\(\left(-\dfrac{2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{3}+\dfrac{4}{7}\right)+\dfrac{4}{5}\\ =-\dfrac{5}{21}:\dfrac{4}{5}+\dfrac{5}{21}\\ =\left(-\dfrac{5}{21}+\dfrac{5}{21}\right):\dfrac{4}{5}\\ =0:\dfrac{4}{5}\\ =0.\)

Sửa cho mk dòng đầu là :4/5 và dòng tiếp theo mk thiếu :4/5

\(\dfrac{5}{6}+\dfrac{1}{3}=\dfrac{5}{6}+\dfrac{2}{6}=\dfrac{5+2}{6}=\dfrac{7}{6}\)

\(A=\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+..+\dfrac{1}{44.49}\right)\left(\dfrac{1-3-5-7-..-49}{89}\right)\\ A=\dfrac{1}{5}\left(\dfrac{5}{4.9}+\dfrac{5}{9.14}+..+\dfrac{5}{44.49}\right)\left(\dfrac{1-3-5-7-...-49}{89}\right)\\ A=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\left(\dfrac{1-3-5-7-...-49}{89}\right)\)

\(A=\dfrac{9}{196}\left(\dfrac{1-3-5-7-...-49}{89}\right)\)

Ta đặt: \(P=1-3-5-7-...-49\\ =1-\left(3+5+7+..+49\right)\\ =1-624\\ =-623\\ \Rightarrow\dfrac{9}{196}.-\dfrac{623}{89}=-\dfrac{9}{28}.\)

Ta có: $A=\left(\genfrac{}{}{0ex}{}{1}{4\cdot 9}+\genfrac{}{}{0ex}{}{1}{9\cdot 14}+\genfrac{}{}{0ex}{}{1}{14\cdot 19}+...+\genfrac{}{}{0ex}{}{1}{44\cdot 49}\right)\cdot \genfrac{}{}{0ex}{}{1-3-5-7-...-49}{89}$

$\iff A=\genfrac{}{}{0ex}{}{1}{5}\cdot \left(\genfrac{}{}{0ex}{}{5}{4\cdot 9}+\genfrac{}{}{0ex}{}{5}{9\cdot 14}+\genfrac{}{}{0ex}{}{5}{14\cdot 19}+...+\genfrac{}{}{0ex}{}{5}{44\cdot 49}\right)\cdot \genfrac{}{}{0ex}{}{1-3-5-7-...-49}{89}$

$\iff A=\genfrac{}{}{0ex}{}{1}{5}\cdot \left(\genfrac{}{}{0ex}{}{1}{4}-\genfrac{}{}{0ex}{}{1}{9}+\genfrac{}{}{0ex}{}{1}{9}-\genfrac{}{}{0ex}{}{1}{14}+\genfrac{}{}{0ex}{}{1}{14}-\genfrac{}{}{0ex}{}{1}{19}+...+\genfrac{}{}{0ex}{}{1}{44}-\genfrac{}{}{0ex}{}{1}{49}\right)\cdot \genfrac{}{}{0ex}{}{1-3-5-7-...-49}{89}$

$\iff A=\genfrac{}{}{0ex}{}{1}{5}\cdot \left(\genfrac{}{}{0ex}{}{1}{4}-\genfrac{}{}{0ex}{}{1}{49}\right)\cdot \genfrac{}{}{0ex}{}{1-3-5-7-...-49}{89}$

$\iff A=\genfrac{}{}{0ex}{}{1}{5}\cdot \left(\genfrac{}{}{0ex}{}{49-4}{4\cdot 49}\right)\cdot \genfrac{}{}{0ex}{}{1-3-5-7-...-49}{89}$

$\iff A=\genfrac{}{}{0ex}{}{1}{5}\cdot \genfrac{}{}{0ex}{}{45}{196}\cdot \genfrac{}{}{0ex}{}{1-3-5-7-...-49}{89}$

$\iff A=\genfrac{}{}{0ex}{}{9}{196}\cdot \genfrac{}{}{0ex}{}{1-3-5-7-...-49}{89}$

$\iff A=\genfrac{}{}{0ex}{}{9}{196}\cdot \genfrac{}{}{0ex}{}{-623}{89}=-\genfrac{}{}{0ex}{}{9}{28}$
 

\(a,\left(\dfrac{31}{35}-\dfrac{4}{7}\right)\times\dfrac{8}{7}:2\\ =\left(\dfrac{31}{35}-\dfrac{4\times5}{35}\right)\times\dfrac{8}{7}:2\\ =\dfrac{11}{35}\times\dfrac{8}{7}:2\\ =\dfrac{88}{245}:2\\ =\dfrac{44}{245}\\ b,\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times\left(1-\dfrac{1}{5}\right)\\ =\left(\dfrac{2-1}{2}\right)\times\left(\dfrac{3-1}{3}\right)\times\left(\dfrac{4-1}{4}\right)\times\left(\dfrac{5-1}{5}\right)\\ =\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\\ =\dfrac{1}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\\ =\dfrac{1}{4}\times\dfrac{4}{5}=\dfrac{1}{5}\)

a, ( \(\dfrac{31}{35}\) - \(\dfrac{4}{7}\)) \(\times\) \(\dfrac{8}{7}\): 2

= \(\left(\dfrac{31}{35}-\dfrac{20}{35}\right)\) \(\times\) \(\dfrac{8}{7}\) : 2

= \(\dfrac{11}{35}\) \(\times\) \(\dfrac{8}{7}\) \(\times\) \(\dfrac{1}{2}\)

