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Bài 1 : Thực hiện phép tính :
a, \(\frac{4}{5}+1\frac{1}{6}\cdot\frac{3}{4}\)
= \(\frac{4}{5}+\frac{7}{6}\cdot\frac{3}{4}\)
= \(\frac{4}{5}+\frac{7}{8}\)
= \(\frac{32+35}{40}=\frac{67}{40}\)
b, \(\frac{2}{3}:\left(\frac{3}{4}\cdot\frac{4}{3}\right)+2\)
\(=\frac{2}{3}:1+2\)
\(=\frac{2}{3}+2=\frac{2+6}{3}=\frac{8}{3}\)
c, \(\frac{1}{2}\times\left(\frac{2}{3}+\frac{3}{5}\cdot\frac{5}{7}\right)+1\frac{1}{3}\)
\(=\frac{1}{2}\cdot\left(\frac{2}{3}+\frac{9}{35}\right)+\frac{4}{3}\)
\(=\frac{1}{2}\cdot\frac{97}{105}+\frac{4}{3}\)
\(=\frac{97}{210}+\frac{4}{3}=\frac{377}{210}\)
Bài 2 : Tìm \(x\inℤ\), biết :
a, \(\frac{2}{3}< \frac{x}{6}\le\frac{10}{3}\)
\(\Leftrightarrow\frac{4}{6}< \frac{x}{6}\le\frac{20}{6}\)
mà \(x\inℤ\Rightarrow\text{x}\in\) {\(5;6;7;8;9;10;11;12;13;14;15;16;17;18;19;20\)}
b, \(\frac{1}{3}+x=1\frac{1}{2}\)
\(\frac{1}{3}+x=\frac{3}{2}\)
\(x=\frac{3}{2}+\frac{\left(-1\right)}{3}\)
\(x=\frac{7}{6}\) (loại vì \(x\notinℤ\))
\(\Rightarrow x\in\varnothing\)
c, \(\frac{1}{7}+x=\frac{25}{14}+\frac{5}{14}\)
\(\frac{1}{7}+x=\frac{15}{7}\)
\(x=\frac{15}{7}+\frac{(-1)}{7}\)
\(x=\frac{14}{7}=2\).
\(\left(4\frac{5}{37}-3\frac{4}{5}+8\frac{15}{29}\right)-\left(3\frac{5}{37}-6\frac{14}{29}\right)\)
\(\left(4\frac{5}{37}-3\frac{4}{5}+8\frac{15}{29}\right)-\left(3\frac{5}{37}-6\frac{14}{29}\right)\)
a.\(\frac{11}{14}-\frac{3}{x-1}=\frac{5}{14}\)
\(\frac{3}{x-1}=\frac{11}{14}-\frac{5}{14}\)
\(\frac{3}{x-1}=\frac{3}{7}\)
\(\Rightarrow x-1=7\)
\(x=7+1\)
Vậy \(x=8\)
b.\(\frac{42}{25}:\frac{2x+1}{5}=\frac{6}{5}\)
\(\frac{2x+1}{5}=\frac{42}{25}:\frac{6}{5}\)
\(\frac{2x+1}{5}=\frac{7}{5}\)
\(\Rightarrow2x+1=7\)
\(2x=7-1\)
\(2x=6\)
\(x=6:2\)
\(x=3\)
a.\(\frac{11}{14}-\frac{3}{x-1}=\frac{5}{14}\)
\(\frac{3}{x-1}=\frac{11}{14}-\frac{5}{14}=\frac{3}{7}\)
\(\Rightarrow x-1=7\)
\(x=7+1=8\)
VẬY, \(x=8\)
b. \(\frac{42}{25}:\frac{2x+1}{5}=\frac{6}{5}\)
\(\frac{2x+1}{5}=\frac{42}{25}:\frac{6}{5}=\frac{7}{5}\)
\(\Rightarrow2x+1=7\)
\(2x=7-1=6\)
\(x=6:2=3\)
VẬY, \(x=3\)
