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\(1\frac{13}{15}.0,75-\left(\frac{11}{20}+\frac{25}{100}\right):\frac{3}{7}\)
= \(\frac{7}{5}-\frac{4}{5}:\frac{3}{7}\)
= \(\frac{7}{5}-\frac{28}{15}\)
= \(-\frac{7}{15}\)
\(1\frac{13}{15}\cdot0,75-\left(\frac{11}{20}+\frac{25}{100}\right):\frac{3}{7}\)
=>\(\frac{28}{15}\cdot\frac{3}{4}-\left(\frac{55}{100}+\frac{25}{100}\right)\cdot\frac{7}{3}\)
=>\(\frac{84}{60}-\frac{80}{100}\cdot\frac{7}{3}\)
=>\(\frac{7}{5}-\frac{4}{5}\cdot\frac{7}{3}=\frac{7}{5}-\frac{28}{15}\)
=>\(\frac{7}{5}-\frac{18}{14}=\frac{98}{60}-\frac{90}{60}=\frac{8}{60}=\frac{4}{30}\)
\(A=\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
\(A=\left(\frac{3}{8}+\frac{-6}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
\(A=\left(\frac{-3}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
\(A=\left(\frac{-36}{24}+\frac{56}{24}\right):\frac{5}{6}+\frac{1}{2}\)
\(A=\frac{5}{6}:\frac{5}{6}+\frac{1}{2}\)
\(A=\frac{5}{6}\times\frac{6}{5}+\frac{1}{2}\)
\(A=1+\frac{1}{2}\)
\(A=\frac{1}{1}+\frac{1}{2}=\frac{2}{2}+\frac{1}{2}\)
\(A=\frac{3}{2}\)
\(a,\frac{5}{16}:0,125-\left(2\frac{1}{4}-0,6\right)\cdot\frac{10}{11}\)
\(=\frac{5}{16}:\frac{1}{8}-\left(\frac{9}{4}-\frac{3}{5}\right)\cdot\frac{10}{11}\)
\(=\frac{5}{2}-\frac{33}{20}\cdot\frac{10}{11}\)
\(=\frac{5}{2}-\frac{3}{2}\)
\(=1\)
\(\frac{5}{16}:0,125-\left(2\frac{1}{4}-0,6\right).\frac{10}{11}\)
\(\Rightarrow\frac{5}{16}:\frac{125}{1000}-\left(\frac{9}{4}-\frac{6}{10}\right).\frac{10}{11}\)
\(\Rightarrow\frac{5}{16}:\frac{1}{8}-\left(\frac{9}{4}-\frac{3}{5}\right).\frac{10}{11}\)
\(\Rightarrow\frac{5}{16}.\frac{8}{1}-\left(\frac{45}{20}-\frac{12}{20}\right).\frac{10}{11}\)
\(\Rightarrow\frac{5}{2}-\frac{33}{20}.\frac{10}{11}\)
\(\Rightarrow\frac{5}{2}-\frac{3.1}{2.1}\)
\(\Rightarrow\frac{5}{2}-\frac{3}{2}=\frac{2}{2}=1\)
\(bai1:a,\frac{3}{7}\cdot\frac{-5}{9}+\frac{4}{9}\cdot\frac{3}{7}-\frac{3}{7}\cdot\frac{8}{9}\)
\(< =>\frac{-15}{63}+\frac{12}{63}-\frac{24}{63}\)
\(< =>\frac{-15+12-24}{63}\)
\(< =>\frac{-3}{7}\)
\(b,1\frac{13}{15}\cdot0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)
\(< =>\frac{28}{15}\cdot\frac{3}{4}-\left(\frac{11}{20}+\frac{1}{4}\right):\frac{7}{5}\)
\(< =>\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)
\(< =>\frac{7}{5}-\frac{4}{7}\)
\(< =>\frac{29}{35}\)
\(bai2:\)
\(a,\frac{-3}{4}\cdot x-\frac{4}{10}=\frac{1}{5}\)
\(< =>\frac{-3}{4}\cdot x=\frac{1}{5}+\frac{4}{10}\)
\(< =>\frac{-3}{4}\cdot x=\frac{3}{5}\)
\(< =>x=\frac{3}{5}:\frac{-3}{4}\)
\(< =>x=\frac{-4}{5}\)
\(b,3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)
\(< =>3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)
\(< =>\left[3\left(x-\frac{1}{3}\right)\right]=\frac{1}{12}< =>x-\frac{1}{3}=\frac{1}{12}:3=\frac{1}{36}=>x=\frac{1}{36}+\frac{1}{3}=>x=\frac{13}{36}\)
\(< =>\left[\frac{1}{3}\cdot x\right]=\frac{1}{12}< =>x=\frac{1}{12}:\frac{1}{3}=>x=\frac{1}{4}\)
Bài 1:
