A = \(\dfrac{10^{20}+3}{10^{21^{ }}+3}\)

B = \(\dfrac{10^{21}+4}{10^{22}+4}\) < 1

\(\Rightarrow\) B < \(\dfrac{10^{21}+4+6}{10^{22}+4+6}\) 

\(\Rightarrow\) B < \(\dfrac{10^{21}+10}{10^{22}+10}\)

\(\Rightarrow\) B < \(\dfrac{10\left(10^{20}+1\right)}{10\left(10^{21}+1\right)}\)

\(\Rightarrow\) B < \(\dfrac{10^{20}+1}{10^{21}+1}\) < \(\dfrac{10^{21}+1+2}{10^{22}+1+2}\)

\(\Rightarrow\) B < \(\dfrac{10^{21}+3}{10^{22}+3}\)

\(\Rightarrow\) B < A

\(\dfrac{21}{36}-\left(-\dfrac{11}{30}\right)=\dfrac{7}{12}+\dfrac{11}{30}=\dfrac{7.5+11.2}{60}=\dfrac{57}{60}=\dfrac{19}{20}\\ ----\\\dfrac{-4}{8}+\left(-\dfrac{3}{10}\right)=\dfrac{-1}{2}-\dfrac{3}{10}=\dfrac{-1.5-3}{10}=\dfrac{-8}{10}=-\dfrac{4}{5}\\ ----\\ \dfrac{7}{12}-\left(-\dfrac{9}{20}\right)=\dfrac{7}{12}+\dfrac{9}{20}=\dfrac{7.5+9.3}{60}=\dfrac{62}{60}=\dfrac{31}{30}\\ ---\\ \dfrac{-2}{5}+\left(-\dfrac{11}{30}\right)=-\dfrac{2}{5}-\dfrac{11}{30}=\dfrac{-2.6-11}{30}=-\dfrac{29}{30}\)

a) \(\dfrac{1}{2}-\left(x+\dfrac{1}{3}\right)=\dfrac{5}{6}\)

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{5}{6}\)

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{-1}{3}\)

\(\Rightarrow x=\dfrac{-1}{3}-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{-2}{3}\)

b)\(\dfrac{3}{4}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\)

\(\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{4}-\dfrac{4}{5}\)

\(\Rightarrow x+\dfrac{1}{2}=\dfrac{-1}{20}\)

\(\Rightarrow x=\dfrac{-1}{20}-\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{-11}{20}\)

c) \(\dfrac{3}{35}-\left(\dfrac{3}{5}+x\right)=\dfrac{2}{7}\)

\(\Rightarrow\dfrac{3}{5}+x=\dfrac{3}{35}-\dfrac{2}{7}\)

\(\Rightarrow\dfrac{3}{5}+x=\dfrac{-1}{5}\)

\(\Rightarrow x=\dfrac{-1}{5}-\dfrac{3}{5}\)

\(\Rightarrow x=\dfrac{-4}{5}\)

d)\(\dfrac{2}{3}.x=\dfrac{4}{27}\)

\(\Rightarrow x=\dfrac{4}{27}:\dfrac{2}{3}\)

\(\Rightarrow x=\dfrac{2}{9}\)

e) \(\dfrac{-3}{5}.x=\dfrac{21}{10}\)

\(\Rightarrow x=\dfrac{21}{10}:\dfrac{-3}{5}\)

\(\Rightarrow x=\dfrac{-7}{2}\)

a: x^2-7x+13=0

Δ=(-7)^2-4*1*13=49-52=-3<0

=>PTVN

b: -5x^2+5x-1.25=0

=>4x^2-4x+1=0

=>(2x-1)^2=0

=>2x-1=0

=>x=1/2

d: 2x^2+3x+1=0

=>(x+1)(2x+1)=0

=>x=-1 hoặc x=-1/2

a) \(5^6:5^5+\left(\dfrac{4}{9}\right)^0=5^{6-5}+1=5+1=6\)

b) \(\left(\dfrac{3}{7}\right)^{21}:\left(1-\dfrac{40}{49}\right)^3\)

