\(A=1-3+5-7+......-2019+2021-2023\)

\(A=\left(1-3\right)+\left(5-7\right)+....+\left(2021-2023\right)\)

\(A=-2+\left(-2\right)+....+\left(-2\right)\left(506 cặp\right)\)

\(A=-2.506\)

\(A=-1012\)

*) A=(1-3)+(5-7)+....+(2021-2023)

<=> A=-2+(-2)+...+(-2)

Dãy A có (2023-1):2+1=1012 số số hạng 

=> Có 506 số (-2)

=> A=(-2).506=-1012

\(B=1+2-3-4+5+6-7-8+......+2017+2018-2019-2020\)

\(B=\left(1+2-3-4\right)+\left(5+6-7-8\right)+.....+\left(2017+2018-2019-2020\right)\)

\(B=-4+\left(-4\right)+.....+\left(-4\right)\left(505 cặp\right)\)

\(B=-4.505\)

\(B=-2020\)

Ta có : x - 1 = 2018 - 1 = 2017 

N = x⁶ - 2017x⁵ - 2017x⁴ - 2017x³ - 2017x² - 2017x - 2017

N = x⁶ - ( x - 1 ).x⁵ - ( x - 1 ).x⁴ - ( x - 1 ).x³ - ( x - 1 ).x² - ( x - 1 ).x - ( x - 1 )

N = x⁶ - x⁶ + x⁵ - x⁵ + x⁴ - x⁴ + x³ - x³ + x² - x² + x - x + 1

N = 1

63  nhé bạn

1 + 2 + 3 * 4 + 6 * 8 

= 1 + 2 + 12 + 48

= 15 + 48

= 63 nhé bạn

a) \(\dfrac{5}{6}-\dfrac{2}{14}\)

\(=\dfrac{5}{6}-\dfrac{1}{7}\)

\(=\dfrac{35}{42}-\dfrac{6}{42}\)

\(=\dfrac{29}{42}\)

b) \(\dfrac{5}{20}-\dfrac{1}{6}\)

\(=\dfrac{1}{4}-\dfrac{1}{6}\)

\(=\dfrac{3}{12}-\dfrac{2}{12}\)

\(=\dfrac{1}{12}\)

c) \(\dfrac{5}{9}-\dfrac{3}{12}\)

\(=\dfrac{5}{9}-\dfrac{1}{4}\)

\(=\dfrac{20}{36}-\dfrac{9}{36}\)

\(=\dfrac{11}{36}\)

d) \(8-\dfrac{4}{6}\)

\(=\dfrac{48}{6}-\dfrac{4}{6}\)

\(=\dfrac{44}{6}=\dfrac{22}{3}\).

\(M=2^0+2^2+2^4+2^6+2^8+...+2^{2018}\)

\(M=2^0+2^2+\left(2^4+2^6+2^8\right)+...+\left(2^{2014}+2^{2016}+2^{2018}\right)\)

\(M=1+4+2^4.\left(1+2^2+2^4\right)+...+2^{2014}.\left(1+2^2+2^4\right)\)

\(M=5+2^4.21+2^{10}.21+...+2^{2014}.21\)

\(M=5+21.\left(2^4+2^{10}+...+2^{2014}\right)\)

vì  \(21.\left(2^4+2^{10}+...+2^{2014}\right)⋮7\)

nên \(M=5+21.\left(2^4+2^{10}+...+2^{2014}\right)\)chia 7 dư 5

S₁ = \(\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+...+\frac{127}{128}\)

2S₁ = 1 + \(\frac{3}{2}+\frac{7}{4}+\frac{15}{8}+\frac{31}{16}+\frac{63}{32}+\frac{127}{64}\)
2S₁ - S₁ = S₁ = 1 + (1 + 1 + 1 + 1 + 1 + 1) - \(\frac{127}{128}\)= 6 + \(\frac{1}{128}\)
=> S = S₁ - 6 = 6 + \(\frac{1}{128}\)- 6 = \(\frac{1}{128}\)

\(S=\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\frac{31}{32}+\frac{63}{64}+\frac{127}{128}-6\)

\(S=\frac{1}{2}+\left(\frac{3}{4}+\frac{7}{8}\right)+\left(\frac{15}{16}+\frac{31}{32}\right)+\left(\frac{63}{64}+\frac{127}{128}\right)-6\)

\(S=\frac{1}{2}+\frac{13}{8}+\frac{61}{32}+\frac{253}{128}-6\)

\(S=\frac{64}{128}+\frac{208}{128}+\frac{244}{128}+\frac{253}{128}-6\)

\(S=\frac{769}{128}-6\)

\(S=\frac{769}{128}-\frac{768}{128}\)

\(S=\frac{1}{128}\)

hok tốt!!

Mơn 2 bạn nhé!!

\(\dfrac{4^5\cdot10\cdot5^6+25^5\cdot2^8}{2^8\cdot5^4+5^7\cdot5^2}\\ =\dfrac{\left(2^2\right)^5\cdot2\cdot5\cdot5^6+\left(5^2\right)^5\cdot2^8}{2^8\cdot5^4+5^7\cdot5^2}\\ =\dfrac{2^{10}\cdot2\cdot5\cdot5^6+5^{10}\cdot2^8}{2^8\cdot5^4+5^7\cdot5^2}\\ =\dfrac{2^{11}\cdot5^7+5^{10}\cdot2^8}{2^8\cdot5^4+5^7\cdot5^2}\\ =\dfrac{2^8\cdot5^7\left(2^3+5^3\right)}{2^5\cdot5^4\left(2^3+5^3\right)}\\ =\dfrac{2^8\cdot5^7}{2^5\cdot5^4}\\ =2^3\cdot5^3\\ =8\cdot125\\ =1000\)

\(B=2\left(-1+2-3+4+...-49+50-51\right)\)

\(=2.\left[\left(2-1\right)+\left(4-3\right)+...\left(50-49\right)-51\right]\)

\(=2.\left(1+1+...+1-51\right)\)

\(=2.\left(25-51\right)=-52\)

\(B=\left(-2\right)+4+\left(-6\right)+8+...+\left(-98\right)+100+\left(-102\right)\)

\(\Rightarrow B=\left[\left(-2\right)+4\right]+\left[\left(-6\right)+8\right]+...+\left[\left(-98\right)+100\right]+\left(-102\right)\) ( 25 cặp số )

\(\Rightarrow B=2+2+...+2-102\) ( 25 số 2 )

\(\Rightarrow B=2.25-102\)

\(\Rightarrow B=50-102\)

\(\Rightarrow B=-52\)

Vậy \(B=-52\)

\(B=\left(-2\right)+4+\left(-6\right)+8+...+\left(-98\right)+100+\left(-102\right)\)(có 51 số hạng)

\(B=\left[\left(-2\right)+4\right]+\left[\left(-6\right)+8\right]+...+\left[\left(-98\right)+100\right]+\left(-102\right)\)(có 25 nhóm và 1 số hạng)

\(B=2+2+...+2+\left(-102\right)\)

\(B=2\cdot50+\left(-102\right)\)

\(B=\left(-52\right)\)