Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
2)Ta có: \(2^{332}< 2^{333}=\left(2^3\right)^{111}=8^{111}\)
\(3^{223}>3^{222}=\left(3^2\right)^{111}=9^{111}\)
Vì \(8^{111}< 9^{111}\) mà \(2^{332}< 8^{111},3^{223}>9^{111}\) nên suy ra \(2^{332}< 3^{223}\)
Vậy \(2^{332}< 3^{223}\)
1) \(A=\dfrac{10^{2013}+1}{10^{2014}+1}\Rightarrow10A=\dfrac{10^{2014}+10}{10^{2014}+1}=\dfrac{10^{2014}+1}{10^{2014}+1}+\dfrac{9}{10^{2014}+1}=1+\dfrac{9}{10^{2014}+1}\)
\(B=\dfrac{10^{2014}+1}{10^{2015}+1}\Rightarrow10B=\dfrac{10^{2015}+10}{10^{2015}+1}=\dfrac{10^{2015}+1}{10^{2015}+1}+\dfrac{9}{10^{2015}+1}=1+\dfrac{9}{10^{2015}+1}\)Vì: \(10^{2014}+1< 10^{2015}+1\Rightarrow\dfrac{9}{10^{2014}+1}>\dfrac{9}{10^{2015}+1}\Rightarrow1+\dfrac{9}{10^{2014}+1}>1+\dfrac{9}{10^{2015}+1}\)
Nên suy ra \(10A>10B\Rightarrow A>B\)
A= 1+2-3-4+5+6-7-8+...+2013+2014
A=(1+2-3-4)+(5+6-7-8)+.....+(2013+2014)
A=(-4)+(-4)+...+(-4)+4027
A=(-4).503+4027
A=-2012+4027
A=2015
B=\(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{2016}{2015}\)
B=\(\dfrac{3.4.5.6.....2016}{2.3.4.5.....2015}=\dfrac{2016}{2}=1008\)
1) \(\left|5x-4\right|=\left|x+2\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-4=x+2\\5x-4=-\left(x+2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-4=x+2\\5x-4=-x-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-x=2+4\\5x+x=-2+4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=6\\6x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(x=\dfrac{1}{3}\) hoặc \(x=\dfrac{3}{2}\)
2) \(\left|x+15\right|=\left|3x-4\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x+15=3x-4\\x+15=-\left(3x-4\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+15=3x-4\\x+15=-3x+4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3x=-4-15\\x+3x=4-15\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x=-19\\4x=-11\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{19}{2}\\x=-\dfrac{11}{4}\end{matrix}\right.\)
Vậy \(x=-\dfrac{11}{4}\) hoặc \(x=\dfrac{19}{2}\)
3) \(\left|\dfrac{5}{4}x-\dfrac{7}{2}\right|-\left|\dfrac{5}{8}x+\dfrac{3}{5}\right|=0\)
\(\Leftrightarrow\left|\dfrac{5}{4}x-\dfrac{7}{2}\right|=\left|\dfrac{5}{8}x+\dfrac{3}{5}\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{4}x-\dfrac{7}{2}=\dfrac{5}{8}x+\dfrac{3}{5}\\\dfrac{5}{4}x-\dfrac{7}{2}=-\left(\dfrac{5}{8}x+\dfrac{3}{5}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{4}x-\dfrac{7}{2}=\dfrac{5}{8}x+\dfrac{3}{5}\\\dfrac{5}{4}x-\dfrac{7}{2}=-\dfrac{5}{8}x-\dfrac{3}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{4}x-\dfrac{5}{8}x=\dfrac{3}{5}+\dfrac{7}{2}\\\dfrac{5}{4}x+\dfrac{5}{8}x=-\dfrac{3}{5}+\dfrac{7}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{8}x=\dfrac{41}{10}\\\dfrac{15}{8}x=\dfrac{29}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{164}{25}\\x=\dfrac{116}{75}\end{matrix}\right.\)
Vậy \(x=\dfrac{116}{75}\) hoặc \(x=\dfrac{164}{25}\)
4) \(\left|2x-6\right|-\left|x+3\right|=0\)
