Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(B=\left[1+\left(-3\right)\right]+\left[5+\left(-7\right)\right]+........+\left[97+\left(-99\right)\right]+101\)
\(=\left(-2\right)+\left(-2\right)+.......+\left(-2\right)+101\)( có 25 số -2)
\(=\left(-2\right).25+101\)
\(=\left(-50\right)+101\)
\(=51\)
\(B=1+\left(-3\right)+5+\left(-7\right)+...+97+\left(-99\right)+101\)
\(\Rightarrow B=\left[1+\left(-3\right)\right]+\left[5+\left(-7\right)\right]+...+\left[97+\left(-99\right)\right]+101\)
\(\Rightarrow B=\left(-2\right)+\left(-2\right)+...+\left(-2\right)+101\)
có 25 số -2
\(\Rightarrow B=\left(-2\right).25+101\)
\(\Rightarrow B=-50+101\)
\(\Rightarrow B=51\)
a) (–38) + 28 = –(38 – 28) = –10.
b) 273 + (–123) = 273 – 123 = 150.
c) 99 + (–100) + 101 = (99 + 101) + (–100) = 200 + (–100) = 100.
Tính A=\(\left(1-\dfrac{1}{100}\right)\left(1-\dfrac{1}{99}\right)...\left(1+\dfrac{1}{100}\right)\)
\(A=\dfrac{99}{100}\cdot\dfrac{98}{99}\cdot...\cdot\dfrac{1}{2}=\dfrac{1}{100}\)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)
b, \(\left(1-\dfrac{1}{100}\right)\left(1-\dfrac{1}{99}\right)...\left(1-\dfrac{1}{2}\right)=\dfrac{99.98...1}{100.99...2}=\dfrac{1}{100}\)
Lời giải:
a. $(-2018)+2018=2018-2018=0$
b) $57+(-93)=-(93-57)=-36$
c) $(-38)+46=46-38=8$
Mk lm đc câu a thôi nhé !
A= 150-(100-99+98-97+...-3+2-1)
từ 1-100 có 100 SH. Ta nhóm 4 số vs nhau như sau : (100-00+98-87)+(...)+(4-3+2-1)
Có tất cả số nhóm là : 100:4=25 nhóm. Mà mỗi nhóm ta tính có kết quả là 2, vậy tao có
A=150-(2.25)
A=150-50
A=100
ĐKXĐ : 101x \(\ge\)0 nên x \(\ge\)0
khi đó : \(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|=101x\)
\(\Leftrightarrow\left(x+\frac{1}{101}\right)+\left(x+\frac{2}{101}\right)+...+\left(x+\frac{100}{101}\right)=101x\)
\(\Leftrightarrow100x+\frac{5050}{101}=101x\Leftrightarrow x=50\)
*ĐK : 101x\(\ge\) 0 => x\(\ge\)0
=> \(x+\frac{1}{101}\ge\frac{1}{101}>0\Rightarrow\left|x+\frac{1}{101}\right|=x+\frac{1}{101}\)
\(x+\frac{2}{101}\ge\frac{2}{101}>0\Rightarrow\left|x+\frac{2}{101}\right|=x+\frac{2}{101}\)
...
\(x+\frac{100}{101}\ge\frac{100}{101}>0\Rightarrow\left|x+\frac{100}{101}\right|=x+\frac{100}{101}\)
Ta có :
\(x+\frac{1}{101}+x+\frac{2}{101}+...+x+\frac{100}{101}=101x\)
<=> \(100x+\left(\frac{1+2+...+100}{101}\right)=101x\)
<=> \(100x+\frac{5050}{101}=101x\)
<=> \(100x-101x=\frac{-5050}{101}\)
<=> -x = -50
=> x = 50 ( t/m x >/ 0)
\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+.....+\left(1+2+3+4+......+100\right)}{\left(1.100+2.99+3.98+.......+99.2+100.1\right).2013}\)
\(=\frac{1.100+2.99+3.98+......+99.2+100.1}{\left(1.100+2.99+3.98+.....+99.2+100.1\right).2013}\)
\(=\frac{1}{2013}\)
\(A=\frac{38}{45}-\left(\frac{8}{45}-\frac{17}{51}-\frac{3}{11}\right)\)
\(A=\frac{38}{45}-\frac{8}{45}+\frac{17}{51}+\frac{3}{11}\)
\(A=\left(\frac{38}{45}-\frac{8}{45}\right)+\frac{1}{3}+\frac{3}{11}\)
\(A=\left(\frac{2}{3}+\frac{1}{3}\right)+\frac{3}{11}\)
\(A=1+\frac{3}{11}\)
\(A=\frac{11}{11}+\frac{3}{11}=\frac{14}{11}\)
\(B=\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right)...\left(\frac{1}{99}+1\right)\)
\(B=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.\frac{6}{5}...\frac{100}{99}\)
\(B=\frac{3.4.5.6...100}{2.3.4.5...99}\)
\(B=\frac{100}{2}\)
\(B=50\)
a)(-38) + 28= -( \(\left|-38\right|\) - \(\left|28\right|\)) = - (38-28)= - 10
b) 272+ (-123)= (\(\left|272\right|\) - \(\left|-123\right|\) ) = 272- 123= 149
c) 99+(-100)+101= -( 100-99) +101= -1+ 101=100
a) (-38)+28 = -(38-28)= -10
b) 272 + (-273)= -(273-272)= -1
c) 99+ (-100)+ 101= (99+101)-100= 200-100= 100