a) 17.13+17.42-17.35

=17.(13+42-35)

=17.20=340

b) [25.(18-4²)-10]:4+6

=(25.2-10):4+6

=40:4+6=16

c) 3⁶:3²+2³.2²-3².3

=3⁴+2⁵-3³

=81+32-27=86

d) B=3.4²-2².3

=3.(16-4)

=3.12=36

e)2022⁰+3.[5².10-(23-13)²]

=1+3.(250-100)

=1+450=451

g) 27.77+24.27-27

=27.(77+24-1)

=27.100=2700

h) 5.2³+7⁹:7⁷-1²⁰²⁰

=40+7²-1

=89-1=88

i) 120:{54[50:2+(3²-2.4)]}

=120:[54(25+1)]

=120:1404=10/117

a)lim\(\dfrac{4-3^n}{2.3^n+2}\)=lim\(\dfrac{4.\dfrac{1^n}{3^n}-\dfrac{3^n}{3^n}}{2.\dfrac{3^n}{3^n}+2\dfrac{1^n}{3^n}}=\)\(lim\dfrac{4.(\dfrac{1}{3})^n-1}{2.1+2.(\dfrac{1}{3})^n}=\dfrac{4.0-1}{\dfrac{2+2.0}{ }}=\dfrac{-1}{2}\)

b) lim\(\dfrac{3^{n+1}-2^n}{2-2.3^n}=lim\dfrac{3^n.3-2^n}{2-2.3^n}=lim\dfrac{3.\dfrac{3^n}{3^n}-\left(\dfrac{2}{3}\right)^n}{2.\left(\dfrac{1}{3}\right)^n-2.\dfrac{3^n}{3^n}}=\dfrac{3.1-0}{2.0-2.1}=\dfrac{-3}{2}\)

a) 

\(\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{4}{12}-\dfrac{3}{12}=\dfrac{1}{12}\)

b)\(\dfrac{1}{4}-\dfrac{1}{5}=\dfrac{5}{20}-\dfrac{4}{20}=\dfrac{1}{20}\)

c) Ta có: \(S=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{n\cdot\left(n+1\right)}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\)

\(=1-\dfrac{1}{n+1}\)

\(=\dfrac{n+1-1}{n+1}=\dfrac{n}{n+1}\)

d)

 \(M=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}\\ =\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{99-97}{97.99}=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\\ =\dfrac{1}{3}-\dfrac{1}{99}=\dfrac{32}{99}\)

12x-33=3².3³

12x    =243

x    =243:12

x    =20,25

k

mơn cậu nhìu

\(3^{x+1}-2.3^x=243\\ \Rightarrow3^x.3-2.3^x=243\\ \Rightarrow3^x=3^5\\ \Rightarrow x=5\)

\(d.3^{x+1}-2.3^x=243\)

mk làm phần a thui nhé

a. A = 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6

A = 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6

A =  1/2 - 1/6

A= 3/6 - 1/6

A = 1/3

\(B=\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\)

\(b=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)

\(b=\frac{1}{2}-\frac{1}{14}\)

\(b=\frac{3}{7}\)

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

\(d=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{10\cdot11}\)

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

\(d=1-\frac{1}{11}\)

\(d=\frac{10}{11}\)

\(e=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)

\(e=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+\frac{1}{14\cdot17}+\frac{1}{17\cdot20}\)

\(e=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{17\cdot20}\right)\)

\(e=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right)\)

\(e=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(e=\frac{1}{3}\cdot\frac{9}{20}=\frac{3}{20}\)

a) 8 . 2ⁿ + 2ⁿ⁺¹ = 2ⁿ . ( 8 + 2 ) = 2ⁿ . 10 = ....0 

b) có vấn đề

c) 4ⁿ⁺³ + 4ⁿ⁺² - 4ⁿ⁺¹ - 4ⁿ = 4ⁿ . ( 4³ + 4² - 4 - 1 ) = 4ⁿ . 75 = 4^n-1 . 4 . 75 = 300 . 4^n-1 \(⋮\)300