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\(=\frac{4}{11}\times\left(\frac{5}{13}+\frac{8}{13}\right)\)
\(=\frac{4}{11}\times1\)
\(=\frac{4}{11}\)
Chúc cậu học tốt
\(=\frac{4}{11}\times\left(\frac{5}{13}+\frac{8}{13}\right)\)
\(=\frac{4}{11}\times1\)
\(=\frac{4}{11}\)
Ta có: \(\frac{x}{13}+\frac{x}{5}+\frac{x}{15}=18\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{13}+\frac{x}{5}+\frac{x}{15}=\frac{x+x+x}{13+5+15}=\frac{3x}{33}=\frac{x}{11}=18\)
=> \(\Rightarrow\frac{x}{11}=18\Rightarrow x=18\cdot11=198\)
Vậy x = 198
a)\(\frac{8}{13}+\frac{4}{9}+\frac{1}{3}+\frac{5}{13}+3\)
\(=\left(\frac{8}{13}+\frac{5}{13}\right)+\left(\frac{4}{9}+\frac{1}{3}\right)+3\)
\(=1+\left(\frac{4}{9}+\frac{3}{9}\right)+3\)
\(=1+\frac{7}{9}+3\)
\(=1+3+\frac{7}{9}\)
\(=4\frac{7}{9}\)
\(=\frac{43}{9}\)
b)\(\frac{8}{5}\cdot\frac{4}{7}+\frac{4}{7}\cdot\frac{2}{5}-1\)
\(=\frac{4}{7}\cdot\left(\frac{8}{5}+\frac{2}{5}\right)-1\)
\(=\frac{4}{7}\cdot2-1\)
\(=\frac{8}{7}-1\)
\(=\frac{1}{7}\)
a) \(\frac{8}{13}+\frac{4}{9}+\frac{1}{3}+\frac{5}{13}+3\)
\(=\left(\frac{8}{13}+\frac{5}{13}\right)+\left(\frac{4}{9}+\frac{3}{9}\right)+3\)
\(=1+\frac{7}{9}+3=4\frac{7}{9}\)
b)\(\frac{8}{5}.\frac{4}{7}+\frac{4}{7}.\frac{2}{5}-1\)
\(=\frac{4}{7}.\left(\frac{8}{5}+\frac{2}{5}\right)-1\)
\(=\frac{4}{7}.2-1\)
\(=\frac{8}{7}-\frac{7}{7}=\frac{1}{7}\)
\(\frac{x+9}{13-x}=\frac{6}{5}\)
\(\Rightarrow\orbr{\begin{cases}x+9=6\\13-x=5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=8\end{cases}}\)
\(\frac{120-x}{30}=3\frac{1}{2}\)
\(\Rightarrow\frac{120-x}{30}=\frac{7}{2}\)
\(\Rightarrow\frac{120-x}{30}=\frac{105}{30}\)
\(\Rightarrow120-x=105\)
\(\Rightarrow x=120-105\)
\(\Rightarrow x=15\)