Bài 3:

a: a*S=a^2+a^3+...+a^2023

=>(a-1)*S=a^2023-a

=>\(S=\dfrac{a^{2023}-a}{a-1}\)

b: a*B=a^2-a^3+...-a^2023

=>(a+1)B=a-a^2023

=>\(B=\dfrac{a-a^{2023}}{a+1}\)

Bài 1:

a, 3\(\dfrac{2}{5}\) - \(\dfrac{1}{2}\) 

= \(\dfrac{17}{5}\) - \(\dfrac{1}{2}\) 

= \(\dfrac{34}{10}\) - \(\dfrac{5}{10}\)

= \(\dfrac{29}{10}\)

b, \(\dfrac{4}{5}\) + \(\dfrac{1}{5}\) x \(\dfrac{3}{4}\)

= \(\dfrac{4\times4}{5\times4}\) + \(\dfrac{1\times3}{5\times4}\)

= \(\dfrac{16}{20}\) + \(\dfrac{3}{20}\)

= \(\dfrac{19}{20}\)

c, 4\(\dfrac{4}{9}\) : 2\(\dfrac{2}{3}\) + 3\(\dfrac{1}{6}\)

= \(\dfrac{40}{9}\) : \(\dfrac{8}{3}\) + \(\dfrac{19}{6}\)

= \(\dfrac{5}{3}\) + \(\dfrac{19}{6}\)

= \(\dfrac{10}{6}\) + \(\dfrac{19}{6}\)

= \(\dfrac{29}{6}\)

Bài 2:

   3\(\dfrac{2}{5}\) + 2\(\dfrac{1}{5}\) 

= \(\dfrac{17}{5}\) + \(\dfrac{11}{5}\)

= \(\dfrac{28}{5}\)

b, 7\(\dfrac{1}{6}\) : 5\(\dfrac{2}{3}\)

= \(\dfrac{43}{6}\) : \(\dfrac{17}{3}\)

= \(\dfrac{43}{34}\)

Bài 1

a) 3 2/5 - 1/2

= 17/5 - 1/2

= 34/10 - 5/10

= 29/10

b) 4/5 + 1/5 × 3/4

= 4/5 + 3/20

= 16/20 + 3/20

= 19/20

c) 3 1/2 × 1 1/7

= 7/2 × 8/7

= 4

d) 4 1/6 : 2 1/3

= 25/6 : 7/3

= 25/14

Bài 2

a) 3 × 1/2 + 1/4 × 1/3

= 3/2 + 1/12

= 18/12 + 1/12

= 19/12

b) 1 4/5 - 2/3 : 2 1/3

= 9/5 - 2/3 : 7/3

= 9/5 - 2/7

= 63/35 - 10/35

= 53/35

bai EZ​ qua

????

:))))

2 câu c,d làm tương tự  

5:

a: \(3^{2n}=\left(3^2\right)^n=9^n\)

\(\left(2^{3n}\right)=\left(2^3\right)^n=8^n\)

=>\(3^{2n}>2^{3n}\)

b: \(199^{20}=\left(199^4\right)^5=1568239201^5\)

\(2003^{15}=\left(2003^3\right)^5=8036054027^5\)

mà \(1568239201< 8036054027\)

nên \(199^{20}< 2003^{15}\)

4: \(100< 5^{2x-1}< 5^6\)

mà \(25< 100< 125\)

nên \(125< 5^{2x-1}< 5^6\)

=>3<2x-1<6

=>4<2x<7

=>2<x<7/2

mà x nguyên

nên x=3

#)Giải :

\(A=\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}+\frac{5}{6}-\frac{6}{7}-\frac{5}{6}+\frac{4}{5}-\frac{3}{4}+\frac{2}{3}-\frac{1}{2}\)

\(A=\left(\frac{1}{2}-\frac{1}{2}\right)+\left(-\frac{2}{3}+\frac{2}{3}\right)+\left(\frac{3}{4}-\frac{3}{4}\right)+\left(-\frac{4}{5}+\frac{4}{5}\right)+\left(\frac{5}{6}-\frac{5}{6}\right)-\frac{6}{7}\)

\(A=0+0+0+0+0-\frac{6}{7}\)

\(A=-\frac{6}{7}\)

troi,zậy mà khó

Ta có \(a-b=5\Rightarrow\left(a-b\right)^2=25\Rightarrow a^2+b^2=25+2ab=25+2\cdot2=29\) (Do ab=2) 

\(B=3\left[\left(a^2+b^2\right)^2-2a^2b^2\right]+2\left[\left(a-b\right)\left(a^4+b^4+a^3b^2+a^2b^3\right)\right]\) 

= \(3\left[29^2-2\cdot4\right]+2\left\{5\left[\left(a^2+b^2\right)^2-2a^2b^2+ab\left(a^2+b^2\right)\right]\right\}\)

= 3\(\cdot833+10\left[29^2-2\cdot4+2\cdot29\right]\) \(=2499+10\cdot891=11409\)