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B1a)\(11\frac34-\left(6\frac56-4\frac12\right)+1\frac23\)
=\(11\frac34-6\frac56+4\frac12+1\frac23\)
=\(\left(11-6+4+1\right)+\left(\frac34-\frac56+\frac12+\frac23\right)\)
=\(10+\left(\frac{9}{12}-\frac{10}{12}+\frac{6}{12}+\frac{8}{12}\right)\)
=\(10+\left(-\frac{1}{12}+\frac{6}{12}+\frac{8}{12}\right)\)
=10+\(\frac{13}{12}\)
=\(\frac{120}{12}+\frac{13}{12}\)
=\(\frac{133}{12}\)
b)\(2\frac{17}{20}-1\frac{11}{5}+6\frac{9}{20}:3\)
= \(\frac{57}{20}-\frac{16}{5}+\frac{129}{20}\times\frac13\)
=\(\frac{57}{20}-\frac{16}{5}+\frac{129}{60}\)
=\(\frac{171}{60}-\frac{192}{60}+\frac{129}{60}\)
=\(\frac{108}{60}\)
=\(\frac95\)
\(1.a,\frac{18}{24}:\frac{5}{2}+\frac{7}{-10}\)
\(=\frac{18}{24}:\frac{5}{2}+\frac{-7}{10}\)
\(=\frac{18}{24}\cdot\frac{2}{5}+\frac{-7}{10}\)
\(=\frac{3}{4}\cdot\frac{2}{5}+\frac{-7}{10}\)
\(=\frac{3}{2}\cdot\frac{1}{5}+\frac{-7}{10}\)
\(=\frac{3}{10}+\frac{-7}{10}=\frac{-4}{10}=\frac{-2}{5}\)
\(\left[\frac{12}{199}-\frac{23}{200}+\frac{34}{201}\right]\cdot\left[\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right]\)
\(=\left[\frac{12}{199}-\frac{23}{200}+\frac{34}{201}\right]\cdot\left[\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right]\)
\(=\left[\frac{12}{199}-\frac{23}{200}+\frac{34}{201}\right]\cdot\left[\frac{3-2-1}{6}\right]\)
\(=\left[\frac{12}{199}-\frac{23}{200}+\frac{34}{301}\right]\cdot0=0\)
2. \(a,x+5=15\Leftrightarrow x=15-5=10\)
\(b,x-\frac{5}{3}=\frac{1}{9}\)
\(\Leftrightarrow x=\frac{1}{9}+\frac{5}{3}\)
\(\Leftrightarrow x=\frac{1}{9}+\frac{15}{9}=\frac{16}{9}\)
c, Sửa lại đề một xíu :
\(\frac{x}{15}=\frac{-2}{3}+\frac{3}{5}\)
\(\Leftrightarrow\frac{x}{15}=\frac{-10}{15}+\frac{9}{15}\)
\(\Leftrightarrow\frac{x}{15}=\frac{-1}{15}\)
\(\Leftrightarrow x=-1\)
\(d,\frac{x}{9}< \frac{7}{x}< \frac{x}{6}(x\inℤ)\)
\(\frac{x\cdot x}{9\cdot x}< \frac{7\cdot9}{9\cdot x}< \frac{7\cdot6}{6\cdot x}\)
\(\Leftrightarrow\frac{x^2}{9x}< \frac{63}{9x}< \frac{42}{6x}\)
Tự làm nốt :>
a) \(\left(\frac{-1}{6}+\frac{5}{-12}\right)+\frac{7}{12}=\left(\frac{-2}{12}+\frac{-5}{12}\right)+\frac{7}{12}=\left(\frac{-7}{12}\right)+\frac{7}{12}=0\)
b)\(\frac{7}{36}-\frac{8}{-9}+\frac{-2}{3}=\frac{7}{36}+\frac{32}{36}-\frac{24}{36}=\frac{15}{36}=\frac{5}{12}\)
c) \(\frac{3}{5}-\frac{2}{5}.\frac{10}{12}=\frac{3}{5}-\frac{2}{5}.\frac{5}{6}=\frac{3}{5}-\frac{1}{3}=\frac{9}{15}-\frac{5}{15}=\frac{4}{15}\)
d) \(\frac{2}{\left(-3\right)^2}+\frac{5}{-13}-\frac{-3}{4}=\frac{2}{9}-\frac{5}{13}+\frac{3}{4}=\frac{8}{36}-\frac{15}{36}+\frac{27}{36}=\frac{5}{9}\)
Hình như mik chưa tính nhưng vế sau là = 0 nên bnj ko cần tính vế trước đâu
( \(\frac{12}{199}\)+ \(\frac{23}{200}\)- \(\frac{34}{201}\)) x ( \(\frac{1}{2}\)- \(\frac{1}{3}\)- \(\frac{1}{6}\))
= ( \(\frac{12}{199}\)+ \(\frac{23}{200}\)- \(\frac{34}{201}\)) x ( \(\frac{3}{6}\)- \(\frac{2}{6}\)- \(\frac{1}{6}\))
=( \(\frac{12}{199}\) + \(\frac{23}{200}\) - \(\frac{34}{201}\)) x 0
= 0
Học tốt ^-^
vì\(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}=\frac{3}{6}-\frac{2}{6}-\frac{1}{6}=\frac{3-2-1}{6}=\frac{0}{6}=0\)
=> \(\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\cdot\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)=\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\cdot0=0\)
\(\frac37+\frac{5}{13}+\frac{4}{13}\)
=\(\frac37+\left(\frac{5}{13}+\frac{4}{13}\right)\)
=\(\frac37+\frac{9}{13}\)
=\(\frac{39}{91}+\frac{63}{91}\)
=\(\frac{102}{91}\)
\(\left(\frac38+\frac{-3}{4}+\frac{7}{12}\right):\frac56+\frac12\)
=\(\left(\frac{9}{24}+\frac{-18}{24}+\frac{7}{24}\right)\times\frac65+\frac12\)
=\(\frac{5}{24}\times\frac65+\frac12\)
=\(\frac14+\frac12\)
=\(\frac14+\frac24\)
=\(\frac34\)