Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) B = \(\dfrac{4}{3}.\dfrac{5}{4}....\dfrac{21}{20}=\dfrac{1}{3}.1.....\dfrac{21}{1}=\dfrac{21}{3}=7\)
b) Em chịu, chưa học số âm :)
a) -32 -4.(x-5) = 0
<=>4.(x-5)=-32
<=>x-5=(-32):4
<=>x+5=-8
<=>x=-8+5
<=>x=-3
Vậy x=-3
b) 13.(x-5)=-169
<=>x-5=(-169):13
<=>x-5=-13
<=>x=-13+5
<=>x=-8
vậy x=-8
c) (-2).x+5 = (-3).(-3)+8
<=>(-2).x+5=17
<=>(-2).x=17-5
<=>(-2).x=12
<=>x=12:(-2)
<=>x=-6
Vậy x=-6
d) (-8).x = (-10).(-2)-4
<=>(-8).x=16
<=>x=16:(-8)
<=>x=-2
vậy x=-2
e) (-9).x+3 = (-2).(-7)+16
<=>(-9).x+3=30
<=>(-9).x=30-3
<=>(-9).x=27
<=>x=27:(-9)
<=>x=-3
Vậy x=-3
\(A=\frac{\left[\left(25-1\right):1+1\right]\left(25+1\right)}{2}=325.\)
\(B=\frac{\left[\left(51-3\right):2+1\right]\left(51+3\right)}{2}=675\)
\(C=\frac{\left[\left(81-1\right):4+1\right]\left(81+1\right)}{2}=861\)
ta có
\(S_2=\left(1-3\right)+\left(5-7\right)+..+\left(1997-1999\right)+2001\)
ha y \(S_2=-2-2-2..+2001=-2.500+2001=1001\)
\(S_3=\left(1-2-3+4\right)+\left(5-6-7+8\right)+..+\left(1997-1998-1999+2002\right)\)
hay \(S_3=0+0+..+0=0\)
\(S_2=\left(1-3\right)+\left(5-7\right)+...+\left(1997-1999\right)+2001\)
\(=\left(-2\right)+\left(-2\right)+....+\left(-2\right)+2001=\left(-2\right).500+2001=-1000+2001=1001\)
\(S_3=\left(0+1-2-3\right)+\left(4+5-6-7\right)+...+\left(1996+1997-1998-1999\right)+2000\)
\(=-4+\left(-4\right)+...+\left(-4\right)+2000=\left(-4\right).500+2000=0\)
`x/8 = 3/4 + (-5/8)`
`=>x/8 =6/8+ (-5/8)`
`=>x/8= 1/8`
`=>x=1`
\(\dfrac{x}{8}=\dfrac{3}{4}+\dfrac{-5}{8}\)
\(\dfrac{x}{8}=\dfrac{6}{8}+\dfrac{5}{8}\)
\(\Rightarrow x=6+5\)
\(x=11\)