97/198

k cho mk nha

dễ mà bạn

giúp mình với

P = \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{99}\right)\)

= \(\frac{1}{2}.\frac{2}{3}.....\frac{98}{99}\)

= \(\frac{1}{99}\)
 

Tại sao lại bằng 1/ 99 luôn? 

P =  1 − 2 1 1 − 3 1 ... 1 − 99 1 =  2 1 . 3 2 ..... 99 98 =  99 1 

Xin hãy giúp mình

a,=3/2*4/3*....100/99

=3*4*5*....*100/2*3*...*99

=100/2=50

b, nhân lên băng:

1*2*3*...*99/2*3*...*100=1/100

2A=1-1/2+1/2^2-...+1/2^98-1/2^99

=>3A=1-1/2^100

=>\(A=\dfrac{2^{100}-1}{3\cdot2^{100}}\)

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

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

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