{2009.2010}\)
Chúc bạn học tốt!!!
a)\(\dfrac{a}{b}< 1\Leftrightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(\dfrac{2009}{2010}< 1\)
\(\Leftrightarrow\dfrac{2009}{2010}< \dfrac{2009+1}{2010+1}\Leftrightarrow\dfrac{2009}{2010}< \dfrac{2010}{2011}\)
b)
\(\dfrac{2008}{2008.2009}=\dfrac{1}{2009}\)
\(\dfrac{2009}{2009.2010}=\dfrac{1}{2010}\)
\(\dfrac{1}{2009}>\dfrac{1}{2010}\Leftrightarrow\dfrac{2008}{2008.2009}>\dfrac{2009}{2009.2010}\)
d)
\(\dfrac{1}{3^{400}}=\dfrac{1}{\left(3^4\right)^{100}}=\dfrac{1}{81^{100}}\)
\(\dfrac{1}{4^{300}}=\dfrac{1}{\left(4^3\right)^{100}}=\dfrac{1}{64^{100}}\)
\(81^{100}>64^{100}\Leftrightarrow\dfrac{1}{81^{100}}< \dfrac{1}{64^{100}}\)