\(\dfrac{2022-1}{3^2\cdot3^2}\)

\(=\dfrac{2021}{3^{2+2}}\)

\(=\dfrac{2021}{3^4}\)

\(=\dfrac{2021}{81}\)

Bạn nên gõ đề đầy đủ và bằng công thức toán để mọi người hỗ trợ tốt hơn.

\(A=\dfrac{7}{1.9}+\dfrac{7}{9.17}+\dfrac{7}{17.25}+...+\dfrac{7}{81.89}\)

\(\dfrac{8}{7}A=\dfrac{8}{1.9}+\dfrac{8}{9.17}+\dfrac{8}{17.25}+...+\dfrac{8}{81.89}\)

\(\dfrac{8}{7}A=1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{25}+...+\dfrac{1}{81}-\dfrac{1}{89}\)

\(\dfrac{8}{7}A=1-\dfrac{1}{89}=\dfrac{88}{89}\Rightarrow A=\dfrac{88}{89}:\dfrac{8}{7}=\dfrac{77}{89}\)

\(B=\dfrac{5^2}{1.4}+\dfrac{3^2}{4.7}+\dfrac{3^2}{7.10}+...+\dfrac{3^2}{37.40}\)

\(B=\dfrac{25}{1.4}+\dfrac{9}{4.7}+\dfrac{9}{7.10}+...+\dfrac{9}{37.40}\)

\(\dfrac{1}{3}B=\dfrac{25}{12}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{37.40}\)

\(\dfrac{1}{3}B=\dfrac{25}{12}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{37}-\dfrac{1}{40}\)

\(\dfrac{1}{3}B=\dfrac{25}{12}+\dfrac{1}{4}-\dfrac{1}{40}=\dfrac{277}{120}\Rightarrow B=\dfrac{277}{120}:\dfrac{1}{3}=\dfrac{277}{40}\)

\(A=\dfrac{7}{1.9}+\dfrac{7}{9.17}+\dfrac{7}{17.25}+...+\dfrac{7}{81.89}\)

\(=7\left(\dfrac{8}{1.9}+\dfrac{8}{9.17}+\dfrac{8}{17.25}+...+\dfrac{8}{81.89}\right)\)

\(=7\left(1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{25}+\dfrac{1}{25}+...+\dfrac{1}{81}-\dfrac{1}{89}\right)\)

\(=7.\left(1-\dfrac{1}{89}\right)=7.\dfrac{88}{89}=\dfrac{616}{89}\)

\(=\left(-3\right)^2\left(135+130\right)+5=9.265+5=2385+5=2390\)

a)-21+50-29-2016

=-2016

b) 36:9+9.8-1

=4+72-1

=75

\(18-\left(x-4\right)=3^2\)

=>x-4=18-9=9

=>x=9+4=13

98.42 - 50[(18 - 2mũ3) : 2 + 3mũ2 ]

= 98.42 - 50[(18 - 8) : 2 + 9]

= 98.42 - 50[10 : 2 + 9]

= 98.42 - 50[ 5 + 9 ]

= 98.42 - 50.14

=  28812.100

=  2881200

\(\Rightarrow3S=3^2+3^3+3^4+...+3^{2023}\)

trừ vế với vế ta được :

\(3S-S=3^{2023}-3\)

\(\Rightarrow2S=3^{2023}-3\)

\(\Rightarrow S=\dfrac{3^{2023}-3}{2}\)

A=120:{60:[(3mux2 + 4mux2)-5]}

A=120:{60[(9+16)-5]}

A=120:{60[25-5]}

A=120:{60.20}

A=120:1200

A=0,1

ko cos mays tính à

A=120:{60:[(3mux2 + 4mux2)-5]}

A=120:{60[(9+16)-5]}

A=120:{60[25-5]}

A=120:{60.20}

A=120:1200

A=0,1