a)\(\frac{3}{7}.\frac{-5}{9}+\frac{4}{9}.\frac{3}{7}-\frac{3}{7}.\frac{8}{9}\) b,\(1\frac{13}{15}.0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)
\(=\frac{3}{7}.(\frac{-5}{9}+\frac{4}{9}-\frac{8}{9})\) \(=\frac{28}{15}.\frac{3}{4}-\left(\frac{11}{20}+\frac{5}{20}\right):\frac{7}{5}\)
\(=\frac{3}{7}.\frac{-9}{9}\) \(=\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)
\(=\frac{-3}{7}\) \(=\frac{7}{5}-\frac{4}{7}\)
\(=\frac{29}{35}\)
Bài 2:
a)\(\frac{-3}{4}x-\frac{4}{10}=\frac{1}{5}\) b,\(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)
\(\frac{-3}{4}x\) \(=\frac{1}{5}+\frac{4}{10}\) \(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)
\(\frac{-3}{4}x\) \(=\frac{3}{5}\) \(\left(x.3-\frac{1}{3}.3\right)+\frac{1}{3}x=\frac{1}{12}\)
\(x\) \(=\frac{3}{5}:\frac{-3}{4}\) \(\left(x.3-1\right)+\frac{1}{3}x=\frac{1}{12}\)
\(x\) \(=\frac{4}{-5}\) \(x.\left(3+\frac{1}{3}\right)-1=\frac{1}{12}\)
\(x.\left(3+\frac{1}{3}\right)=\frac{1}{12}+1\)
\(x.\frac{10}{3}=\frac{13}{12}\)
\(x=\frac{13}{12}:\frac{10}{3}\)
\(x=\frac{13}{40}\)
a, \(\left(\dfrac{3}{8}+\dfrac{-3}{4}+\dfrac{7}{12}\right)\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\left(\dfrac{-3}{8}+\dfrac{7}{12}\right)\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\dfrac{5}{24}.\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\dfrac{25}{144}+\dfrac{1}{2}\)
\(=\dfrac{97}{144}\)
b, \(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+25\%\right):\dfrac{7}{5}\)
\(=\dfrac{28}{15}.0,75-\dfrac{4}{5}:\dfrac{7}{5}\)
\(=\dfrac{7}{5}-\dfrac{4}{7}\)
\(=\dfrac{29}{35}\)
Đặt \(A=\frac{2,75-2,2+\frac{11}{7}+\frac{11}{13}}{0,75-0,6+\frac{3}{7}+\frac{11}{13}}-x-\frac{1}{9}=\frac{2}{3}+\frac{2}{15}\frac{2}{35}+\frac{2}{63}\)
\(\Rightarrow A=\frac{0,55+\frac{11}{7}+\frac{11}{13}}{0,15+\frac{3}{7}+\frac{11}{13}}-x-\frac{1}{9}=\frac{42}{63}+\frac{14}{105}+\frac{6}{105}+\frac{2}{63}\)
\(\Rightarrow A=\frac{\frac{11}{20}+\frac{11}{7}+\frac{11}{13}}{\frac{3}{20}+\frac{3}{7}+\frac{11}{13}}-x-\frac{1}{9}=\frac{44}{63}+\frac{20}{105}\)
\(\Rightarrow A=\frac{\frac{1001}{1820}+\frac{2860}{1820}+\frac{1540}{1820}}{\frac{273}{1820}+\frac{780}{1820}+\frac{1540}{1820}}-x-\frac{1}{9}=\frac{44}{63}+\frac{4}{21}\)
\(\Rightarrow A=\frac{\frac{5401}{1820}}{\frac{2593}{1820}}-x-\frac{1}{9}=\frac{44}{63}+\frac{12}{63}\)
\(\Rightarrow A=\frac{5401}{1820}:\frac{2593}{1820}-x-\frac{1}{9}=\frac{56}{63}\)
\(\Rightarrow A=\frac{5401}{1820}.\frac{1820}{2593}-x-\frac{1}{9}=\frac{8}{9}\)
\(\Rightarrow A=\frac{5401}{2593}-x-\frac{1}{9}=\frac{8}{9}\)
\(\Rightarrow\frac{5401}{2593}-x=\frac{8}{9}+\frac{1}{9}\)
\(\Rightarrow\frac{5401}{2593}-x=1\)
\(\Rightarrow x=\frac{5401}{2593}-1\)
\(\Rightarrow x=\frac{2808}{2593}\)
Vậy \(x=\frac{2808}{2593}\)
Chúc bn học tốt
\(1\frac{13}{15}\) . 0,75 - ( 11/20 +25% ) :3/7
=\(\frac{28}{15}\) . 3/4 - ( 11/20 +25/100 ) :3/7
=28/15. 3/4- 4/5 : 3/7
=7/5 - 28/15
=-7/15
Thấy đúng thì bấm đúng cho mh nha!!!