Tính :
a) \(\left(-38\right)+28\)
b) \(272+\left(-123\right)\)
c) \(99+\left(-100\right)+101\)
Tính:
\(B=1+\left(-3\right)+5+\left(-7\right)+.......+97+\left(-99\right)+101\)
\(B=\left[1+\left(-3\right)\right]+\left[5+\left(-7\right)\right]+........+\left[97+\left(-99\right)\right]+101\)
\(=\left(-2\right)+\left(-2\right)+.......+\left(-2\right)+101\)( 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
\(\Rightarrow B=\left(-2\right).25+101\)
\(\Rightarrow B=-50+101\)
\(\Rightarrow B=51\)
Trả lời:
\(B=1+\left(-3\right)+5+\left(-7\right)+...+97+\left(-99\right)+101\)(Có 51 số)
\(B=\left[1+\left(-3\right)\right]+\left[5+\left(-7\right)\right]+...+\left[97+\left(-99\right)\right]+101\)
\(B=\left(-2\right)+\left(-2\right)+...+\left(-2\right)+101\)( Có 25 số (-2) )
\(B=\left(-2\right)\times25+101\)
\(B=\left(-50\right)+101\)
\(B=51\)
Vậy\(B=51\)
Hok tốt!
Good girl
Tính:
a) (-38 ) + 28
b) 273 + (-123)
c) 99 + (-100) + 101
a) (–38) + 28 = –(38 – 28) = –10.
b) 273 + (–123) = 273 – 123 = 150.
c) 99 + (–100) + 101 = (99 + 101) + (–100) = 200 + (–100) = 100.
Tính A=\(\left(1-\dfrac{1}{100}\right)\left(1-\dfrac{1}{99}\right)...\left(1+\dfrac{1}{100}\right)\)
\(A=\dfrac{99}{100}\cdot\dfrac{98}{99}\cdot...\cdot\dfrac{1}{2}=\dfrac{1}{100}\)
5. Tìm x biết:
a, \(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+...+\left|x+10\right|=11x+1\)
b, \(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|=101x\)
Tính hợp lí:
a) (100-9)(99-9)(98-9)...(1-9).
b) \(\left(1-\dfrac{1}{100}\right)\left(1-\dfrac{1}{99}\right)...\left(1-\dfrac{1}{2}\right)\)
c) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)
b, \(\left(1-\dfrac{1}{100}\right)\left(1-\dfrac{1}{99}\right)...\left(1-\dfrac{1}{2}\right)=\dfrac{99.98...1}{100.99...2}=\dfrac{1}{100}\)
Tính:
a) \(\left(-2018\right)+2018;\) b) \(57+\left(-93\right);\) c) \(\left(-38\right)+46.\)
Lời giải:
a. $(-2018)+2018=2018-2018=0$
b) $57+(-93)=-(93-57)=-36$
c) $(-38)+46=46-38=8$
Tính:
a) (−2018)+2018 =0
b) 5+(−93) = -88
c) (−38)+46 = 8
a) (- 2018) + 2018 = - (2018 – 2018) = 0
b) 57 + (- 93) = (93 – 57) = 38
c) (- 38) + 46 = - (38 – 48) = 8
$\left(100+\frac{99}{2}+\frac{98}{3}+...+\frac{1}{100}\right)\div \left(\frac{1}{2}+\frac{1}{3}+..+\frac{1}{101}\right)-2
Tính hợp lý
a) \(\left(\frac{2}{5}\right)^6.\left(\frac{25}{4}\right)^2\)
b) \(\frac{100}{123}:\left(\frac{3}{4}+\frac{7}{12}\right)+\frac{23}{123}:\left(\frac{9}{5}-\frac{7}{15}\right)\)
\(a,\left(\frac{2}{5}\right)^6.\left(\frac{25}{4}\right)^2=\left(\frac{2}{2.3}\right)^6.\left(\frac{5}{2}\right)^4\)
\(=\frac{1}{3^6}.\frac{5^4}{2^4}=\frac{5^4}{3^6.2^4}\)
\(b,\frac{100}{123}:\left(\frac{3}{4}+\frac{7}{12}\right)+\frac{23}{123}:\left(\frac{9}{5}-\frac{7}{15}\right)\)
\(=\frac{100}{123}:\left(\frac{9+7}{12}\right)+\frac{23}{123}:\left(\frac{27-7}{15}\right)\)
\(=\frac{100}{123}:\frac{16}{12}+\frac{23}{123}:\frac{20}{15}\)
\(=\frac{100.12}{123.16}+\frac{23.15}{123.20}\)
\(=\frac{5.5.4.3.4}{41.3.4.4}+\frac{23.3.5}{41.3.4.5}\)
\(=\frac{25}{41}+\frac{23}{164}=\frac{25.4+23}{164}\)
\(=\frac{123}{164}=\frac{3}{4}\)
Tìm x, biết
\(\frac{1}{x\left(x+1\right)}\)+ \(\frac{1}{\left(x+1\right)\left(x+2\right)}\)+ ... + \(\frac{1}{\left(x+99\right)\left(x+100\right)}\)= \(\frac{100}{101}\)
=> ĐK: \(x\ne\left\{0;-1;-2;...;-99;-100\right\}\)
Đây là dạng dãy số đặc biệt, bạn có thể giải như sau:
Ta có:
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+99\right)\left(x+100\right)}=\frac{100}{101}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+99}-\frac{1}{x+100}=\frac{100}{101}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+100}=\frac{100}{101}\)
\(\Leftrightarrow\frac{x+100-x}{x.\left(x+100\right)}=\frac{100}{101}\)
\(\Leftrightarrow\frac{100}{x^2+100x}=\frac{100}{101}\)
\(\Leftrightarrow x^2+100x=101\)
\(\Leftrightarrow x^2+100x-101=0\)
\(\Leftrightarrow x^2+101x-x-101=0\)
\(\Leftrightarrow x\left(x+101\right)-\left(x+101\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+101\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+101=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\left(n\right)\\x=-101\left(n\right)\end{cases}}\)
Vậy: S={1;-101)