99 x 95 + 99 x 3 + 99 + 99
Giải pt : ( x-1/99 +x-99 )+( x-3/97 + x-97/3 )+ (x-5/95 + x-95/5 )=6
\(\Leftrightarrow\left(\dfrac{x-1}{99}-1\right)+\left(\dfrac{x-99}{1}-1\right)+\left(\dfrac{x-3}{97}-1\right)+\left(\dfrac{x-97}{3}-1\right)+\left(\dfrac{x-5}{95}-1\right)+\left(\dfrac{x-95}{5}-1\right)=0\)
=>x-100=0
=>x=100
(x-1/99+x-99)+(x-3/97+x-7/93)+(x-5/95+x-95/5)=6
\(\frac{x-99-1}{99}-\frac{x-99-1}{98}-\frac{x-99-1}{97}-\frac{x-99-1}{96}-\frac{x-99-1}{95}-\frac{x-99-1}{94}\)=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 ~
a,(x-1/99+x-99)+(x-3/97+x-7/93)+(x-5/95+x-95/5)=6
(x-1)/99+(X-2)/98+(X-5)/95=3+1/99+1/98+1/95
Giải từng bước dùm mình nhé
'Mình tick cho
-x-94+62-99=95+62-99
-x - 94 + 62 - 99 = 95 + 62 - 99.
=> -x - 94 + 62 - 99 = 157 - 99.
=> -x - 94 + 62 - 99 = 58.
=> -x - 94 + 62 = 58 + 99
=> -x - 94 + 62 = 157.
=> -x - 94 = 157 - 62.
=> -x - 94 = 95.
=> -x = 95 + 94
=> -x = 189.
Vậy -x = 189.
#Học tốt.
\(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=3+\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)\(\frac{1}{95}\)
\(\frac{1}{95}\frac{1}{95}\)
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:
\(\frac{x-1}{99}\)+\(\frac{x-2}{98}+\frac{x-5}{95}=3+\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)
giải pt
\(\left(\dfrac{x-1}{99}+x-99\right)+\left(\dfrac{x-3}{97}+\dfrac{x-7}{93}\right)+\left(\dfrac{x-5}{95}+\dfrac{x-95}{5}\right)=6\)
\(\Leftrightarrow\left(\dfrac{x-1}{99}-1\right)+\left(\dfrac{x-99}{1}-1\right)+\left(\dfrac{x-3}{97}-1\right)+\left(\dfrac{x-7}{93}-1\right)+\left(\dfrac{x-5}{95}-1\right)+\left(\dfrac{x-95}{5}-1\right)=0\)=>x-100=0
hay x=100