\(\frac{x+1}{99}+\frac{x+2}{99}=\frac{x+10}{99}+\frac{x+20}{99}\)
\(\frac{x+1}{99}+\frac{x+2}{99}=\frac{x+10}{99}+\frac{x+20}{99}\)
X=-100
bạn cộng 2 vế vs 2 r nhóm là xong
\(\frac{x+1}{99}+\frac{x+2}{99}=\frac{x+10}{99}+\frac{x+20}{99}\)
Làm chi tiết giúp mk nha
phương trình này nhìn từ đầu cũng bik vô nghiệm ko có x
\(\frac{x+1}{99}+\frac{x+2}{99}=\frac{x+10}{99}+\frac{x+20}{99}\)
Nhân 2 vế cho 99 ta được:
\(99.\left(\frac{x+1}{99}+\frac{x+2}{99}\right)=99.\left(\frac{x+10}{99}+\frac{x+20}{99}\right)\)
=>x+1+x+2=x+10+x+20
=>2x+3=2x+30
=>0x=27 (vô lí)
Vậy ko tìm dc x
\(\frac{x+1}{99}+\frac{x+2}{99}=\frac{x+10}{90}+\frac{x+20}{80}\)
khẳng định là sai còn nếu đúng thì quy đồng lên làm đảm bảo số lớn khủng
\(\frac{x-99-1}{99}-\frac{x-99-1}{98}-\frac{x-99-1}{97}-\frac{x-99-1}{96}-\frac{x-99-1}{95}-\frac{x-99-1}{94}\)=0
Ta có :
\(\frac{x-99-1}{99}-\frac{x-99-1}{98}-\frac{x-99-1}{97}-\frac{x-99-1}{96}-\frac{x-99-1}{95}-\frac{x-99-1}{94}=0\)
\(\Leftrightarrow\)\(\frac{x-100}{99}-\frac{x-100}{98}-\frac{x-100}{97}-\frac{x-100}{96}-\frac{x-100}{95}-\frac{x-100}{94}=0\)
\(\Leftrightarrow\)\(\left(x-100\right)\left(\frac{1}{99}-\frac{1}{98}-\frac{1}{97}-\frac{1}{96}-\frac{1}{95}-\frac{1}{94}\right)=0\)
Vì \(\frac{1}{99}-\frac{1}{98}-\frac{1}{97}-\frac{1}{96}-\frac{1}{95}-\frac{1}{94}\ne0\)
Nên \(x-100=0\)
\(\Rightarrow\)\(x=100\)
Vậy \(x=100\)
Bài làm mang tính chất tham khảo vì em mới lớp 7 ~
\(\frac{x+1}{99}+\frac{x+2}{99}+\frac{x+3}{99}+\frac{x+4}{99}=-4\)
Ai giải giúp mình với mình ktick cho :D
\(\frac{x+1}{99}+\frac{x+2}{99}+\frac{x+3}{99}+\frac{x+4}{99}=-4\)
=>\(\frac{\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+\left(x+4\right)}{99}=-4\)
=> (x+1)+(x+2)+(x+3)+(x+4)=-4.99=-396
=>4x+10=-396
4x=-406
x=-406:4=-101,5
Tính \(T=\left(\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}\right)X\left(\frac{1}{99}+\frac{2}{98}+...+\frac{98}{2}\right)-\left(\frac{1}{99}+\frac{2}{98}+..+\frac{99}{1}\right)X\left(\frac{2}{98}+\frac{3}{97}+...+\frac{98}{2}\right)\)
Tìm x:
\(\frac{x-1}{99}\)+\(\frac{x-2}{98}+\frac{x-5}{95}=3+\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)
\(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=3+\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)\(\frac{1}{95}\)
\(\frac{1}{95}\frac{1}{95}\)
là sao ???
\(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=3+\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)
\(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=\frac{2765070}{921690}+\frac{9310}{921690}+\frac{9405}{921690}+\frac{9702}{921690}\)
\(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=\frac{2793487}{921690}\)
\(BCNN\left(99,98,95\right)=921690\Rightarrow x=101\)
tìm x
a, (x+20):99 =(1004+1)x\(\frac{2}{99}\)
b, \(\frac{x+140}{x}\)+260= 21+65x4
(x + 20) : 99 = (1004 + 1) x 2/99
=> 99x + 1980 = 1005 x 2/99
=> 99x = 99495/2 - 1980
=> 99x = 95535/2
=> x = 965/2
b. \(\frac{x+140}{x}\) + 260 = 21 + 65 x 4
=> \(\frac{x+140}{x}\)= 21 + 260 - 260
=> \(\frac{x+140}{x}\)= 21
=> 21x = x + 140
=> 20x = 140
=> x = 7