Tìm STN x biết:
\(\frac{4}{7}< \frac{x}{10}< \frac{5}{7}\)
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Ta có: \(\frac{4}{7}< \frac{x}{10}< \frac{5}{7}\)
Ta lại có: \(\frac{8}{14}< \frac{x}{14}< \frac{x}{10}< \frac{10}{14}\)
\(\Rightarrow\frac{8}{14}< \frac{9}{14}< \frac{9}{10}< \frac{10}{14}\)
Vậy giá trị của x là 9.
Ta có:
\(\frac{4}{7}< \frac{x}{10}< \frac{5}{7}\)
\(\Rightarrow\frac{40}{70}< \frac{7x}{70}< \frac{50}{70}\)
\(\Rightarrow40< 7x< 50\)
Vì \(7x⋮7\)
\(\Rightarrow7x\in\left\{42;49\right\}\)
\(\Rightarrow x\in\left\{6;7\right\}\)
Vậy: x = 6 hoặc x = 7
Bài 1:
\(A=\left(\frac{-5}{11}+\frac{7}{22}-\frac{4}{33}-\frac{5}{44}\right):\left(38\frac{1}{122}-39\frac{7}{22}\right)\)
\(=\frac{-49}{132}:\left(-\frac{879}{671}\right)=\frac{2989}{105408}\)
Bài 2:
\(\frac{4}{5}-\left(\frac{-1}{8}\right)=\frac{7}{8}-x\)
<=> \(\frac{7}{8}-x=\frac{27}{40}\)
<=> \(x=\frac{7}{8}-\frac{27}{40}=\frac{1}{5}\)
Vậy...
a) \(y+2\frac{3}{4}=5\frac{2}{3}\)
\(y+\frac{11}{4}=\frac{17}{3}\)
y = 35/12
b) \(y-1\frac{4}{5}=3\frac{2}{7}\)
y - 9/5 = 23/7
y = 178/35
\(a,y+2\frac{3}{4}=5\frac{2}{3}\)
\(\Rightarrow y+\frac{11}{4}=\frac{17}{3}\)
\(\Rightarrow y=\frac{17}{3}-\frac{11}{4}\)
\(\Rightarrow y=\frac{35}{12}\)
\(b,y-1\frac{4}{5}=3\frac{2}{7}\)
\(\Rightarrow y=3\frac{2}{7}-1\frac{4}{5}\)
\(\Rightarrow y=\frac{52}{35}\)
a) \(\left(\frac{2}{3}x-\frac{4}{9}\right)\left(\frac{1}{2}-\frac{3}{7}:x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}x-\frac{4}{9}=0\\\frac{1}{2}-\frac{3}{7}:x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{6}{7}\end{cases}}\)
Vậy \(x\in\left\{\frac{2}{3};\frac{6}{7}\right\}\)
b)
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1+\frac{x+329}{5}+4=4\)
\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Mà \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\ne0\)
\(\Rightarrow x+329=0\)
\(\Rightarrow x=-329\)
Vậy \(x=-329\)
Ta có: \(\frac{x+1}{2014}+\frac{x+2}{2013}+\frac{x+3}{2012}=\frac{x+4}{2011}+\frac{x+5}{2010}+\frac{x+6}{2009}\)
\(\Rightarrow\frac{x+1}{2014}+1+\frac{x+2}{2013}+1+\frac{x+3}{2012}+1=\frac{x+4}{2011}+1+\frac{x+5}{2010}+1+\frac{x+6}{2009}+1\)
\(\Rightarrow\frac{2015+x}{2014}+\frac{2015+x}{2013}+\frac{2015+x}{2012}=\frac{2015+x}{2011}+\frac{2015+x}{2010}+\frac{2015+x}{2009}\)
\(\Rightarrow\left(2015+x\right)\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}\right)=0\)
=> 2015 + x = 0
=> x = -2015
quy đồng lên
\(\frac{40}{70}< \frac{x}{70}< \frac{50}{70}\)
x thuộc {41;42;43;44;45;46;47;48;49}
\(\frac{41}{70};\frac{42}{70}\frac{43}{70}\frac{44}{70}\frac{45}{70}\frac{46}{70}\frac{47}{70}\frac{48}{70}\frac{49}{70}\)