Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{13}{90}\)+ \(\frac{1}{4}\)+ \(\frac{77}{90}\)
= \(\left(\frac{13}{90}+\frac{77}{90}\right)+\frac{1}{4}\)
= 1 + \(\frac{1}{4}\)
= \(\frac{5}{4}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{56}=\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}...\frac{1}{7x8}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)\(-\frac{1}{8}=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
b,
Lời giải:
$A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}$
$2\times A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}$
$2\times A-A=(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32})-(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64})$
$A=1-\frac{1}{64}=\frac{63}{64}$
1/2 + 2/3 - 3/4 * 4/5 * 5/6 * 6/7 * 7/8 * 8/9 *...* 29/30 : 6/10
= 1/2 + 2/3 - 1/10 : 6/10
= 1/2 + 2/3 - 1/6
= 1
\(\dfrac{3}{8}+\dfrac{9}{2}\times\dfrac{12}{16}-\dfrac{9}{4}:3\)
= \(\dfrac{3}{4}\times\dfrac{1}{2}+\dfrac{9}{2}\times\dfrac{3}{4}-\dfrac{3}{4}\times3:3\)
= \(\dfrac{3}{4}\times\dfrac{1}{2}+\dfrac{9}{2}\times\dfrac{3}{4}-\dfrac{3}{4}\times1\)
= \(\dfrac{3}{4}\times\left(\dfrac{1}{2}+\dfrac{9}{2}-1\right)\)
= \(\dfrac{3}{4}\times\left(5-1\right)\)
= \(\dfrac{3}{4}\times4\)
= 3
a.
1+2+3+...+99+100
=(1+100)*100/2
=5050
b.12+14+16+...+38
=(12+38)*[(38-12)/2+1]/2
=350
=\(\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}\)
=\(\frac{15}{16}\)
k mk nhé bạn
8/16 + 4/16 + 2/16 + 1/16 = 15/16