Tìm x biết (x+1) + (x+2) + (x+3) + (x+4) + .......+ (x+98) + (x+99)=9900
Tìm x biết ( x + 1 ) + ( x + 2 ) + .......+ ( x + 98 ) + ( x + 99) = 9900
x + 1 + x + 2 + ... + x + 98 + x + 99 = 9900
( x + x + ... + x ) + ( 1 + 2 + ... + 98 + 99 ) = 9900
Số số hạng là : ( 99 - 1 ) : 1 + 1 = 99 ( số )
Tổng là : ( 99 + 1 ) x 99 : 2 = 4950
99x + 4950 = 9900
99x = 4950
x = 50
Vậy,......
( x + 1 ) + ( x + 2 ) + .......+ ( x + 98 ) + ( x + 99 ) = 9900
x + 1 + x + 2 + ... + x + 98 + x + 99 = 9900
( x + x + ... + x + x ) + ( 1 + 2 + ... + 98 + 99 ) = 9900
99x + 4950 = 9900
99x = 4950
x = 50
Vậy x = 50
2*x+5*x-3*x=125:4+27:3
(x+1)+(x+2)+...+(x+98)+(x+99)=9900
( x + 1 ) + ( x + 2 ) + ... + ( x + 98 ) + ( x + 99 ) = 9900
( x + x + ... + x + x ) + ( 1 + 2 + ... + 98 + 99 ) = 9900
99.x + \(\frac{\left(99+1\right).99}{2}\)= 9900
99.x + 4950 = 9900
99.x = 9900 - 4950
99.x = 4950
x = 4950 : 99
x = 50
2 . x + 5 . x - 3 . x = 125 : 4 + 27 : 3
x . ( 2 + 5 - 3 ) = 31,25 + 9
x . 4 = 40,25
x = 40,25 : 4
x = 10,0625
tìm x
(x+1)+(x+2)+......+(x+98)+(x+99)=9900
(x+1)+(x+2)+...+(x+98)+(x+99)=9900
x.99+(1+2+3+...+98+99)=9900
x.99+[(99-1):1+1].(99+1):2=9900
x.99+99.100:2
x.99+99.50=9900
x.99+4950=9900
x.99=9900-4950
x.99=4950
x=4950:99
x=50
chúc bạn học tốt nha
\(\left(x+1\right)+\left(x+2\right)+...+\left(x+98\right)+\left(x+99\right)=9900\)
\(\left(x+x+x+...+x+x\right)+\left(1+2+3+...+99\right)=9900\)
\(\left(99\cdot x\right)+\left(100\times99\div2\right)=9900\)
\(99x+4950=9900\)
\(99x=9900-4950\)
\(x=4950\div99\)
\(x=50\)
=> :(x+x+x+...+x)(co 99so hang)+(1+2+3+...+99)=9900
=> :x.99+(99+1).99:2=4950=9900
=>:x.99+4950=9900
=>:x.99=9900-4950=4950
=>x=4950:90=55
vay x=55
(x+1)+(x+2)+(x+3)+...+(x+98)+(x+99)=9900
\(\Rightarrow99x+\left(1+2+3+...+98+99\right)=9900\)(vì có 99 số hạng nha)
\(\Rightarrow99x+4950=9900\)
\(\Rightarrow99x=4950\)
\(\Rightarrow x=50\)
(x+1)+(x+2)+...+(x+98)+(x+99)=9900
(x+1)+(x+2)+...+(x+98)+(x+99)=9900
x+x+x+x+...x+(1+2+...+98+99)=9900
50x+2500=9900
=>50x=7400
vậy x=148
(x+1)+(x+2)+...+(x+98)+(x+99)=9900
giải chi tiết giúp mình hen mình cảm ơn
\(\left(x+1\right)+\left(x+2\right)+...+\left(x+99\right)=9900\)
\(\Leftrightarrow99x+\left(\frac{99-1}{1}+1\right)=9900\)
\(\Leftrightarrow99x=9900-99\)
\(\Leftrightarrow x=99\)
k mk nha
(x + 1) +( x + 2) + ... + (x + 99 )= 9900
=>99x +(99-1/1 + 1 )=9900
=>99x=9900-99
=>x=90
ấn chậm quá
@@
làm lại nha . làm theo này chăc chắn hơn
(x+1)+(x+2)+...+(x+98)+(x+99)=9900
=>99x + ( 1+2+3+...+99)=9900
=>99x + 4950 = 9900
=> 99x = 4950
=> x= 50
vậy ...
Chắc chắn #
tìm x biết (x + 1) /100+ (x + 2)/99 =(x + 3)/98 + (x + 4)/97
Ta có :
\(\frac{x+1}{100}+\frac{x+2}{99}=\frac{x+3}{98}+\frac{x+4}{97}\)
\(\Leftrightarrow\)\(\left(\frac{x+1}{100}+1\right)+\left(\frac{x+2}{99}+1\right)=\left(\frac{x+3}{98}+1\right)+\left(\frac{x+4}{97}+1\right)\)
\(\Leftrightarrow\)\(\frac{x+101}{100}+\frac{x+101}{99}=\frac{x+101}{98}+\frac{x+101}{97}\)
\(\Leftrightarrow\)\(\frac{x+101}{100}+\frac{x+101}{99}-\frac{x+101}{98}-\frac{x+101}{97}=0\)
\(\Leftrightarrow\)\(\left(x+101\right)\left(\frac{1}{100}+\frac{1}{99}-\frac{1}{98}-\frac{1}{97}\right)=0\)
Vì \(\frac{1}{100}+\frac{1}{99}-\frac{1}{98}-\frac{1}{97}\ne0\)
Nên \(x+101=0\)
\(\Rightarrow\)\(x=-101\)
Vậy \(x=-101\)
Chúc bạn học tốt ~
Tìm x biết : (x-1)/ 99 + (x-2) /98 + (x-3) /97 + (x-4) /96 + (x-5) / 95 = 5
\(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-3}{97}+\frac{x-4}{96}+\frac{x-5}{95}=5\)
\(\Rightarrow\left(\frac{x-1}{99}-1\right)+\left(\frac{x-2}{98}-1\right)+\left(\frac{x-3}{97}-1\right)+\left(\frac{x-4}{96}-1\right)+\left(\frac{x-5}{95}-1\right)\)\(=5-1-1-1-1-1\)
\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{98}+\frac{x-100}{97}+\frac{x-100}{96}+\frac{x-100}{95}=0\)
\(\Rightarrow\left(x-100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}+\frac{1}{95}\right)=0\)
Mà \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}+\frac{1}{95}\ne0\)
\(\Rightarrow x-100=0\)
\(\Rightarrow x=100\)
Vậy x=100
Chúc bạn học tốt
Tìm x, biết:
a) 1 1/5 . x + 2/3 . x = - 56/125
b) x+1/99 + x+2/98 + x+3/97 + x+4/96 = - 4
Ta có \(1\frac{1}{5}x+\frac{2}{3}x=-\frac{56}{125}\)
<=> \(\frac{6}{5}x+\frac{2}{3}x=-\frac{56}{125}\)
<=> \(\frac{28}{15}x=-\frac{56}{125}\)
<=> \(x=-\frac{2}{15}\)
b) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)
<=> \(\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=0\)
<=> \(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)
<=> \(\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)
<=> x + 100 = 0
<=> x = -100
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