Đặt tổng trên = A

\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{16}+...+\frac{1}{152}\)

\(Ax2=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{16}+...+\frac{1}{76}\)

\(Ax2-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{16}+...+\frac{1}{76}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{16}+...+\frac{1}{152}\right)\)

\(A=1-\frac{1}{152}\)

\(A=\frac{151}{152}\)

\(\text{Đặt tổng trên = A}\)

\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{16}+...+\frac{1}{152}\)

\(Ax2=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{16}+...+\frac{1}{76}\)

\(Ax2-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{16}+...+\frac{1}{76}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{16}+...+\frac{1}{152}\right)\)

\(A=1-\frac{1}{152}\)

$A=\frac{151}{152}$

Đặt A=1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256

A=1/2¹+1/2²+1/2³+1/2⁴+1/2⁵+1/2⁶+1/2⁷+1/2⁸

1/2A=1/2²+1/2³+1/2⁴+1/2⁵+1/2⁶+1/2⁷+1/2⁸+1/2⁹

A-1/2A=(1/2+1/2²+1/2³+1/2⁴+1/2⁵+1/2⁶+1/2⁷+1/2⁸)-(1/2²+1/2³+1/2⁴+1/2⁵+1/2⁶+1/2⁷+1/2⁸+1/2⁹)

1/2A=1/2-1/2⁹

A=2(1/2-1/2⁹)

A=1-1/2⁸

A=2⁸-1

\(C=\frac{1}{2}+\frac{1}{4}+\frac{1}{35}+\frac{1}{65}+\frac{1}{104}+\frac{1}{152}=\frac{431}{532}\)

Ai giúp mình làm bài này với. Mình cảm ơn nhiều.

a) 153^2+99.153+47^2

= 153^2+2.47.153+47^2

= (153+47)^2

=200^2

=40000

b) 126^2-152.126+5776

= 126^2-2.76.126+76^2

= (126-76)^2

= 50^2

= 2500

c)3^8.5^8-(15^4-1).(15^4+1)

= 15^8-[(15^4)^2-1^2]

= 15^8-15^8+1

=1

d) (2+1).(2^2+1).(2^4+1)...(2^32+1)+1

= 1.(2+1).(2^2+1).(2^4+1)...(2^32+1)+1

= (2-1).(2+1).(2+1).(2^4+1)...(2^32+1)+1

= (2^2-1).(2^2+1).(2^4+1)...(2^32+1)+1

= (2^4-1).(2^4+1)...(2^32+1)+1

= (2^8-1)...(2^32+1)+1

= (2^32-1).(2^32+1)+1

= 2^64-1+1

= 2^64

1, 3/5 + -4/15 = \(\dfrac{9}{15}\)+ \(\dfrac{-4}{15}\)= \(\dfrac{5}{15}\)= \(\dfrac{1}{3}\)

2, -1/3 + 2/5 + 2/15= \(\dfrac{-5}{15}+\dfrac{6}{15}+\dfrac{2}{15}\)=\(\dfrac{3}{15}=\dfrac{1}{5}\)

3, -3/5 + 7/21 + -4/5 + 7/5= \(\dfrac{1}{3}\)

\(1.\dfrac{3}{5}+\dfrac{-4}{15}=\dfrac{9}{15}+\dfrac{-4}{15}=\dfrac{5}{15}=\dfrac{1}{3}\)

\(2.\dfrac{-1}{3}+\dfrac{2}{5}+\dfrac{2}{15}=\dfrac{-5}{15}+\dfrac{6}{15}+\dfrac{2}{15}=\dfrac{3}{15}=\dfrac{1}{5}\)

\(3.\dfrac{-3}{5}+\dfrac{7}{21}+\dfrac{-4}{5}+\dfrac{7}{5}=\left(\dfrac{-3}{5}+\dfrac{-4}{5}+\dfrac{7}{5}\right)+\dfrac{7}{21}=0+\dfrac{7}{21}=\dfrac{1}{3}\)

\(1,\dfrac{3}{5}+\dfrac{-4}{15}=\dfrac{9}{15}+\dfrac{-4}{15}\\ =\dfrac{9-4}{15}=\dfrac{1}{3}\\ 2,\dfrac{-1}{3}+\dfrac{2}{5}+\dfrac{2}{15}=\dfrac{-5}{15}+\dfrac{6}{15}+\dfrac{2}{15}\\ =\dfrac{3}{15}=\dfrac{1}{5}\)

^{Đã Làm !?}

a: \(=\dfrac{7}{2}\left(-\dfrac{3}{4}+\dfrac{5}{13}-\dfrac{9}{4}-\dfrac{8}{13}\right)=\dfrac{7}{2}\cdot\left(-3-\dfrac{3}{13}\right)=\dfrac{7}{2}\cdot\dfrac{-42}{13}=\dfrac{-147}{13}\)

b: \(=-12+\dfrac{8}{9}-\dfrac{5}{18}=\dfrac{-216}{18}+\dfrac{16}{18}-\dfrac{5}{18}=\dfrac{-205}{18}\)

c: \(=\dfrac{45}{4}-\dfrac{19}{7}-\dfrac{21}{4}=6-\dfrac{19}{7}=\dfrac{23}{7}\)

d: \(=\dfrac{-1}{4}\left(\dfrac{152}{11}+\dfrac{68}{11}\right)=\dfrac{-1}{4}\cdot20=-5\)

a,

\(A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{900}\right)\\ =\left(1-\frac{1}{2}\right)\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1+\frac{1}{3}\right)...\left(1-\frac{1}{30}\right)\left(1+\frac{1}{30}\right)\\ =\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot...\cdot\frac{29}{30}\cdot\frac{31}{30}\\ =\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{29}{30}\cdot\frac{3}{2}\cdot\frac{4}{3}\cdot...\cdot\frac{31}{30}\\ =\frac{1\cdot2\cdot...\cdot29}{2\cdot3\cdot...\cdot30}\cdot\frac{3\cdot4\cdot...\cdot31}{2\cdot3\cdot...\cdot30}\\ =\frac{1}{30}\cdot\frac{31}{2}=\frac{31}{60}\)

b,

\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{98\cdot99\cdot100}\right)\\ =\frac{1}{2}\left(\frac{3-1}{1\cdot2\cdot3}+\frac{4-2}{2\cdot3\cdot4}+...+\frac{100-98}{98\cdot99\cdot100}\right)\\ =\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\\ =\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)\\ =\frac{1}{2}\cdot\frac{4450-1}{9900}=\frac{1}{2}\cdot\frac{4449}{9900}=\frac{4449}{19800}=\frac{1483}{6600}\)

c, (Chịu :V)

d,

\(D=\frac{1}{3}\left(\frac{3}{1\cdot2\cdot3\cdot4}+\frac{3}{2\cdot3\cdot4\cdot5}+...+\frac{3}{27\cdot28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{4-1}{1\cdot2\cdot3\cdot4}+\frac{5-2}{2\cdot3\cdot4\cdot5}+...+\frac{30-27}{27\cdot28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{1}{1\cdot2\cdot3}-\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4}-\frac{1}{3\cdot4\cdot5}+...+\frac{1}{27\cdot28\cdot29}-\frac{1}{28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{1}{6}-\frac{1}{24630}\right)\\ =\frac{228}{4105}\)

Chúc bạn học tốt nha.