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3/ Chu vi hình chữ nhật:
\(\left(\dfrac{1}{4}+\dfrac{3}{10}\right)\cdot2=\dfrac{11}{10}\) (chưa biết đơn vị)
Diện tích hình chữ nhật:
\(\dfrac{1}{4}\cdot\dfrac{3}{10}=\dfrac{11}{20}\) (chưa biết đơn vị)
\(c=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+...+\dfrac{1}{27.28.29.30}\)
\(3C=\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}+...+\dfrac{1}{27.28.29}-\dfrac{1}{28.29.30}\)
\(c=\dfrac{1}{1.2.3}-\dfrac{1}{28.29.30}\)\(=\)\(\dfrac{1}{6}-\dfrac{1}{24360}\)
\(C=\) \(\dfrac{4059}{24360}\)
Ta có:
\(C=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+...+\dfrac{1}{27.28.29.30}\)
\(\Rightarrow3C=\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+...+\dfrac{3}{27.28.29.30}\)
\(\Rightarrow3C=\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+...+\dfrac{1}{27.28.29}-\dfrac{1}{28.29.30}\)
\(\Rightarrow3C=\dfrac{1}{1.2.3}-\dfrac{1}{28.29.30}\)
\(\Rightarrow3C=\dfrac{1353}{8120}\)
\(\Rightarrow C=\dfrac{1353}{\dfrac{8120}{3}}=\dfrac{451}{8120}\)
Vậy \(C=\dfrac{451}{8120}\)
\(\dfrac{x-7}{y-6}=\dfrac{7}{6}\)
\(\Leftrightarrow6\left(x-7\right)=7\left(y-6\right)\)
\(6x-42=7y-42\)
\(6x=7y\Leftrightarrow x=\dfrac{7}{6}y\)
\(x=-4:\left(7-6\right).7=-28\)
\(y=-28-4=-24\)
b tương tự
Giải:b)
\(\dfrac{x-7}{y-6}=\dfrac{7}{6}\) nên \(6\left(x-7\right)=7\left(y-6\right)\)
Do đó \(6x-42=7y-42\) nên \(6x=7y\)
Suy ra \(6x-6y=y\) hay \(6\left(x-y\right)=y\)
Nên 6.(-4) = y
Vậy y = -24, x = \(\dfrac{7.\left(-24\right)}{6}\)= -28
c)
\(\dfrac{x+3}{y+5}=\dfrac{3}{5}\) nên \(5\left(x+3\right)=3\left(y+5\right)\)
Do đó \(5x+15=3y+15\) nên \(5x=3y\)
Suy ra \(5x+5y=3y+5y\)
\(5\left(x+y\right)=8y\)
\(5.16=8y\)
Nên \(y=\dfrac{5.16}{8}=\dfrac{80}{8}=10\)
Vậy y = 10, x = 16 - 10 =6
a, \(x+\dfrac{2}{3}=0,2\)
\(\Rightarrow x+\dfrac{2}{3}=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{5}-\dfrac{2}{3}\)
\(\Rightarrow x=\dfrac{-7}{10}\)
b, \(\dfrac{17}{7}-\dfrac{6}{5}x=\dfrac{17}{4}\)
\(\Rightarrow\dfrac{6}{5}x=\dfrac{17}{7}-\dfrac{17}{4}\)
\(\Rightarrow\dfrac{6}{5}x=\dfrac{-51}{28}\)
\(\Rightarrow x=\dfrac{-51}{28}:\dfrac{6}{5}\)
\(\Rightarrow x=\dfrac{-85}{56}\)
1\(\dfrac{5}{7}\) . 0,75 - \(\dfrac{6}{7}.1\dfrac{1}{3}+\dfrac{6}{7}\)
= \(\dfrac{12}{7}.0,75-\dfrac{6}{7}.\dfrac{4}{3}+\dfrac{6}{7}\)
= \(\dfrac{6}{7}.2.0,75-\dfrac{6}{7}.\dfrac{4}{3}+\dfrac{6}{7}\)
= \(\dfrac{6}{7}.\left(2.0,75-\dfrac{4}{3}+1\right)\)
= \(\dfrac{6}{7}.\dfrac{7}{6}\)
=1