Tìm X
( X-\(\frac{2}{3}\)) x3 = \(\frac{3}{4}\)
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Đặt \(\frac{x_1-1}{5}=\frac{x_2-2}{4}=\frac{x_3-3}{3}=\frac{x_4-4}{2}=\frac{x_5-5}{1}=k\)
Áp dụng TC DTSBN ta có :
\(k=\frac{\left(x_1-1\right)+\left(x_2-2\right)+\left(x_3-3\right)+\left(x_4-4\right)+\left(x_5-5\right)}{5+4+3+2+1}\)
\(=\frac{x_1+x_2+x_3+x_4+x_5-15}{15}=\frac{30-15}{15}=1\)
\(\frac{x_1-1}{5}=1\Rightarrow x_1=6;\frac{x_2-2}{4}=1\Rightarrow x_2=6;\frac{x_3-3}{3}=1\Rightarrow x_3=6;\frac{x_4-4}{2}=1\Rightarrow x_4=6;\frac{x^5-5}{2}=1\Rightarrow x_5=6\)
Vậy \(x_1=x_2=x_3=x_4=x_5=6\)
-\(\left(x-\frac{2}{3}\right).3=\frac{3}{4}\)
-\(\left(x-\frac{2}{3}\right)=\frac{3}{4}:3\)
-\(x-\frac{2}{3}=\frac{1}{4}\)
\(-x=\frac{1}{4}+\frac{2}{3}\)
\(-x=\frac{11}{12}\)
\(=>x=-\frac{11}{12}\)
<br class="Apple-interchange-newline"><div id="inner-editor"></div>(x−23 ).3=34
-(x−23 )=34 :3
-x−23 =14
−x=14 +23
−x=1112
x.\(2\frac{3}{14}\)+ x.\(3\frac{2}{7}\)= 5 - \(2\frac{3}{4}\)
x.\(\frac{31}{14}\)+ x.\(\frac{23}{7}\)=5 -\(\frac{11}{4}\)
x (\(\frac{31}{14}\)+ \(\frac{23}{7}\) = \(\frac{20}{4}\)- \(\frac{11}{4}\)
x ( \(\frac{31}{14}\)+ \(\frac{46}{14}\))=\(\frac{9}{4}\)
x\(\frac{11}{2}\) =\(\frac{9}{4}\)
x =\(\frac{9}{4}\):\(\frac{11}{2}\)
x =\(\frac{9}{4}\). \(\frac{2}{11}\)
x =\(\frac{9}{2.11}\)
x =\(\frac{9}{22}\)
\(\left(x-\frac{2}{3}\right).3=\frac{3}{4}\)
\(\left(x-\frac{2}{3}\right)=\frac{3}{4}:3\)
\(x-\frac{2}{3}=\frac{1}{4}\)
\(x=\frac{1}{4}+\frac{2}{3}\)
\(x=\frac{11}{12}\)
x-2/3=3/4:3
x-2/3=1/4
x=1/4+2/3
x=11/12