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D
1
19 tháng 6 2017
a) Ta có : $1.3+2.4+3.5+...+99.101+100.102$
$=(2-1)(2+1)+(3-1)(3+1)+(4-1)(4+1)+...+(100-1)(100+1)+(101-1)(101+1)$
$=2^2-1+3^2-1+4^2-1+...+100^2-1+101^2-1$
$=(2^2+3^2+4^2+...+100^2+101^2)-100$
b) $1.100+2.99+3.98+...+99.2+100.1$
$=1.100+2.(100-1)+3.(100-2)+...+99.(100-98)+100.(100-99)$
$=100(1+2+3+...+99+100)-(1.2+2.3+...+99.100)$
$=100.\dfrac{101.100}{2}-\dfrac{99.100.101}{3}=171700$
VD
10 tháng 6 2016
S = 1.2.3.4 + 2.3.4.5 + 3.4.5.6+...97.98.99.100
5S = (1.2.3.4+2.3.4.5+3.4.5.6+ ... + 97.98.99.100).5
5S = 1.2.3.4.(5-0) + 2.3.4.5.(6-1)+ 3.4.5.6(7-2)+......+ 97.98.99.100.(101-96)
5S = (1.2.3.4.5 + 2.3.4.5.6 + 3.4.5.6.7 + ....+ 97.98.99.100.101) - (0.1.2.3.4 + 1.2.3.4.5 + 2.3.4.5.6+.....+96.97.98.99.100)
5S = 97.98.99.100.101
S= 97.98.99.100.101/5
S=1901009880
10 tháng 6 2016
S=1*2*3*4+2*3*4*5+....+97*98*99*100
5S=1.2.3.4.5+2.3.4.5.5+...+97.98.99.100.5
5S=1.2.3.4.(5-0)+2.3.4.5.(6-1)+...+97.98.99.100.(101-96)
5S=1.2.3.4.5-0.1.2.3.4+2.3.4.5.6-1.2.3.4.5+...+97.98.99.100.101-96.97.98.99.100
5S=(1.2.3.4.5+2.3.4.5.6+...+97.98.99.100.101)-(0.1.2.3.4+1.2.3.4.5+...+96.97.98.99.100)
5S=97.98.99.100.101
S=9505049400:5=1901009880.
C
2
20 tháng 9 2016
\(1^2-2^2+3^2-4^2+...+97^2-98^2+99^2-100^2=\left(1-2\right)\left(1+2\right)+\left(3-4\right)\left(3+4\right)+...+\left(97-98\right)\left(97+98\right)+\left(99-100\right)\left(99+100\right)\)\(=-\left(1+2+3+4+...+97+98+99+100\right)\)
\(=-\left(\frac{101\times100}{2}\right)=-5050\)
23 tháng 1 2021
a) Ta có: \(\dfrac{x-3}{5}=6-\dfrac{1-2x}{3}\)
\(\Leftrightarrow\dfrac{3\left(x-3\right)}{15}=\dfrac{90}{15}-\dfrac{5\left(1-2x\right)}{15}\)
\(\Leftrightarrow3x-9=90-5+10x\)
\(\Leftrightarrow3x-9=10x+85\)
\(\Leftrightarrow3x-10x=85+9\)
\(\Leftrightarrow-7x=94\)
hay \(x=-\dfrac{94}{7}\)
Vậy: \(S=\left\{-\dfrac{94}{7}\right\}\)
b) Ta có: \(\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\)
\(\Leftrightarrow\dfrac{2\left(3x-2\right)}{12}-\dfrac{60}{12}=\dfrac{3\left(3-2x-14\right)}{12}\)
\(\Leftrightarrow6x-4-60=9-6x-42\)
\(\Leftrightarrow6x-64=-6x-33\)
\(\Leftrightarrow6x+6x=-33+64\)
\(\Leftrightarrow12x=31\)
hay \(x=\dfrac{31}{12}\)
Vậy: \(S=\left\{\dfrac{31}{12}\right\}\)
c) Ta có: \(3\left(x-1\right)+3=5x\)
\(\Leftrightarrow3x-3+3=5x\)
\(\Leftrightarrow3x-5x=0\)
\(\Leftrightarrow-2x=0\)
hay x=0
Vậy: S={0}
d) Ta có: \(\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\)
\(\Leftrightarrow\dfrac{x+1}{100}+1+\dfrac{x+2}{99}+1=\dfrac{x+3}{98}+1+\dfrac{x+4}{97}+1\)
\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}=\dfrac{x+101}{98}+\dfrac{x+101}{97}\)
\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}-\dfrac{x+101}{98}-\dfrac{x+101}{97}=0\)
\(\Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\right)=0\)
mà \(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\ne0\)
nên x+101=0
hay x=-101
Vậy: S={-101}
23 tháng 1 2021
a) \(\dfrac{x-3}{5}=6-\dfrac{1-2x}{3}\\ \Leftrightarrow\dfrac{3\left(x-3\right)}{15}=\dfrac{90-5\left(1-2x\right)}{15}\\ \Leftrightarrow3x-9=90-5+10x\\ \Leftrightarrow3x-10x=90-5+9\\ \Leftrightarrow-7x=94\\ \Leftrightarrow x=\dfrac{-94}{7}\)
Vậy \(x=\dfrac{-94}{7}\) là nghiệm của pt
