\(a,13\dfrac{3}{5}-\left(8\dfrac{3}{5}-4\dfrac{3}{4}\right)\)

\(=\dfrac{68}{5}-\dfrac{43}{5}+\dfrac{19}{4}\)

\(=5+\dfrac{19}{4}\)

\(=\dfrac{20}{4}+\dfrac{19}{4}=\dfrac{39}{4}\)

\(------\)

\(b,11\dfrac{1}{4}-\left(2\dfrac{5}{7}+5\dfrac{1}{4}\right)\)

\(=\dfrac{45}{4}-\dfrac{19}{7}-\dfrac{21}{4}\)

\(=\left(\dfrac{45}{4}-\dfrac{21}{4}\right)-\dfrac{19}{7}\)

\(=6-\dfrac{19}{7}\)

\(=\dfrac{42}{7}-\dfrac{19}{7}=\dfrac{23}{7}\)

a: Ta có: \(\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{72}\right)\)

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

=0

a)

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

\(=\dfrac{13}{5}+\dfrac{49}{10}\\ =\dfrac{26}{10}+\dfrac{49}{10}\\ =\dfrac{15}{2}\)

b)

\(=\dfrac{52}{4}-\dfrac{11}{3}:\dfrac{7}{6}\)

\(=\dfrac{52}{4}-\dfrac{22}{7}\\ =\dfrac{69}{7}\)

a) $2\dfrac35 + 1\dfrac25 . 3\dfrac12$

$= \dfrac{13}5 + \dfrac75.\dfrac72$

$= \dfrac{26}{10} + \dfrac{49}{10}$

$=\dfrac{15}2$.

b) $4\dfrac34 - 3\dfrac23 : 1\dfrac16$

$= \dfrac{19}4 - \dfrac{11}3 : \dfrac76$

$= \dfrac{19}4 - \dfrac{11}3 . \dfrac67$

$= \dfrac{19}4 - \dfrac{22}7$

$= \dfrac{45}{28}$.

a) \(A=2^{2010}-2^{2009}-2^{2008}-...-2-1\)

\(A=2^{2010}\left(2^{2009}+2^{2008}+...+2+1\right)\)

Đặt \(\text{A = 1 + 2 + . . . + 2^{2008} + 2^{2009}}\)

\(\text{⇒ 2 A = 2 + 2 2 + . . + 2^{2010}}\)

⇒ \(A=2^{2010}-1\)

⇒ \(A=2^{2010}-\left(2^{2010}-1\right)\)

⇒ \(A=1\)

b) \(B=2072\)

c) \(\dfrac{4949}{19800}\)

Xin lỗi mình không có nhiều thời gian để giải thích trên đây á nên tạm gửi ảnh mình tạo nhé . Học tốt !

Lời giải:
a.

$x=\frac{-5}{6}-\frac{2}{3}=\frac{-3}{2}$

b.

$\frac{2}{3}x=\frac{1}{10}-\frac{1}{2}=\frac{-2}{5}$

$x=\frac{-2}{5}: \frac{2}{3}=\frac{-3}{5}$

c.

$\frac{7}{8}x=\frac{2}{9}-\frac{1}{3}=\frac{-1}{9}$
$x=\frac{-1}{9}: \frac{7}{8}=\frac{-8}{63}$

d.

$\frac{5}{7}: x=\frac{1}{6}-\frac{4}{5}=\frac{-19}{30}$

$x=\frac{5}{7}: \frac{-19}{30}=\frac{-150}{133}$

e.

$(\frac{2}{5}-1\frac{2}{3}):x=\frac{2}{5}+\frac{3}{5}=1$

$\frac{-19}{15}: x=1$

$x=\frac{-19}{15}:1 =\frac{-19}{15}$

f.

$(-\frac{3}{4}+x).2\frac{2}{3}=1$

$\frac{-3}{4}+x=1: 2\frac{2}{3}=\frac{3}{8}$

$x=\frac{3}{8}+\frac{3}{4}=\frac{9}{8}$

a) \(\dfrac{2}{3}\times\dfrac{1}{4}-\dfrac{1}{3}\times\dfrac{1}{2}=\dfrac{2}{12}-\dfrac{1}{6}=\dfrac{1}{6}-\dfrac{1}{6}=\dfrac{0}{6}=0\)

b) \(\dfrac{8}{5}\times\dfrac{1}{4}-\dfrac{2}{5}\times\dfrac{1}{2}-\dfrac{1}{2}\times\dfrac{1}{5}=\dfrac{8}{20}-\dfrac{2}{10}-\dfrac{1}{10}=\dfrac{4}{10}-\dfrac{2}{10}-\dfrac{1}{10}=\dfrac{4-2-1}{10}=\dfrac{1}{10}\)

CHÚC EM HỌC TỐT NHÉ

a. 1/6 - 1/6 = 0 (hoặc 2/12 - 1/6= 2/12 - 2/12 = 0)

b. 4/5 - 1/2 x ( 2/5 - 1/5 ) = 4/5 - 1/2 x 1/5 

                                        = 4/5 x 2/10

                                        = 4/25

A= 1/3 + 1/3^2 + ... + 1/3^8

3A= 3. (1/3+ 1/3^2+ ... + 1/3^8)

3A=1+ 1/3 + 1/3^2+ ... +1/3^7

=> 3A - A= (1 + 1/3 + 1/3^2 + ... + 1/3^7) - (1/3 + 1/3^2+ ... + 1/3^8)

=> 2A= 1 - 1/ 3^8

2A= 6560/6561

A= 6560/6561 : 2

A= 3280/6561

Tính P = Chúc bạn học tốt nhé

P=1+1/3+1/6+1/10+…..+1/2021×2022÷2

P/2=1/2+1/6+1/12+1/20+…..+1/2021×2022

P/2=1/1×2+1/2×3+1/3×4+…….+1/2021×2022

P/2=1-1/2+1/2-1/3+1/3-1/4+….+1/2021-1/2022=1-1/2022=2021/2022

P=2021/1011

Chúc bn học tốt

\(E=\dfrac{\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{2002}-1\right)\left(\dfrac{1}{2003}-1\right)}{\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot...\cdot\dfrac{9999}{10000}}\)

\(=\dfrac{\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2002}\right)\left(1-\dfrac{1}{2003}\right)}{\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{100^2}\right)}\)

\(=\dfrac{\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2002}\right)\left(1-\dfrac{1}{2003}\right)}{\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{100}\right)\left(1+\dfrac{1}{100}\right)}\)

\(=\dfrac{\dfrac{100}{101}\cdot\dfrac{101}{102}\cdot...\cdot\dfrac{2002}{2003}}{\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1+\dfrac{1}{100}\right)}\)

\(=\dfrac{100}{2003}:\left(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\right)\)

\(=\dfrac{100}{2003}:\left(\dfrac{101}{2}\right)=\dfrac{100}{2003}\cdot\dfrac{2}{101}=\dfrac{200}{202303}\)