(3/2 + 7/6 + 13/12 + ... + 421/420 ) - x =20/21
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3/2+7/6+13/12+21/20+...+133/132
=1+ 1/2 + 1+ 1/6 + 1+ 1/12 + 1+ 1/20+......+ 1+ 1/132
=(1+ 1 + 1+.....+ 1) + ( 1/2+ 1/6+1/12+ 1/20+....+ 1/132)
= 11+ ( 1/1×2+ 1/2×3+1/3×4+1/4×5+.....+1/11×12)
=11+ ( 1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+......+1/11-1/12)
=11+(1-1/12)
= 11+11/12
=143/12
3/2+7/6+13/12+21/20+...+133/132 =1+ 1/2 + 1+ 1/6 + 1+ 1/12 + 1+ 1/20+......+ 1+ 1/132 =(1+ 1 + 1+.....+ 1) + ( 1/2+ 1/6+1/12+ 1/20+....+ 1/132) = 11+ ( 1/1×2+ 1/2×3+1/3×4+1/4×5+.....+1/11×12) =11+ ( 1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+......+1/11-1/12) =11+(1-1/12) = 11+11/12 =143/12
5.\(-\dfrac{3}{7}+\dfrac{5}{13}+\dfrac{-4}{12}=-\dfrac{103}{273}\)
b.\(-\dfrac{5}{21}+\dfrac{-2}{21}+\dfrac{8}{24}=\dfrac{-5-2}{21}+\dfrac{8}{24}=-\dfrac{7}{21}+\dfrac{8}{24}=-\dfrac{1}{3}+\dfrac{8}{24}=0\)
c.\(\dfrac{5}{13}+\dfrac{-5}{7}+\dfrac{-20}{41}+\dfrac{8}{13}+\dfrac{-21}{41}=\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\left(\dfrac{-20}{41}+\dfrac{-21}{41}\right)+-\dfrac{5}{7}=1-1-\dfrac{5}{7}=-\dfrac{5}{7}\)
c) Ta có: \(\dfrac{3}{5}+\dfrac{-5}{20}+\dfrac{30}{75}+\dfrac{-7}{4}\)
\(=\dfrac{3}{5}+\dfrac{2}{5}+\dfrac{-1}{4}+\dfrac{-7}{4}\)
\(=1-2=-1\)
Giải:
a)-1/12+4/3=-1/12+16/12=15/12=5/4
b)(-4/14-3/15)-(1/5-20/35-(-1)).7
=-17/35-22/35.7
=-17/35-22/5
=-171/35
c)3/5+-5/20+30/75+-7/4
=3/5+-1/4+2/5+-7/4
=(3/5+2/5)+(-1/4+-7/4)
=1+-2
=-1
d)5/6.-12/14+7/13
=-5/7+7/13
=-16/91
e)2/-9-5/-36-1/4
=-1/12-1/4
=-1/3
f)2/23+-5/12+7/18+21/23+-7/12
=(2/23+21/23)+(-5/12+-7/12)+7/18
=1+-1+7/18
=7/18
a: =>x=7-20=-13
b: =>x=-18+12=-6
c: =>x=9 hoặc x=-6
d: =>x=0 hoặc x=4
e: =>6-x=13-3+14=24
=>x=-18
Câu g và h đề thiếu rồi bạn