Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
\(=\dfrac{-3-39}{32}+\dfrac{-6-11}{17}+\dfrac{-1}{6}=-\dfrac{21}{16}+\dfrac{-1}{6}-1=-\dfrac{119}{48}\)
Bài 2:
\(\Leftrightarrow x:5=-\dfrac{13}{20}\)
hay x=-13/4
a) (-25)+72-80
=47-80
=-33
b) (-3) . 5 + 28 - (-15)
=-15 +28+15
=(15-15)+28
=0+28
=28
c) (-42) : 7 + (-14) - (-3) . 4
=(-6)-14-(-12)
=-20+12
=-8
d) 15 - 6 : (-3) + (-7) . (-2)
=15-(-2)+14
=17+14
=21
a) (-2) . ( x+7 ) + (-5) = 7
<=>(-2).(x+7)=7+5
<=>x+7=12:(-2)
<=>x+7=-6
<=>x=(-6)-7
<=>x=-13
Vậy x=-13
b)(x+4) : (-7) = 14
<=>x+4=14 x (-7)
<=>x+4=-98
<=>x=-98-4
<=>x=-102
Vậy x= -102
c) 72 : ( x+5) - 4 = -12
<=>72:(x+5)=(-12)+4
<=>x+5=72:(-8)
<=>x+5=-9
<=>x=-9-5
<=>x=-14
Vậy x= -14
d) (x+3) : (-6 ) + 12 = 8
<=>(x+3) :(-6)=8-12
<=>x+3=(-4)x(-6)
<=>x+3=24
<=>x=24-3
<=>x=21
Vậy x= 21
\(105-\left[\left(2x+7\right)-13\right]=\left(-15\right)^{10}:\left(9^5.5^8\right)\\ 105-\left[\left(2x+7\right)-13\right]=25\\ \left(2x+7\right)-13=105-25\\ \left(2x+7\right)-13=80\\ 2x+7=80+13\\ 2x+7=93\\ 2x=93-7\\ 2x=86\\ x=\dfrac{86}{2}\\ x=43\)
\(105-\left[\left(2x+7\right)-13\right]=\left(-15\right)^{10}:\left(9^5.5^8\right)\\ 105-\left[\left(2x+7\right)-13\right]=15^{10}:3^{10}:5^8\\ 105-\left[\left(2x+7\right)-13\right]=5^{10}:5^8\\ 105-\left[\left(2x+7\right)-13\right]=25\\ \left(2x+7\right)-13=105-25\\ \left(2x+7\right)-13=80\\ 2x+7=80+13\\ 2x+7=93\\ 2x=93-7\\ 2x=86\\ x=86:2\\ x=43\)
Ta có: \(S=1-2-3+4+5-6-7+8+9-...-1998-1999+2000+2001\)
\(\Leftrightarrow S=\left(1-2\right)-\left(3-4\right)+\left(5-6\right)-\left(7-8\right)+...-\left(1999-2000\right)+2001\)
\(\Leftrightarrow S=\left(-1\right)-\left(-1\right)+\left(-1\right)-\left(-1\right)+...-\left(-1\right)+2001\) ( có 500 chữ số \(-1\))
\(\Leftrightarrow S=2001\)
a)\(S=1+5+5^2+...+5^{10}\)
\(5S=5+5^2+5^3+...+5^{11}\)
\(5S-S\)hay 4S\(=5^{11}-1\)
\(\Rightarrow S=\left(5^{11}-1\right):4\)
b)\(S=1+7+7^2+...+7^{10}\)
\(7S=7+7^2+7^3+...+7^{11}\)
\(7S-S\)hay 6S\(=7^{11}-1\)
\(\Rightarrow S=\left(7^{11}-1\right):6\)
Học tốt nha!!!