= \(\dfrac{44}{35}\) \(\times\) \(\dfrac{4}{7}\)

= \(\dfrac{44}{245}\)

b, ( 1 - \(\dfrac{1}{2}\)) \(\times\) ( 1 - \(\dfrac{1}{3}\)) \(\times\) ( 1 - \(\dfrac{1}{4}\)) \(\times\) ( 1 - \(\dfrac{1}{5}\))

= \(\dfrac{1}{2}\) \(\times\) \(\dfrac{2}{3}\) \(\times\) \(\dfrac{3}{4}\) \(\times\) \(\dfrac{4}{5}\)

= \(\dfrac{1}{5}\) \(\times\) \(\dfrac{2\times3\times4}{2\times3\times4}\)

= \(\dfrac{1}{5}\)

3/4 x 4/5 x 5/3 x 8/3 = 8/3

\(\dfrac{3}{4}:\dfrac{5}{4}:\dfrac{3}{5}:\dfrac{3}{8}=\dfrac{3}{5}:\dfrac{3}{5}:\dfrac{3}{8}=1:\dfrac{3}{8}=\dfrac{8}{3}\)

a: Ta có: \(\dfrac{2}{\sqrt{3}+1}+\dfrac{2}{2-\sqrt{3}}\)

\(=\sqrt{3}-1+2+\sqrt{3}\)

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

b: Ta có: \(\dfrac{4}{\sqrt{5}+2}+\dfrac{2}{3+\sqrt{5}}\)

\(=4\sqrt{5}-8+\dfrac{3}{2}-\dfrac{\sqrt{5}}{2}\)

\(=-\dfrac{13}{2}+\dfrac{7}{2}\sqrt{5}\)

\(a,\dfrac{7}{12}+\dfrac{3}{4}\times\dfrac{2}{9}=\dfrac{7}{12}+\dfrac{1}{6}=\dfrac{7}{12}+\dfrac{2}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)

\(b,\dfrac{8}{9}-\dfrac{4}{15}:\dfrac{2}{5}=\dfrac{8}{9}-\dfrac{4}{15}\times\dfrac{5}{2}=\dfrac{8}{9}-\dfrac{2}{3}=\dfrac{8}{9}-\dfrac{6}{9}=\dfrac{2}{9}\)

Đáp án a là 0,75 

Đáp án b: là 0.86222222222

\(\text{∘ Ans}\)

\(\downarrow\)

\(A=\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-...-\dfrac{1}{6}-\dfrac{1}{2}\)

`=`\(\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)\)

`=`\(\dfrac{8}{9}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}\right)\)

`=`\(\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)

`=`\(\dfrac{8}{9}-\left[1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-...-\left(\dfrac{1}{8}-\dfrac{1}{8}\right)-\dfrac{1}{9}\right]\)

`=`\(\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)\)

`=`\(\dfrac{8}{9}-\dfrac{8}{9}=0\)

Vậy, ` A = 0.`

\(A=\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)=\)

\(A=\dfrac{8}{9}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{7.8}+\dfrac{1}{8.9}\right)=\)

\(A=\dfrac{8}{9}-\left(\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{9-8}{8.9}\right)\)

\(A=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+..+\dfrac{1}{8}-\dfrac{1}{9}\right)\)

\(A=\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)=0\)

a) \(4,5:\left[\left(\dfrac{9-10}{6}\right)-\dfrac{9}{5}+\dfrac{12}{5}\right]-\dfrac{1}{7}\)

\(=4,5:\left(\dfrac{-1}{6}-\dfrac{-3}{5}\right)-\dfrac{1}{7}\)

=\(4,5:\left(\dfrac{-5+18}{30}\right)-\dfrac{1}{7}\)

=\(4,5:\dfrac{13}{30}-\dfrac{1}{7}\)=\(\dfrac{135}{13}-\dfrac{1}{7}=\dfrac{932}{91}\)

b) \(\dfrac{13}{3}:\left(\dfrac{1}{4}+\dfrac{5}{4}\right)-\dfrac{20}{3}\)

=\(\dfrac{13}{3}.\dfrac{2}{3}-\dfrac{20}{3}\)=\(\dfrac{26}{9}-\dfrac{20}{3}=\dfrac{26}{9}-\dfrac{60}{9}=\dfrac{-34}{9}\)

c) \(5.\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+.....+\dfrac{1}{91.94}\right)\)

\(=5.\left[\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{91}-\dfrac{1}{94}\right)\right]\)

\(=5.\left[\dfrac{1}{3}.\left(1-\dfrac{1}{94}\right)\right]\)

=\(5.\left(\dfrac{1}{3}.\dfrac{93}{94}\right)\)

\(=5.\dfrac{31}{94}=\dfrac{155}{94}\)

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