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: \(\frac{1}{9}\)

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

 =   \(\left(1^2+2^2+3^2+...+2012^2\right).0\)

=                               \(0\)

b) (-284) . 172 + (-284) . (-72)

= (-284) . [172 + (-72)]

= (-284) . 100

= -28400

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

= \(\frac{1}{9}\)

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 = 5.(2² . 3² )⁹. (2²)⁶ - 2. (2².3)¹⁴.3⁴ / 5.2²⁸.3¹⁸ - 7.2²⁹.3¹⁸

A = 5.2¹⁸.3¹⁸.2¹² - 2. 2^28. 3¹⁴. 3⁴ / 5. 2²⁸. 3¹⁸ - 7. 2²⁹. 3¹⁸

A = 5.2³⁰.3¹⁸ - 2²⁹. 3¹⁸ /  5.2 ²⁸.3¹⁸ - 7.2²⁹. 3¹⁸

A = 2²⁹. 3¹⁸(5.2-1) / 2²⁸. 3¹⁸ (5 -7.2)

A = 2.9 / -9 = -2

Vậy A = -2

b,

B =81.( 12 - 12/7- 12/289 - 12/85 /  4 - 4/7 - 4/289 - 4/85 :

5 + 5/13 + 5/169 + 5/91 /  6 + 6/13 + 6/169 + 5/91 ) x 158158158

B = 81.[ 12.( 1 - 1/7 - 1/289 - 1/85) / 4.( 1 - 1/7 - 1/289 - 1/85 :

5.(1 + 1/13 + 1/169 + 1/91) / 6.(1+ 1/13 +1/169 + 1/91)] . 2/9

B = 81.[ 3 : 5/6 ] .2/9

B = 81.18/5 . 2/9

B = 1458/5 . 2/9

B = 324/5.

Vậy B = 324/5

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