\(\Leftrightarrow\left|2x-6\right|=\left|x+3\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=x+3\\2x-6=-\left(x+3\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=x+3\\2x-6=-x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3+6\\2x+x=-3+6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\3x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)
Vậy \(x=1\) hoặc \(x=9\)
A= 1+(\(\dfrac{1}{2014}\)+1)+(\(\dfrac{2}{2013}\)+1)+...+(\(\dfrac{2013}{2}\)+1)
= \(\dfrac{2015}{2015}\)+(\(\dfrac{1}{2014}\)+1)+(\(\dfrac{2}{2013}\)+1)+...+(\(\dfrac{2013}{2}\)+1)
= 2015.(\(\dfrac{1}{2015}\)+\(\dfrac{1}{2014}\)+\(\dfrac{1}{2013}\)+...+\(\dfrac{1}{2}\))=2015.B
\(\Rightarrow\) \(\dfrac{A}{B}\)=2015
2/
a/ \(\dfrac{7}{10}=\dfrac{7.15}{10.15}=\dfrac{105}{150}\)
\(\dfrac{11}{15}=\dfrac{11.10}{15.10}=\dfrac{110}{150}\)
-Vì \(\dfrac{105}{150}< \dfrac{110}{150}\)(105<110)nên \(\dfrac{7}{10}< \dfrac{11}{15}\)
b/ \(\dfrac{-1}{8}=\dfrac{-1.3}{8.3}=\dfrac{-3}{24}\)
-Vì \(\dfrac{-3}{24}>\dfrac{-5}{24}\left(-3>-5\right)\)nên\(\dfrac{-1}{8}>\dfrac{-5}{24}\)
c/\(\dfrac{25}{100}=\dfrac{25:25}{100:25}=\dfrac{1}{4}\)
\(\dfrac{10}{40}=\dfrac{10:10}{40:10}=\dfrac{1}{4}\)
-Vì \(\dfrac{1}{4}=\dfrac{1}{4}\)nên\(\dfrac{25}{100}=\dfrac{10}{40}\)
a/ \(\dfrac{7}{10}< \dfrac{11}{15}\)
c/ \(\dfrac{25}{100}=\dfrac{10}{40}\)
1) \(\dfrac{2}{x+1}=\dfrac{x+1}{8}\Leftrightarrow\left(x+1\right)\left(x+1\right)=2.8\Leftrightarrow\left(x+1\right)^2=16\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\) vậy \(x=3;x=-5\)
2) thiếu quế phải nha
3) \(\dfrac{x-4}{x-7}=\left(\dfrac{-3}{5}\right)^2\Leftrightarrow\dfrac{x-4}{x-7}=\dfrac{9}{25}\Leftrightarrow9.\left(x-7\right)=25.\left(x-4\right)\)
\(\Leftrightarrow9x-63=25x-100\Leftrightarrow25x-9x=-63+100\)
\(\Leftrightarrow16x=37\Leftrightarrow x=\dfrac{37}{16}\) vậy \(x=\dfrac{37}{16}\)
4) ta có : \(x+y=20\Leftrightarrow y=20-x\)
\(\dfrac{3+x}{7+y}=\dfrac{3}{7}\Leftrightarrow7\left(3+x\right)=3\left(7+y\right)\Leftrightarrow21+7x=21+3y\)
\(\Leftrightarrow7x=3y\Leftrightarrow7x-3y=0\Leftrightarrow7x-3\left(20-x\right)=0\)
\(\Leftrightarrow7x-60+3x=0\Leftrightarrow10x=60\Leftrightarrow x=6\)
\(\Rightarrow6+y=20\Leftrightarrow y=14\) vậy \(x=6;y=14\)
\(\dfrac{23+x}{40-x}=\dfrac{-3}{4}\Leftrightarrow4\left(23+x\right)=-3\left(40-x\right)\)
\(\Leftrightarrow92+4x=-120+3x\Leftrightarrow4x-3x=-120-92\)
\(\Leftrightarrow x=-212\) vậy \(x=-212\)
Lời giải:
$M=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{2014}{5^{2014}}$
$5M=1+\frac{2}{5}+\frac{3}{5^2}+...+\frac{2014}{5^{2013}}$
$\Rightarrow 4M=5M-M=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2013}}-\frac{2014}{5^{2014}}$
$4M+\frac{2014}{5^{2014}}=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2013}}$
$5(4M+\frac{2014}{5^{2014}})=5+1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2012}}$
$\Rightarrow 4(4M+\frac{2014}{5^{2014}})=5-\frac{1}{5^{2013}}$
$M=\frac{5}{16}-\frac{1}{16.5^{2013}-\frac{2014}{4.5^{2014}}$
Thế bạn có làm được không Võ Nguyễn Anh Thư? Trả lời thì trả lời câu hỏi ý, trả lời cái đấy để làm gì?
Ace Legona, Hoàng Thị Ngọc Anh, ... giúp mình câu này với!