tập trung sẽ bt làm

bài 6 \(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+\ldots+\frac{2}{80\times82}+\frac{2}{82\times84}\)

= \(\frac12-\frac14+\frac14-\frac16+\frac16-\cdots+\frac{1}{80}-\frac{1}{82}+\frac{1}{82}-\frac{1}{84}\)

= \(\frac12-\left(\frac14-\frac14\right)-\left(\frac16-\frac16\right)-\cdots-\left(\frac{1}{82}-\frac{1}{82}\right)-\frac{1}{84}\)

=\(\frac12-0-0-\frac{1}{84}\)

=\(\frac12\) - \(\frac{1}{84}\)

=\(\frac{42}{84}-\frac{1}{84}\)

=\(\frac{41}{84}\)

Bài 7

a)\(\frac{19}{4}=4\frac34\)

b)\(\frac{27}{5}=5\frac25\)

c)\(\frac{56}{8}=7\frac08\)

a) (1,5 . 1,9 - x - 0,5) : 0,25 = 7,5 : 0,125

=> (2,85 - x - 0,5) : 0,25 = 60

=> (2,85 - 0,5) - x = 60 . 0,25

=> 2,35 - x = 15

=> x = 2,35 - 15

=> x = -12,65

Vậy x = -12,65

b) \(1-\left(5\frac{2}{9}+x-7\frac{7}{18}\right)\div2\frac{1}{6}=0\)

\(\Rightarrow\left(5\frac{2}{9}-7\frac{7}{18}+x\right)\div2\frac{1}{6}=1-0\)

\(\Rightarrow\left(\frac{47}{9}-\frac{133}{18}+x\right)\div2\frac{1}{6}=1\)

\(\Rightarrow\frac{-13}{6}+x=2\frac{1}{6}\)

\(\Rightarrow x=2\frac{1}{6}-\frac{-13}{6}\)

\(\Rightarrow x=\frac{13}{6}+\frac{13}{6}\)

\(\Rightarrow x=\frac{26}{6}\)

\(\Rightarrow x=\frac{13}{3}\)

Vậy \(x=\frac{13}{3}\)

c) \(35\left(2\frac{1}{5}-x\right)=32\)

\(\Rightarrow2\frac{1}{5}-x=32\div35\)

\(\Rightarrow\frac{11}{5}-x=\frac{32}{35}\)

\(\Rightarrow x=\frac{11}{5}-\frac{32}{35}\)

\(\Rightarrow x=\frac{9}{7}\)

Vậy \(x=\frac{9}{7}\)

d) \(\frac{4}{3}+\left(x\div2\frac{2}{3}-0,5\right).1\frac{35}{55}=0,6\)

\(\Rightarrow\left(x\div\frac{8}{3}-\frac{1}{2}\right).\frac{18}{11}=\frac{3}{5}-\frac{4}{3}\)

\(\Rightarrow\left(x\div\frac{8}{3}-\frac{1}{2}\right).\frac{18}{11}=\frac{-11}{15}\)

\(\Rightarrow x\div\frac{8}{3}-\frac{1}{2}=\frac{-11}{15}\div\frac{18}{11}\)

\(\Rightarrow x\div\frac{8}{3}-\frac{1}{2}=\frac{-121}{270}\)

\(\Rightarrow x\div\frac{8}{3}=\frac{-121}{270}+\frac{1}{2}\)

\(\Rightarrow x\div\frac{8}{3}=\frac{7}{135}\)

\(\Rightarrow x=\frac{7}{135}.\frac{8}{3}\)

\(\Rightarrow x=\frac{56}{405}\)

Vậy \(x=\frac{56}{405}\)

e) \(1\frac{1}{3}.2\frac{2}{4}\div\frac{5}{6}.1\frac{1}{11}=11-5\div x\)

\(\Rightarrow\frac{4}{3}.\frac{5}{2}\div\frac{5}{6}.\frac{12}{11}=11-5\div x\)

\(\Rightarrow\frac{10}{3}\div\frac{5}{6}.\frac{12}{11}=11-5\div x\)

\(\Rightarrow4.\frac{12}{11}=11-5\div x\)

\(\Rightarrow11-5\div x=\frac{48}{11}\)

\(\Rightarrow5\div x=11-\frac{48}{11}\)

\(\Rightarrow5\div x=\frac{73}{11}\)

\(\Rightarrow x=5\div\frac{73}{11}\)

\(\Rightarrow x=\frac{55}{73}\)