\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^3\)

\(=\left(\dfrac{3}{7}\right)^{21}:\left[\left(\dfrac{3}{7}\right)^2\right]^3\)

\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^6\)

\(=\left(\dfrac{3}{7}\right)^{21-6}=\left(\dfrac{3}{7}\right)^{15}\)

c) \(\left(\dfrac{2}{3}\right)^3-\left(\dfrac{-52}{3}\right)^0+\dfrac{4}{9}\)

\(=\dfrac{8}{27}-1+\dfrac{4}{9}\)

\(=\dfrac{8-27+12}{27}=-\dfrac{7}{27}\)

\(a)5^6:5^5+\left(\dfrac{4}{9}\right)^0=5^1+1=6\)

\(b,\left(\dfrac{3}{7}\right)^{21}:\left(1-\dfrac{40}{49}\right)^3\)

\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{49-40}{49}\right)^3\)

\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^3=\left(\dfrac{3}{7}\right)^{21}:[\left(\dfrac{3}{7}\right)^2]^3\)

\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^6=\left(\dfrac{3}{7}\right)^{21-6}\)

\(=\left(\dfrac{3}{7}\right)^{15}\)

\(c,3.\left(\dfrac{2}{3}\right)^3-\left(\dfrac{-52}{3}\right)^0+\dfrac{4}{9}\)

\(=3.\dfrac{8}{27}-1+\dfrac{4}{9}\)

\(=\dfrac{8}{9}-1+\dfrac{4}{9}\)

\(=\dfrac{8-9+4}{9}=\dfrac{1}{3}\)

a) 56:55+(49)0=56−5+1=5+1=656:55+(94​)0=56−5+1=5+1=6

b) (37)21:(1−4049)3(73​)21:(1−4940​)3

=(37)21:(949)3=(73​)21:(499​)3

=(37)21:[(37)2]3=(73​)21:[(73​)2]3

=(37)21:(37)6=(73​)21:(73​)6

=(37)21−6=(37)15=(73​)21−6=(73​)15

c) (23)3−(−523)0+49(

a) \(\dfrac{-15}{4}:\dfrac{21}{-10}=\dfrac{-15}{4}.\dfrac{-10}{21}=\dfrac{25}{14}\)

b) \(\dfrac{-7}{14}:\left(-0,14\right)=\dfrac{-7}{14}.\dfrac{-50}{7}=\dfrac{25}{7}\)

c) \(\left(\dfrac{-11}{15}\right):1\dfrac{1}{10}=\dfrac{-11}{15}.\dfrac{10}{11}=\dfrac{-2}{3}\)

d) \(2\dfrac{1}{7}:1\dfrac{1}{14}=\dfrac{15}{7}.\dfrac{14}{15}=2\)

\(a.-\dfrac{15}{4}:\left(\dfrac{21}{-10}\right)\)

\(=-\dfrac{15}{4}\cdot\left(-\dfrac{10}{21}\right)\)

\(=\dfrac{25}{14}\)

\(b.-\dfrac{7}{14}:\left(-0,14\right)\)

\(=-\dfrac{1}{2}:\left(-\dfrac{7}{50}\right)\)

\(=\dfrac{25}{7}\)

\(c.\left(-\dfrac{11}{15}\right):\left(1\dfrac{1}{10}\right)\)

\(=\left(-\dfrac{11}{15}\right):\dfrac{11}{10}\)

\(=-\dfrac{2}{3}\)

\(d.\left(2\dfrac{1}{7}\right):\left(1\dfrac{1}{14}\right)\)

\(=\dfrac{15}{7}:\dfrac{15}{14}\)

