3/2x4/3x5/4x....x100/99

50

GIẢI RA ấy mà

Ta có:

B=1/2-1/2^2-1/2^3-...-1/2^100

B/2=1/2^2-1/2^3-1/2^4-....-1/2^101

B/2-B=1/2^101-1/2

=>B=(1/2^101-1/2).2

Vậy:B=(1/2^101-1/2).2

Bạn nào làm đc nhanh mình sẽ k

Bạn làm nhanh và đúng nhất mình sẽ k

Làm trc cho 2 câu cuối

c) \(a^2-b^2-4a+4b\)

\(=\left(a+b\right)\left(a-b\right)-4\left(a-b\right)\)

\(=\left(a-b\right)\left[\left(a+b\right)-4\right]\)

d) \(a^2+2ab+b^2-2a-2b+1\)

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

\(=\left(a+b\right)\left[\left(a+b\right)-2\right]+1\)

\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(C=\frac{1}{100}-\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{99-98}{98.99}+\frac{100-99}{99.100}\right)\)

\(C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{2}{100}-1=-\frac{49}{50}\)

bạn k trước mk mới kb

=1/125

\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{99}\right)\)

\(=\frac{3}{2}\times\frac{4}{3}\times...\times\frac{100}{99}\)

\(=\frac{100}{2}=50\)

A = 100/2 = 50

- Nhận xét:

Với mọi số dương n, ta có:

1 - n² = ( 1 - n ) + ( n - n² )

          = ( 1 - n ) + n ( 1 - n )

          = ( 1 - n )( 1 + n ).

Do đó:

\(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right).....\) \(\left(\frac{1}{99^2}-1\right)\)

= \(\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}.\frac{1-4^2}{4^2}.....\frac{1-99^2}{99^2}\)

= \(\frac{\left(1-2\right)\left(1+2\right)}{2^2}.\frac{\left(1-3\right)\left(1+3\right)}{3^2}.\)\(\frac{\left(1-4\right)\left(1+4\right)}{4^2}.....\frac{\left(1-99\right)\left(1+99\right)}{99^2}\)

= \(\frac{\left(-1\right).3}{2^2}.\frac{\left(-2\right).4}{3^2}.\frac{\left(-3\right).5}{4^2}.....\frac{\left(-98\right).100}{99^2}\)

= \(\frac{\left(-1\right).3.\left(-2\right).4.\left(-3\right).5.....\left(-98\right).100}{\left(2.2\right)\left(3.3\right)\left(4.4\right).....\left(99.99\right)}\)

= \(\frac{\left[\left(-1\right).\left(-2\right).\left(-3\right).....\left(-98\right)\right]\left(3.4.5.....100\right)}{\left(2.3.4.....99\right)\left(2.3.4.....99\right)}\)

= \(\frac{\left(1.2.3.....98\right).\left(3.4.5.....100\right)}{\left(2.3.4.....99\right)\left(2.3.4.....99\right)}\)

= \(\frac{1.100}{99.2}\)

= \(\frac{50}{99}\).

\(\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{99^2}-1\right)\)

\(=\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right).\left(\frac{1}{16}-1\right)...\left(\frac{1}{9801}-1\right)\)

\(=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}...\frac{-9800}{9801}=\frac{-\left(1.3\right)}{2.2}.\frac{-\left(2.4\right)}{3.3}.\frac{-\left(3.5\right)}{4.4}...\frac{-\left(98.100\right)}{99.99}\)

\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{98.100}{99.99}=\frac{1.3.2.4.3.5...97.99.98.100}{2.2.3.3.4.4...99.99}\)

\(=\frac{\left(1.2.3.4....98\right).\left(3.4.5.6...100\right)}{\left(2.3.4.5...99\right).\left(2.3.4.5...99\right)}=\frac{100}{99.2}=\frac{50}{99}\)

1. 1-2+3-4+5-6-.....+99-100

=(1-2)+(3-4)+(5-6)+...+(99-100)                              (50 cặp)

=(-1)+(-1)+(-1)+...+(-1)                                          (50 số -1)

=(-1).50

=-50

2.1+3-5-7+9+11-.....-397-399

=(1+3-5-7)+(9+11-13-15)+....+(387+389-391-393)+395-397-399 (99 cặp)

=(-8)+(-8)+(-8)+...+(-8)+(-401)(có 99 có -8)

=(-8).99+(-401)

=(-792)+(-401)

=-1193

3. 1-2-3+4+5-6-7+...+96+97-98-99+100

=(1-2-3+4)+(5-6-7+8)+...+(93-94-95+96)+(97-98-99+100)            (25 cặp)

=0+0+0+...+0

=0

4. A=2¹⁰⁰-2⁹⁹-2⁹⁸-.....-2²-2-1

2A=2¹⁰¹-2¹⁰⁰-2⁹⁹-....-2³-2²-2

2A-A=A=2¹⁰¹-2¹⁰⁰-2¹⁰⁰+1

A=2¹⁰¹-2.2¹⁰⁰+1

A=2¹⁰¹-2¹⁰¹+1

A=1

(1/2+1/3+1/4+...+1/100)/(99/1+98/2+97/3+...+1/99)

=(1/2+1/3+1/4+...+1/100)/(1+100/2+100/3+100/4+....+100/99)

=(1/2+1/3+1/4+...+1/100)/100*(1/100+1/99+1/98+...+1/4+1/3+1/2)

=1/100

chỗ 99/1+99/2+99/3+...+1/99 hình như đề bài sai đề bài đúng hình như là trên đã sửa rồi

bạn lm sai rùi