1)A\(\frac{-2}{4}\)+\(\frac{2}{7}\)_\(\frac{5}{28}\)
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a, \(A=\frac{2}{5}+\frac{-1}{6}-\frac{3}{4}-\frac{-2}{3}\)
\(A=\left(\frac{2}{5}-\frac{3}{4}\right)+\left(\frac{-1}{6}-\frac{-2}{3}\right)\)
\(A=\left(\frac{8}{20}-\frac{15}{20}\right)+\left(\frac{-3}{18}-\frac{-12}{18}\right)\)
\(A=\frac{-7}{20}+\frac{1}{2}\)
\(\Rightarrow A=\frac{-7}{20}+\frac{10}{20}=\frac{3}{20}\)
b, \(B=\frac{7}{10}-\frac{-3}{4}+\frac{-5}{6}-\frac{1}{5}+\frac{-2}{3}\)
\(B=\left(\frac{7}{10}-\frac{1}{5}\right)+\left(\frac{-5}{6}+\frac{-2}{3}\right)-\frac{-3}{4}\)
\(B=\left(\frac{7}{10}-\frac{2}{10}\right)+\left(\frac{-5}{6}+\frac{-4}{6}\right)-\frac{-3}{4}\)
\(B=\frac{1}{2}+\frac{-3}{2}-\frac{-3}{4}\)
\(B=\frac{2}{4}+\frac{-6}{4}-\frac{-3}{4}\)
\(\Rightarrow B=\frac{2+-6+3}{4}=\frac{-1}{4}\)
c, \(C=\frac{\left(\frac{1}{2}-0,75\right)\times\left(0,2-\frac{2}{5}\right)}{\frac{5}{9}-1\frac{1}{12}}\)
\(C=\frac{\left(\frac{1}{2}-\frac{3}{4}\right)\times\left(\frac{1}{5}-\frac{2}{5}\right)}{\frac{5}{9}-\frac{1\times12+1}{12}}\)
\(C=\frac{\left(\frac{2}{4}-\frac{3}{4}\right)\times\left(\frac{-1}{5}\right)}{\frac{5}{9}-\frac{13}{12}}\)
\(C=\frac{\left(\frac{-1}{4}\right)\times\left(\frac{-1}{5}\right)}{\frac{60}{108}-\frac{117}{108}}\)
\(C=\frac{\frac{1}{20}}{\frac{-19}{36}}=\frac{1}{20}\div\frac{-19}{36}=\frac{1}{20}\times\frac{36}{-19}\)
\(\Rightarrow C=\frac{36}{-380}=\frac{-9}{95}\)
d, \(D=\frac{\frac{2}{3}+\frac{2}{7}-\frac{1}{4}}{-1-\frac{3}{7}+\frac{3}{28}}\)
\(D=\frac{\frac{56}{84}+\frac{24}{84}-\frac{21}{84}}{\frac{-10}{7}+\frac{3}{28}}\)
\(D=\frac{\frac{59}{84}}{\frac{-40}{28}+\frac{2}{28}}=\frac{59}{84}\div\frac{-37}{28}=\frac{59}{84}\times\frac{28}{-37}\)
\(\Rightarrow D=\frac{1652}{-3108}=\frac{-59}{111}\)
a) \(\frac{14}{21}+1-\left|\frac{1}{3}-1\right|\)
\(=\frac{2}{3}+1-\frac{2}{3}\)
\(=1+\left(\frac{2}{3}-\frac{1}{3}\right)\)
\(=1\)
b) \(\frac{1}{3}-\left|\frac{-1}{4}+\frac{5}{6}\right|-\left|\frac{-7}{12}\right|\)
\(=\frac{1}{3}-\frac{7}{12}-\frac{7}{12}\)
\(=-\frac{5}{6}\)
Bài 2 :
1) \(x-70=-45\) 2) \(\frac{4}{7}:x=\frac{12}{28}\)