\(=2\)

a) \(\dfrac{-15}{4}:\dfrac{21}{-10}=\dfrac{-15}{4}.\dfrac{-10}{21}=\dfrac{25}{14}\)

b) \(\dfrac{-7}{14}:\left(-0,14\right)=\dfrac{-7}{14}:\dfrac{-7}{50}=\dfrac{-7}{14}.\dfrac{50}{-7}=\dfrac{25}{7}\)

c) \(\dfrac{-11}{15}:1.\dfrac{1}{10}=\dfrac{-11}{15}.\dfrac{1}{10}=\dfrac{-11}{150}\)

d) \(\dfrac{21}{7}:1.\dfrac{1}{14}=3.\dfrac{1}{14}=\dfrac{3}{14}\)

Ta có b/c=3/4 nên b/3=c/4 =>b/12=c/16(1)

mà  a/1=b/4=>a/3=b/12(2)

 Từ (1)và (2) suy ra :

a/3=b/12=c/16=>4a/12=b/12=c/16

áp dụng tính chất dãy tỉ số bằng nhau ta có :

4a/12=b/12=c/16=4a+b-c/12+12-16=8/8=1

=>a/3=1=>a=1.3=3

=>b/12=1=>b=1.12=12

=>c/16=1=>c=1.16=16

Vậy a=3,b=12,c=16

theo đề bài ta có :\(\frac{b}{3}=\frac{c}{4}\)và\(\frac{3}{4}=\frac{b}{c}\)

từ trên \(\Rightarrow\frac{b}{12}=\frac{c}{16}\)

\(\frac{a}{1}=\frac{b}{4}\Rightarrow\frac{a}{3}=\frac{b}{12}\)

như vậy từ đây áp dụng tính chất dãy tỉ số bằng nhau ,ta có:

\(\frac{4a}{12}=\frac{b}{12}=\frac{c}{16}=\frac{4a+b-c}{12+12-16}=\frac{8}{8}hay1\)

\(\frac{a}{3}=1\Rightarrow a=3\)

\(\frac{b}{12}=1\Rightarrow b=12\)

\(\frac{c}{16}=1\Rightarrow c=16\)

vậy ta có a=3  ; b=12 ; c=16

\(A=1+3+3^2+.....+3^{11}\)

\(A=\left(1+3+3^2\right)+....+\left(3^9+3^{10}+3^{11}\right)\)

\(A=\left(3^0.1+3^0.3+3^0.3^2\right)+....+\left(3^9.1+3^9.3+3^9.3^2\right)\)

\(A=1.\left(1+3+3^2\right)+....+3^9\left(1+3+3^2\right)\)

\(A=1.13+....+3^9.13\)

\(A=13.\left(1+....+3^9\right)⋮13\left(đpcm\right)\)

Cảm ơn bạn nhé!

\(A=1+3+3^2+3^3+...+3^{11}\)

\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^9+3^{10}+3^{11}\right)\)

\(=13+13.3^3+...+13.3^9\)

\(=13.\left(1+3^3+...+3^9\right)⋮13\)

\(\Rightarrow\)\(đpcm\)

CM: A ⋮ 5

A = 1 + 4 + 4² + 4³ + ... + 4⁶⁰

A = (1 + 4) + (4² + 4³) + ... + (4⁵⁹ + 4⁶⁰)

A = 5 + 4² . (1 + 4) + ... + 4⁵⁹ . (1 + 4)

A = 5 + 4² . 5 + ... + 4⁵⁹ . 5

A = 5 . (1 + 4² + ... + 4⁵⁹)  ⋮ 5

Vậy A ⋮ 5

CM: A ⋮ 21

A = 1 + 4 + 4² + 4³ + ... + 4⁶⁰

A = (1 + 4 + 4²⁾ + (4³ + 4⁴ + 4⁵) + ... + (4⁵⁸ + 4⁵⁹ + 4⁶⁰)

A = 21 + 4³ . (1 + 4 + 4²) + ... + 4⁵⁸ . (1 + 4 + 4²)

A = 21 + 4³ . 21 + ... + 4⁵⁸ . 21

A = 21 . (1 + 4³ + ... + 4⁵⁸)  ⋮ 21

Vậy A ⋮ 21