Tham khảo:

\(\dfrac{4}{7}-\dfrac{1}{3}< \dfrac{1}{2}-\dfrac{2}{9}\)

\(\dfrac{125}{100}-\dfrac{48}{64}< \dfrac{6}{7}-\dfrac{14}{28}\)

\(\dfrac{7}{5}-\dfrac{22}{25}=\dfrac{1}{5}+\dfrac{4}{10}\)\(\dfrac{45}{75}+\dfrac{22}{33}>\dfrac{4}{5}+\dfrac{4}{6}\)

8:

\(A=\dfrac{20^{10}-1+2}{20^{10}-1}=1+\dfrac{2}{20^{10}-1}\)

\(B=\dfrac{20^{10}-3+2}{20^{10}-3}=1+\dfrac{2}{20^{10}-3}\)

mà 20^10-1>20^10-3

nên A<B

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`Answer:`

\(A=\left(1^{100}+2^{100}+...+10^{100}\right)\left(5^{10}-25^5\right)\)

\(=\left(1^{100}+2^{100}+...+10^{100}\right)[5^{10}-\left(5^2\right)^5]\)

\(=\left(1^{100}+2^{100}+...+10^{100}\right)\left(5^{10}-5^{10}\right)\)

\(=\left(1^{100}+2^{100}+...+10^{100}\right).0\)

\(=0\)

\(B=\left(2^5+3^5+4^5\right)\left(1^2+2^2+...+100^2\right)\left(4^{10}-2^{20}\right)\)

\(=\left(2^5+3^5+4^5\right)\left(1^2+2^2+...+100^2\right)\left(2^{2.10}-2^{20}\right)\)

\(=\left(2^5+3^5+4^5\right)\left(1^2+2^2+...+100^2\right).0\)

\(=0\)

\(5^{200}=\left(5^2\right)^{100}=25^{100}\)

\(3< 25=>3^{100}< 25^{100}=>3^{100}< 5^{200}\)

\(\frac{75^{20}}{45^{10}.25^{15}}=\frac{25^{20}.3^{20}}{3^{10}.3^{10}.5^{10}.25^{15}}=\frac{25^{20}}{25^5.25^{15}}=1\)

\(=>75^{20}=45^{10}.25^{15}\left(dpcm\right)\)

P/S:nếu a=b=>a:b=1 mk làm theo cách đó cho nhanh mà bn ghi sai đề r

a) (125 - x) + 25 = 98

125 - x = 98 - 25

125 - x = 73

x = 125 - 73

x = 52

b) 100 : (x + 20) = 68 : 34

100 : (x + 20) = 2

x + 20 = 100 : 2

x + 20 = 50

x = 50 - 20

x = 30

c) (3x + 12) . 5 = 25 . 4 - 10

(3x + 12) . 5 = 100 - 10

(3x + 12) . 5 = 90

3x + 12 = 90 : 5

3x + 12 = 18

3x = 18 - 12

3x = 6

x = 6 : 2

x = 3

d) 210 : (2x - 3) - 20 = 10

210 : (2x - 3) = 10 + 20

210 : (2x - 3) = 30

2x - 3 = 210 : 30

2x - 3 = 70

2x = 70 + 3

2x = 73

x = 73/2

a, ( 125 - x ) + 25 = 98

⇒ 125 - x = 73

⇒ x = 52.

Vậy..

b, 100 : ( x + 20 ) = 68 : 34

⇒ 100 : ( x + 20 ) = 2

⇒ x + 20 = 50

⇒ x = 30

Vậy...

c, ( 3 . x + 12 ) . 5 = 25 . 4 - 10

⇒ ( 3 . x + 12 ) . 5 = 90

⇒ 3 .x + 12 = 18

⇒ 3x = 6

⇒ x = 2.

Vậy...

d, 210 : ( 2 .x - 3 ) - 20 = 10

⇒ ( 2 .x - 3 ) - 20 = 21

⇒ 2x - 3 = 41

⇒ 2x = 44

⇒ x = 22.

Vậy..

Cho mk chỉnh lại câu cuối:

 d, 210 : ( 2 .x - 3 ) - 20 = 10

⇒ 210 : ( 2 .x - 3 ) = 30

⇒ 2 . x - 3 = 7

⇒ 2x = 10

⇒ x = 5.

Vậy..

đáp số = 0 đó các bạn!

quá dễ!

100+100+200-600+400-90-64+50+4-80+20-6+3+3-50+10+200-90-90-20+123-100-20-3+50-49-1+60-30-30+741-700-40-1=108

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mệt quá!

*\(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left[\frac{1}{14}+\frac{1}{7}-\left(-\frac{3}{35}\right)\right].\frac{4}{3}}=\frac{\left(\frac{18}{60}-\frac{16}{60}-\frac{21}{60}\right).\frac{5}{19}}{\left(\frac{5}{70}+\frac{10}{70}+\frac{6}{70}\right).\frac{4}{3}}=\frac{\frac{-19}{60}.\frac{5}{19}}{\frac{21}{70}.\frac{4}{3}}=\frac{\frac{-1}{12}}{\frac{14}{35}}=-\frac{1}{12}.\frac{35}{14}=\frac{-35}{168}\)

*\(\frac{\left(1+2+3+...+100\right).\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(6,3.12-21.3,6\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(\frac{63}{10}.12-21.\frac{18}{5}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(\frac{378}{5}-\frac{378}{5}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).0}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}=0\)