Vậy \(x=\frac{55}{73}\)

a) (1,5 * 1,9 - x - 0,5) : 0,25 = 7,5 : 0,125

(2,85 - x - 0,5) : 0,25 = 60

(2,85 - x - 0,5) = 60 x 0,25

(2,85 - x - 0,5) = 15

2,35 - x = 15

x = 2,35 - 15

x = -12,65

a. \(\frac{-3}{2}-2x+\frac{3}{4}=-22\)2

=> \(-2x=-22+\frac{3}{2}-\frac{3}{4}\)

=> \(-2x=\frac{-85}{4}\)

=> \(x=\frac{-85}{4}:\left(-2\right)\)

=> \(x=\frac{85}{8}\)

b. \(\left(\frac{-2}{3}x-\frac{3}{5}\right).\left(\frac{3}{-2}-\frac{10}{3}\right)=\frac{2}{5}\)

=> \(\left(\frac{-2}{3}x-\frac{3}{5}\right).\frac{-29}{6}=\frac{2}{5}\)

=> \(\frac{-2}{3}x-\frac{3}{5}=\frac{2}{5}:\left(\frac{-29}{6}\right)\)

=> \(\frac{-2}{3}x-\frac{3}{5}=\frac{-12}{145}\)

=> \(\frac{-2}{3}x=\frac{-12}{145}+\frac{3}{5}\)

=> \(\frac{-2}{3}x=\frac{15}{29}\)

=> x = \(\frac{15}{29}:\frac{-2}{3}\)

=> x = \(\frac{-45}{58}\)

a)

\(x.\frac{7}{9}=\frac{2}{3}+2\frac{1}{2}\)

\(x.\frac{7}{9}=\frac{19}{6}\)

\(x=\frac{19}{6}:\frac{7}{9}\)

\(x=\frac{57}{14}\)

b) \(\frac{5}{7}+x:\frac{9}{4}=\frac{4}{3}\)

\(x:\frac{9}{4}=\frac{4}{3}-\frac{5}{7}\)

\(x:\frac{9}{4}=\frac{13}{21}\)

\(x=\frac{13}{21}.\frac{9}{4}\)

\(x=\frac{39}{28}\)

A,27/14

a) \(\frac{11}{15}+\frac{5}{7}+\frac{2}{7}+\frac{4}{15}=\left(\frac{11}{15}+\frac{4}{15}\right)+\left(\frac{5}{7}+\frac{2}{7}\right)\)

\(=2\)

b) \(\frac{5}{9}\times\frac{1}{2}\times\frac{5}{9}\times\frac{6}{4}=\frac{25}{81}\times\frac{3}{4}=\frac{25}{108}\)

c) \(\frac{7}{8}\div\frac{1}{2}+\frac{9}{8}\div\frac{1}{2}=\left(\frac{7}{8}+\frac{9}{8}\right)\div\frac{1}{2}\)

\(=2\div\frac{1}{2}=4\)

d) \(\frac{17}{10}+\frac{1}{2}-\frac{7}{10}=\left(\frac{17}{10}-\frac{7}{10}\right)+\frac{1}{2}\)

\(=1+\frac{1}{2}=\frac{3}{2}\)

a) \(\frac{11}{15}+\frac{5}{7}+\frac{2}{7}+\frac{4}{15}\)

\(=\left(\frac{11}{15}+\frac{4}{15}\right)+\left(\frac{5}{7}+\frac{2}{7}\right)\)

\(=1+1\)

\(=2\)

b) \(\frac{5}{9}.\frac{1}{2}.\frac{5}{9}.\frac{6}{4}\)

\(=\left(\frac{5}{9}\right)^2\left(\frac{1}{2}.\frac{6}{4}\right)\)

\(=\frac{25}{81}.\frac{3}{4}\)

\(=\frac{25}{108}\)

c) \(\frac{7}{8}:\frac{1}{2}+\frac{9}{8}:\frac{1}{2}\)

\(=\frac{7}{8}.2+\frac{9}{8}.2\)

\(=2\left(\frac{7}{8}+\frac{9}{8}\right)\)

\(=2.\frac{16}{8}\)

\(=2.2\)

\(=4\)

d) \(\frac{17}{10}+\frac{1}{2}-\frac{7}{10}\)

\(=\left(\frac{17}{10}-\frac{7}{10}\right)+\frac{1}{2}\)

\(=1+\frac{1}{2}\)

\(=\frac{2}{2}+\frac{1}{2}\)

\(=\frac{3}{2}\)