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(x+1)+(x+2)+(x+3)+...+(x+100)
=(x+x+...)+(1+2+3+4+...+100)
=x.100+(1+100x100:2)
=x.100+5050=5750
x.100=5750-5050
x.100=700
x=700:100
vậy x=7
( x + 1 ) + ( x + 2 ) + ... + ( x + 100 ) = 5750
<=>( x + x + ... + x ) + ( 1 + 2 + ... + 100 ) = 5750
100 số x
<=> x.100 + ( 100 + 1 ) .100 : 2 = 5750
<=> x . 100 + 5050 = 5750
=> x . 100 = 700
=> x = 7
a) \(\frac{28\times7-45\times7+7\times18}{45\times14}\)
\(=\frac{7\left(28-45+7\right)}{45\times14}\)
\(=\frac{7\times\left(-10\right)}{45\times14}=\frac{-1}{9}\)
b) \(\frac{12.3-2.6}{4.5.6}\)
\(=\frac{2.6.3-2.6}{4.5.6}\)
\(=\frac{2.6\left(3-1\right)}{2.2.5.6}\)
\(=\frac{2.6.2}{2.2.5.6}\)\(=\frac{1}{5}\)
`(15-x)+(x-12)=7-(-5+x)`
`=>15-x+x-12=7+5-x`
`=>3=12-x`
`=>x=12-3`
`=>x=9`
Vậy `x=9`
\(\frac{2}{2.4}+\frac{2}{4.6}+....+\frac{2}{x\left(x+2\right)}=\frac{4}{9}\)
<=> \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{x}-\frac{1}{x+2}=\frac{4}{9}\)
<=> \(\frac{1}{2}-\frac{1}{x+2}=\frac{4}{9}\)
<=> \(\frac{1}{x+2}=\frac{1}{18}\)
=> \(x+2=18\)
<=> \(x=16\)
Vậy...
mình cảm ơn ! hỏi nhìu quá :v
g) 23.15-[115-(12-5)2]
= 8.15-[115-49]
= 120 - 66
= 54
h) {9+[150-(12-4)2]}.12
={9+[150-64]}.12
={9+86}.12
=95.12
=1140
g)=120-[115-49]
=54
h)={9+[150-64]}.12
={9+86}.12
=95.12
=1140
a)
\(\dfrac{6}{13}\cdot\dfrac{8}{7}\cdot\dfrac{-26}{3}\cdot\dfrac{-7}{8}\)
\(=\dfrac{6}{13}\cdot\dfrac{-26}{3}\cdot\dfrac{8}{7}\cdot\dfrac{-7}{8}\)
\(=-4\cdot\left(-1\right)\\ =4\)
b)
\(\dfrac{6}{11}+\dfrac{11}{3}\cdot\dfrac{3}{22}\)
\(=\dfrac{6}{11}+\dfrac{1}{2}\\ =\dfrac{12}{22}+\dfrac{11}{22}\\ =\dfrac{23}{22}\)
Ta có : 25^5 = (5^2)^5 = 5^10 < 5^9
=> -1/5^9 > -1/5^10
Hay -1/5^9 > 1/25^5
(x + 1) + (x + 2) +......+ (x + 100) = 5750
(x + x + ... + x) + (1 + 2 + 3 + ... + 100) = 5750
x . 100 + (1 + 2 + 3 + ... + 100) = 5750
x . 100 + ([1 + 100] . 100 : 2) = 5750
x . 100 + 5050 = 5750
x . 100 = 5750 - 5050
x . 100 = 700
=> x = 700 : 100 = 7
100x=5750-(1+2+3+...+100) => x=0,7