a: =5-78*32

=5-2496

=-2491

b: \(=6\left(9-6\right)=6\cdot3=18\)

c: \(=46\cdot\dfrac{\left(123-42\right)}{81}=46\)

d: \(=181+3-84+8\cdot25\)

=100+200

=300

e: \(=64\cdot35+140\cdot84-1=2240-1+11760\)

=14000-1

=13999

f: \(=3^3+25\cdot8-1=26+200=226\)

g: \(=3+2^4+1=16+4=20\)

h: \(=36:4\cdot3+2\cdot25-1=27+50-1=27+49=76\)

\(S_2=3-3^2+3^3-3^4+...-3^{2020}\)

\(3S_2=3^2-3^3+3^4-3^5+...-3^{2021}\)

\(3S_2+S_2=3^2-3^3+3^4-3^5+...-3^{2021}+3-3^2+3^3-3^4+...-3^{2020}\)

\(4S_2=\left(3^2-3^2\right)+\left(3^3-3^3\right)+\left(3^4-3^4\right)+...+\left(3^{2020}-3^{2020}\right)+3-3^{2021}\)

\(4S_2=3-3^{2021}\)

\(S_2=\frac{3-3^{2021}}{4}\)

Xem mình đúng không?

Bài 1 :

\(M=\dfrac{30-2^{20}}{2^{18}}=\dfrac{2.15-2^{20}}{2^{18}}=\dfrac{15}{2^{17}}-2^2=\dfrac{15}{2^{17}}-4< 0\left(\dfrac{15}{2^{17}}< 1\right)\)

\(N=\dfrac{3^5}{1^{2021}+2^3}=\dfrac{3^5}{9}=\dfrac{3^5}{3^2}=3^3=27\)

\(\Rightarrow M< N\)

Bài 3 :

a) \(t^2+5t-8\) khi \(t=2\)

\(=5^2+2.5-8\)

\(=25+10-8\)

\(=27\)

b) \(\left(a+b\right)^2-\left(b-a\right)^3+2021\left(1\right)\)

\(\left\{{}\begin{matrix}a=5\\b=a+1=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=11\\b-a=1\end{matrix}\right.\)

\(\left(1\right)=11^2-1^3+2021=121-1+2021=2141\)

c) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)

\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)

\(\left(1\right)=1^3=1\)

Bài 2 :

a) \(...=2^3\left(19-14\right)+1=8.5+1=41\)

b) \(...=100-\left[60:\left(5^2-15\right)\right]=100-\left[60:10\right]=100-6=94\)

c) \(...=160:\left[17+\left(9.5-\left(14+2^3\right)\right)\right]=160:\left[17+\left(45-22\right)\right]=160:\left[17+23\right]=160:40=4\)

d) \(...=798+100\left[16-2\left(25-22\right)\right]=798+100\left[16-2.3\right]=798+100.10=798+1000=1798\)

A=(1+3+3²)+(3³+3⁴+3⁵)+...+(3²⁰¹⁹+3²⁰²⁰+3²⁰²¹)                                                  A=(1+3+3²)+3³.(1+3+3²)+...+3²⁰¹⁹.(1+3+3²)

A=13+3³.13+...+3²⁰¹⁹.13

A=13.(1+3³+...+3²⁰¹⁹)chia hết cho 13

=>A  chia hết cho 13

a, \(S=3^0+3^2+3^4+3^6+...+3^{2020}\)

\(\Leftrightarrow3^2S=3^2+3^4+3^6+3^8+...+3^{2022}\)

\(\Leftrightarrow3^2S-S=3^{2022}-3^0\)

\(\Leftrightarrow9S-S=3^{2022}-1\)

\(\Leftrightarrow8S=3^{2022}-1\Leftrightarrow S=\frac{3^{2022}-1}{8}\)

b,\(S=3^0+3^2+3^4+3^6+...+3^{2020}\)

\(=\left(3^0+3^2+3^4\right)+\left(3^6+3^8+3^{10}\right)+...+\left(3^{2016}+3^{2018}+3^{2020}\right)\)

\(=\left(1+3^2+3^4\right)+3^6\left(1+3^2+3^4\right)+...+3^{2016}\left(1+3^2+3^4\right)\)

\(=\left(1+3^2+3^4\right)\left(1+3^6+...+3^{2016}\right)\)

\(=91\left(1+3^6+...+3^{2016}\right)=13.7\left(1+3^6+...+3^{2016}\right)⋮7\)

=> đpcm

Tham khảo :

a, $S=3^0+3^2+3^4+3^6+...+3^{2020}$

$\iff 3^{2}S=3^2+3^4+3^6+3^8+...+3^{2022}$

$\iff 3^{2}S-S=3^{2022}-3^0$

$\iff 9S-S=3^{2022}-1$

$\iff 8S=3^{2022}-1\iff S=\frac{3^{2022}-1}{8}$

b,$S=3^0+3^2+3^4+3^6+...+3^{2020}$

$=\left(3^0+3^2+3^4\right)+\left(3^6+3^8+3^{10}\right)+...+\left(3^{2016}+3^{2018}+3^{2020}\right)$

$=\left(1+3^2+3^4\right)+3^6\left(1+3^2+3^4\right)+...+3^{2016}\left(1+3^2+3^4\right)$

$=\left(1+3^2+3^4\right)\left(1+3^6+...+3^{2016}\right)$

$=91\left(1+3^6+...+3^{2016}\right)=13.7\left(1+3^6+...+3^{2016}\right)⋮7$

=> (đpcm)

Khải đẹp zai k

a) \(100-\left(3.5^2-2.3^3\right)\)

\(100-\left(75-52\right)=100-23=77\)

b)\(4.5^2-3.2^3+3^9:3^7\)

\(100-24+9=85\)

c) \(\left(392:7\right).2^3-2^3+2020^0\)

\(8\left(56-1\right)+1=441\)

d) \(\left(6^2+7^2+8^2+9^2+10^2\right)-\left(1^2+2^2+3^2+4^2+5^2\right)=275\)

(lười - thông cảm :(

Đây bạn ơi a,\(100-\left(3.5^2-2.3^3\right)=100-\left(75-54\right)\))=100-19=81

b,\(4.5^2-3.2^3+\frac{3^9}{3^7}=100-24+9=85\)

c,\(\left(392:7\right).2^3-2^3+2020^0=56.2^3-2^3+1=441\)

d,\(=\left(6^2-1^2\right)+\left(7^2-2^2\right)+\left(8^2-3^3\right)+\left(9^2-4^2\right)+\left(10^2-5^2\right)\)

=\(5.7+5.9+5.11+5.13+5.15=5\left(7+9+11+13+15\right)\)

=\(5.55=275\)