a) \(\left(1^2+2^2+3^2+....+2012^2\right).\left(91-273:3\right)\)

\(=\left(1^2+2^2+3^2+...+2012^2\right).\left(91-91\right)\)

\(=0\)

b) \(\left(-284\right).172+\left(-284\right).\left(-72\right)=\left(-284\right).\left(172+-72\right)\)

                                                                             \(=\left(-284\right).100\)

                                                                               \(=-28400\)

c) \(\frac{1}{5}+\frac{-1}{6}+\frac{1}{7}+\frac{-1}{8}+\frac{1}{9}+\frac{1}{8}+\frac{-1}{7}+\frac{1}{6}+\frac{-1}{5}\)

\(=\left(\frac{1}{5}+\frac{-1}{5}\right)+\left(\frac{1}{6}+\frac{-1}{6}\right)+\left(\frac{1}{7}+\frac{-1}{7}\right)+\left(\frac{1}{8}+\frac{-1}{8}\right)+\frac{1}{9}\)

\(=0+0+0+0+\frac{1}{19}\)

= 0

c) Mình nhầm: \(\frac{1}{9}\)

Gọi tổng dãy số hạng trên là A

A = 1 + \(\frac{1}{2}\)+ 1 + \(\frac{1}{6}\)+ 1 + \(\frac{1}{12}\)+ ... + 1 + \(\frac{1}{90}\)+ 1 + \(\frac{1}{110}\)

Mà từ \(\frac{1}{2}\)đén \(\frac{1}{110}\) có 10 số

A = 1 x 10 + \(\frac{1}{2}\)+( \(\frac{1}{2}\)- \(\frac{1}{3}\)) + ( \(\frac{1}{3}\)-\(\frac{1}{4}\)) + (\(\frac{1}{4}\)-\(\frac{1}{5}\)) + ... + \(\frac{1}{11}\) 

A = 10 + \(\frac{1}{2}\)+ \(\frac{1}{2}\)+ \(\frac{1}{11}\)= \(\frac{112}{11}\)

a) $\frac{4}{{25}}:\frac{4}{3} = \frac{4}{{25}} \times \frac{3}{4} = \frac{3}{{25}}$ 

b) $\frac{3}{{14}}:\frac{6}{7} = \frac{3}{{14}} \times \frac{7}{6} = \frac{{3 \times 7}}{{14 \times 6}} = \frac{{3 \times 7}}{{7 \times 2 \times 3 \times 2}} = \frac{1}{4}$

c) $\frac{{12}}{{15}}:2 = \frac{{12}}{{15}} \times \frac{1}{2} = \frac{{12 \times 1}}{{15 \times 2}} = \frac{{6 \times 2 \times 1}}{{15 \times 2}} = \frac{6}{{15}}$           

d) $\frac{{21}}{8}:6 = \frac{{21}}{8} \times \frac{1}{6} = \frac{{21 \times 1}}{{8 \times 6}} = \frac{{7 \times 3 \times 1}}{{8 \times 3 \times 2}} = \frac{7}{{16}}$

a: \(=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}+\dfrac{-\dfrac{1}{4}\cdot\dfrac{-2}{3}-\dfrac{3}{4}:\dfrac{1}{6}}{\dfrac{3}{2}\cdot\left(\dfrac{-2}{3}-\dfrac{3}{4}\cdot\dfrac{-2}{3}\right)}\)

\(=\dfrac{3}{4}+\dfrac{\dfrac{2}{12}-\dfrac{9}{2}}{\dfrac{3}{2}\cdot\dfrac{-1}{6}}=\dfrac{3}{4}+\dfrac{-13}{3}:\dfrac{-3}{12}\)

\(=\dfrac{3}{4}+\dfrac{13}{3}\cdot\dfrac{12}{3}=\dfrac{3}{4}+\dfrac{156}{9}=\dfrac{217}{12}\)

b: \(A=158\left(\dfrac{12\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}{4\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}:\dfrac{5\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}{6\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}\right)\cdot\dfrac{50550505}{711711711}\)

\(=158\cdot\left(3\cdot\dfrac{6}{5}\right)\cdot\dfrac{50550505}{711711711}\)

\(\simeq40.39\)

chịu thôi

B=64,8

\(a)\) \(A=\frac{5\left(2^2.3^2\right)^9.\left(2^2\right)^6-2\left(2^2.3\right)^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)

\(A=\frac{2^{30}.3^{18}.5-2^{29}.3^{18}}{2^{28}.3^{18}.5-2^{29}.3^{18}.7}\)

\(A=\frac{2^{29}.3^{18}\left(2.5-1\right)}{2^{28}.3^{18}\left(5-2.7\right)}\)

\(A=\frac{2\left(10-1\right)}{5-14}\)

\(A=\frac{2.9}{-9}\)

\(A=-2\)

Vậy \(A=-2\)

\(b)\) \(B=81.\left[\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{6}{13}+\frac{6}{169}+\frac{6}{91}}\right].\frac{158158158}{711711711}\)

\(B=81.\left[\frac{12\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{5\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{6\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\right].\frac{158158158}{711711711}\)

\(B=81.\left[\frac{12}{4}:\frac{5}{6}\right].\frac{2}{9}\)

\(B=81.\frac{18}{5}.\frac{2}{9}\)

\(B=\frac{324}{5}\)

Vậy \(B=\frac{324}{5}\)

Chúc bạn học tốt ~ ( mỏi tay qué >_< )