\(\Rightarrow\) \(x=-45+70\) \(\Rightarrow x=\frac{4}{7}:\frac{12}{28}\)
\(\Rightarrow\) \(x=25\) \(\Rightarrow x=\frac{4}{3}\)
Vậy \(x=25\) Vậy \(x=\frac{4}{3}\)
3) Giống câu c) ở bài 1
4) \(x-50=-35\) 5) \(\frac{4}{7}.x=\frac{11}{18}\)
\(\Rightarrow x=-35+50\) \(\Rightarrow x=\frac{11}{28}:\frac{4}{7}\)
\(\Rightarrow x=15\) \(\Rightarrow x=\frac{77}{72}\)
Vậy \(x=15\) Vậy \(x=\frac{77}{72}\)
6) \(\left(\frac{2}{3}x+2,5\right):2\frac{2}{6}=6\)
\(\Rightarrow\)\(\left(\frac{2}{3}x+2,5\right):\frac{14}{6}=6\)
\(\Rightarrow\) \(\frac{2}{3}x+2,5=6.\frac{14}{6}\)
\(\Rightarrow\frac{2}{3}x+2,5=14\)
\(\Rightarrow\frac{2}{3}x=\frac{23}{2}\)
\(\Rightarrow x=\frac{23}{2}:\frac{2}{3}\)
\(\Rightarrow x=\frac{69}{4}\)
Vậy \(x=\frac{69}{4}\)
Bài 1:
1) \(\frac{7}{5}+\frac{-8}{5}=-\frac{1}{5}\)
2) \(-\frac{6}{5}.\frac{15}{24}=-\frac{3}{4}\)
3) \(\left(\frac{2}{3}+1,5\right)-3,5:7\frac{1}{2}=\)\(\frac{13}{6}-\frac{7}{15}=\frac{17}{10}\)
4) \(\frac{5}{8}-\frac{-7}{9}=\frac{5}{8}+\frac{7}{9}=\frac{101}{72}\)
5)\(\frac{-7}{3}.\frac{12}{28}=-1\)
Ta có : \(\left(2^2:\frac{4}{3}-\frac{1}{2}\right).\frac{6}{5}-17\)
=\(=\left(4.\frac{3}{4}-\frac{1}{2}\right).\frac{6}{5}-17\)
\(=\frac{5}{2}.\frac{6}{5}-17\)
\(=3-17=-14\)
Tụi quá mới lớp 5 thui
a: \(=\left(-\dfrac{25}{140}+\dfrac{245}{140}+\dfrac{32}{140}\right)\cdot\dfrac{-69}{20}\)
\(=\dfrac{252}{140}\cdot\dfrac{-69}{20}\)
\(=\dfrac{9}{5}\cdot\dfrac{-69}{20}=\dfrac{-621}{100}\)
b: \(=\left(6-2-\dfrac{4}{5}\right)\cdot\dfrac{25}{8}-\dfrac{8}{5}\cdot4\)
\(=\dfrac{16}{5}\cdot\dfrac{25}{8}-\dfrac{32}{5}=\dfrac{18}{5}\)
c: \(=\left(\dfrac{2}{24}+\dfrac{18}{24}+\dfrac{14}{24}\right):\dfrac{-17}{8}\)
\(=\dfrac{34}{24}\cdot\dfrac{-8}{17}=\dfrac{-1}{3}\cdot2=-\dfrac{2}{3}\)
Bài 2
a. \(-1\frac{2}{3}-|2x-1|:\frac{3}{5}=-2\)
\(|2x-1|:\frac{3}{5}=\frac{5}{3}-2\)
\(|2x-1|:\frac{3}{5}=-\frac{1}{3}\)
\(|2x-1|=-\frac{1}{5}\)
Vì giá trị tuyệt đối luôn \(\ge0\)với mọi x
mà \(-\frac{1}{5}< 0\)
=> \(x\in\varnothing\)
\(A=\frac{-2}{4}+\frac{2}{7}-\frac{5}{28}\)
\(A=\frac{-14}{28}+\frac{8}{28}-\frac{5}{28}\)
\(A=\frac{-14+8-5}{28}\)
\(A=\frac{-11}{28}\)
~Study well~
#KSJ