b) \(\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\\ \Leftrightarrow\dfrac{2\left(3x-2\right)-60}{12}=\dfrac{9-6\left(x+7\right)}{12}\\ \Leftrightarrow6x-4-60=9-6x-42\\ \Leftrightarrow6x+6x=9-42+4+60\\ \Leftrightarrow12x=31\\ \Leftrightarrow x=\dfrac{31}{12}\)
Vậy \(x=\dfrac{31}{12}\) là nghiệm của pt
c) \(3\left(x-1\right)+3=5x\\ \Leftrightarrow3x+3+3=5x\\ \Leftrightarrow5x-3x=3+3\\ \Leftrightarrow2x=6\\ \Leftrightarrow x=3\)
Vậy x = 3 là nghiệm của pt
d) \(\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\\ \Leftrightarrow\left(\dfrac{x+1}{100}+1\right)+\left(\dfrac{x+2}{99}+1\right)=\left(\dfrac{x+3}{98}+1\right)+\left(\dfrac{x+4}{97}+1\right)\\ \Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}-\dfrac{x+101}{98}-\dfrac{x+101}{97}=0\\ \Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\right)=0\\ \Leftrightarrow x+101=0\\ \Leftrightarrow x=-101\)
Vậy x = -101 là nghiệm của pt
e) \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}=-4\\ \Leftrightarrow\left(\dfrac{59-x}{41}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{53-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)=0\\ \Leftrightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}=0\\ \Leftrightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}\right)=0\\ \Leftrightarrow100-x=0\\ \Leftrightarrow x=100\)
Vậy x = 100 là nghiệm của pt
f) \(\dfrac{x-90}{10}+\dfrac{x-76}{12}+\dfrac{x-58}{14}+\dfrac{x-36}{16}+\dfrac{x-15}{17}=15\\ \Leftrightarrow\left(\dfrac{x-90}{10}-1\right)+\left(\dfrac{x-76}{12}-2\right)+\left(\dfrac{x-58}{14}-3\right)+\left(\dfrac{x-36}{16}-4\right)+\left(\dfrac{x-15}{17}-5\right)=0\\ \Leftrightarrow\dfrac{x-100}{10}+\dfrac{x-100}{12}+\dfrac{x-100}{14}+\dfrac{x-100}{16}+\dfrac{x-100}{17}=0\\ \Leftrightarrow\left(x-100\right)\left(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\right)=0\\ \Leftrightarrow x-100=0\\ \Leftrightarrow x=100\)
Vậy x = 100 là nghiệm của pt
TN
17 tháng 6 2017
\(a,A=-1+3-5+7-9+...-2013+2015-2017=\left(-1+3\right)+\left(-5+7\right)+...+\left(-2013+2015\right)-2017\)\(=2+2+..+2-2017\)
\(=2.504-2017=-1009\)
\(b,B=2-4+6-8+...+2014-2016+2018\)\(=2+\left(-4+6\right)+\left(-8+10\right)+...+\left(-2016+2018\right)==2+2+...+2\)\(=2+503.2=1008\)
UI
20 tháng 8 2020
\(=\left(2^2-1^2\right)+\left(4^2-3^2\right)+...+\left(100^2-99^2\right)\)
\(=\left(2+1\right)\left(2-1\right)+\left(4+3\right)\left(4-3\right)+...+\left(100+99\right)\left(100-99\right)\)
\(=1+2+3+4+...+100=\frac{\left(100+1\right).100}{2}=5050\)
B
21 tháng 8 2020
Bài làm :
Ta có :
\(-1^2+2^2-3^2+4^2-5^2+....+100^2\)
\(=\left(2^2-1^2\right)+\left(4^2-3^2\right)+....+\left(100^2-99^2\right)\)
\(=\left(2+1\right)\left(2-1\right)+\left(4+3\right)\left(4-3\right)+....+\left(100+99\right)\left(100-99\right)\)
\(=1+2+3+4+....+100=\frac{\left(100+1\right).100}{2}=5050\)
Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
TD
4
ND
Nguyễn Đức Trí
VIP
10 tháng 8 2023
\(A=-1^2+2^2-3^2+4^2-...-99^2+100^2\)
\(A=\left(2-1\right).\left(1+2\right)+\left(4-3\right).\left(3+4\right)+...\left(+100-99\right).\left(99+100\right)\)
\(A=1.\left(1+2+3+...+99+100\right)\)
\(A=\dfrac{100.\left(100+1\right)}{2}=50.101=5050\)
MT
23 tháng 6 2015
a. M=-1^2+2^2-3^2+4^2-...-99^2+100^2.
M=(2-1)(2+1)+(4-3)(4+3)+...+(100-99)(100+99)
M=3+7+...+199
=>2M=3+7+...+199+3+7+...+199 (198 số)
=(3+199)+(7+195)+...+(199+3) (99 cặp)
=202.99
=19998
=>M=19998:2=9999