S = 1+ 0,5 + 0,5² + 0,5³ + 0,5⁴ + ....

0,5 .S = 0,5 +  0,5² + 0,5³ + 0,5⁴ + ....

=> S - 0,5 S = 1 => s = 1: 0,5 = 2

+...+ mấy nữa, thế thì làm kiểu gì nhỉ?

\(\frac{1}{3}-\frac{3}{4}-\left(-\frac{3}{5}\right)+\frac{1}{72}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}=\left(\frac{1}{3}+\frac{3}{5}+\frac{1}{15}\right)-\left(\frac{3}{4}+\frac{2}{9}+\frac{1}{36}\right)+\frac{1}{72}=1-1+\frac{1}{72}=\frac{1}{72}\)

đúng rồi ! OK!

\(S=0,5^0+0,5^1+0,5^2+...+0,5^5\)

\(0,5S=0,5^1+0,5^2+...+0,5^6\)

  \(S-0,5S=0,5^0+0,5^1+...+0,5^5-0,5^1-0,5^2-...-0,5^6\)

\(0,5S=1-0,5^6\Rightarrow S=\frac{1-0,5^6}{0,5}\)

(1/2)¹⁵⁰⁰=(1/2⁶)²⁵⁰=(1/64)²⁵⁰

Do 1/16>1/64 =>(1/16)²⁵⁰>(1/64)²⁵⁰

Vậy (1/16)²⁵⁰>(1/2)¹⁵⁰⁰

\(\left(\frac{1}{16}\right)^{250}\) và \(\left(\frac{1}{2}\right)^{1500}\)

=> \(\left(\frac{1}{16}\right)^{250}\) và \(\left(\left(\frac{1}{2}\right)^6\right)^{250}\)

=> \(\frac{1}{16}\) và \(\left(\frac{1}{2}\right)^6\)

=> \(\frac{1}{16}\) và \(\frac{1}{64}\)

=>  \(\frac{1}{16}\) >  \(\frac{1}{64}\)  hay  \(\left(\frac{1}{16}\right)^{250}\) >  \(\left(\frac{1}{2}\right)^{1500}\)

(1000 - 1) . (1000 - 2) . (1000 - 3) ... (1000 - 234567)

= (1000 - 1) . (1000 - 2) ... (1000 - 1000) ... (1000 - 234567)

= (1000 - 1) . (1000 - 2) ... 0 ... (1000 - 234567)

= 0

=(1000-1)....(100-1000)...(1000-234567)

=(1000-1)...0....(1000-1234567)

=0

de qua 

ta có: 

40^500000000 = (40⁵)^100000000

50^400000000 = (50⁴)^100000000

=>  40⁵  và 50⁴

=>  40⁵  = (4 . 10)⁵ = 4⁵ . 10⁵  = 1024 . 10⁵

=>  50⁴ = (5 . 10)⁴  = 5⁴  .  10⁴  = 625 . 10⁴

=> 40⁵ >  50⁴  hay 40^500000000 > 50^400000000

40^500000000=40^5.100000000

50^400000000=50^4.100000000

40^5>50^4=> 40^500000000>50^400000000

bạn xem lại đề:

Có \(\frac{3}{2}<\frac{7}{3}\) nhưng \(\frac{3}{2}>\frac{3+7}{2+7}=\frac{10}{9}\)

X x X x X x X x X = X⁵ = 2¹⁰.10⁵ = 40⁵

=> X = 40

=> Y = 40 - 30 = 10

=> Z = 20 : 10 = 2

Ta có: A + B + C = 119 (1)

           B= 2+C

           A - C = C x 10

           A = 10  xC +C

           A = 11x C

Thay B= 2+C; A =11xC vào (1) ta được:

         11xC + 2+C +C = 119

               13 x C +2 =119

                  13 x C = 119 - 2

                   13 x C = 117

                          C  = 117 : 13

                          C  = 9

       =>           B = 2 +C = 9+2 = 11

       =>           A = 11 x 9

       =>           A = 99

A:B=C NEN A=BXC

A-C=BXC-C=C(B-1)=CX10

SUY RA B-1=10

B=10+1

B=11

C+2=B

C+2=11

C=11-2

C=9

A:B=C

A:11=9

A=9X11

A=99

VAY A=99,B=11,C=9

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