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\(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=3+\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)
\(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=\frac{2765070}{921690}+\frac{9310}{921690}+\frac{9405}{921690}+\frac{9702}{921690}\)
\(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=\frac{2793487}{921690}\)
\(BCNN\left(99,98,95\right)=921690\Rightarrow x=101\)
\(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=3+\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)
\(\Leftrightarrow\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=1+\frac{1}{99}+1+\frac{1}{98}+1+\frac{1}{95}\)
\(\Leftrightarrow\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=\frac{100}{99}+\frac{99}{98}+\frac{96}{95}\)
\(\Leftrightarrow\left(\frac{x-1}{99}-\frac{100}{99}\right)+\left(\frac{x-2}{98}-\frac{99}{98}\right)+\left(\frac{x-5}{95}-\frac{96}{95}\right)=0\)
\(\Leftrightarrow\frac{x-101}{99}+\frac{x-101}{98}+\frac{x-101}{95}=0\)
\(\Leftrightarrow\left(x-101\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\right)=0\)
\(\Leftrightarrow x-101=0\)
\(\Leftrightarrow x=101\)
\(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=3+\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)
\(\Leftrightarrow\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=1+\frac{1}{99}+1+\frac{1}{98}+1+\frac{1}{95}\)
\(\Leftrightarrow\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=\frac{100}{99}+\frac{99}{98}+\frac{96}{95}\)
\(\Leftrightarrow\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}-\frac{100}{99}-\frac{99}{98}-\frac{96}{95}=0\)
\(\Leftrightarrow\left(\frac{x-1}{99}-\frac{100}{99}\right)+\left(\frac{x-2}{98}-\frac{99}{98}\right)+\left(\frac{x-5}{95}-\frac{96}{95}\right)=0\)
\(\Leftrightarrow\frac{x-101}{99}+\frac{x-101}{98}+\frac{x-101}{95}=0\)
\(\Leftrightarrow\left(x-101\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\right)=0\)
Do \(\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\ne0\)
Mà \(x-101=0\Leftrightarrow x=101\)
Vậy x = 101
1) \(\left(x-1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}\right)=0\)
mà \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}\ne0\)
\(\Rightarrow x-1=0\Leftrightarrow x=1\)
2) \(\frac{x-1}{99}-1+\frac{x-2}{98}-1+\frac{x-5}{95}-1=\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)
\(\frac{x-100}{99}+\frac{x-100}{98}+\frac{x-100}{95}=\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)
\(\left(x-100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\right)=\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)
x - 100 = 1
x = 101
\(\frac{x+1}{99}+\frac{x+3}{97}+\frac{x+5}{95}=\frac{x+2}{98}+\frac{x+4}{96}+\frac{x+6}{94}\)
\(\left(\frac{x+1}{99}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+5}{95}+1\right)=\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)+\left(\frac{x+6}{94}+1\right)\)
\(\left(\frac{x+1}{99}+\frac{99}{99}\right)+\left(\frac{x+3}{97}+\frac{97}{97}\right)+\left(\frac{x+5}{95}+\frac{95}{95}\right)=\left(\frac{x+2}{98}+\frac{98}{98}\right)+\left(\frac{x+4}{96}+\frac{96}{96}\right)+\left(\frac{\left(x+6\right)}{94}+\frac{94}{94}\right)\)
\(\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}=\frac{x+100}{92}+\frac{x+100}{94}+\frac{x+100}{96}\)
\(\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}-\frac{x+100}{92}-\frac{x+100}{94}-\frac{x+100}{96}=0\)
\(\left(x+100\right).\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{92}-\frac{1}{94}-\frac{1}{96}\right)=0\)
\(Mà\) \(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{92}-\frac{1}{94}-\frac{1}{96}\ne0\)
Nên x+ 100 = 0
x = 0 - 100 = -100
Vậy x= -100
cộng 1 vào mỗi tỉ số,ta được:
\(\left(\frac{x+1}{99}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+5}{95}+1\right)=\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)+\left(\frac{x+6}{94}+1\right)\)\(\Rightarrow\frac{x+1+99}{99}+\frac{x+3+97}{97}+\frac{x+5+95}{95}=\frac{x+2+98}{98}+\frac{x+4+96}{96}+\frac{x+6+94}{94}\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}=\frac{x+100}{98}+\frac{x+100}{96}+\frac{x+100}{94}\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}-\frac{x+100}{98}-\frac{x+100}{96}-\frac{x+100}{94}=0\)
\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{98}-\frac{1}{96}-\frac{1}{94}\right)\)
Vì \(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{98}-\frac{1}{96}-\frac{1}{94}\ne0\)
=>x+100=0
=>x=-100
Vậy x=-100
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)
Theo tính chất dãy tỉ số bằng nhau , ta có :
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)\(=\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}\)
\(\frac{2x-2}{4}+\frac{3y-6}{9}-\frac{z-3}{4}\)\(=\frac{95}{9}\)
=> \(x=\frac{190}{9}\)\(y=\frac{95}{3}\)\(z=\frac{380}{9}\)
\(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{-z+3}{-4}=\frac{2x+3y-z-5}{9}=\frac{90}{9}=10\)
x=;y=;z= tu tinh
a, \(\frac{x+1}{5}+\frac{x+1}{7}=\frac{x+1}{9}\)
\(\Leftrightarrow\frac{x+1}{5}+\frac{x+1}{7}-\frac{x+1}{9}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
b, \(\frac{x+4}{96}+\frac{x+3}{97}=\frac{x+2}{98}+\frac{x+1}{99}\)
\(\Leftrightarrow\left(\frac{x+4}{96}+1\right)+\left(\frac{x+3}{97}+1\right)=\left(\frac{x+2}{98}+1\right)+\left(\frac{x+1}{99}+1\right)\)
\(\Leftrightarrow\frac{x+100}{96}+\frac{x+100}{97}=\frac{x+100}{98}+\frac{x+100}{99}\)
\(\Leftrightarrow\frac{x+100}{96}+\frac{x+100}{97}-\frac{x+100}{98}-\frac{x+100}{99}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{96}+\frac{1}{97}+\frac{1}{98}+\frac{1}{99}\right)=0\)
\(\Leftrightarrow x+100=0\)
\(\Leftrightarrow x=-100\)
a) x + 1/5 + x + 1/7 = x + 1/9
<=> 1/5x + 1/5 + 1/7x + 1/7 = 1/9x + 1/9
<=> (1/5x + 1/7x) + (1/5 + 1/7) = 1/9x + 1/9
<=> 12/35x + 12/35 = 1/9x + 1/9
<=> 12/35x + 12/35 - 1/9x = 1/9
<=> 73/315x + 12/35 = 1/9
<=> 73/315x = 1/9 - 12/35
<=> 73/315x = -73/315
<=> x = 73/315 : -73/315 = -1
=> x = -1
b) làm tương tự
ta có
x+y+y+z+z+x=\(\frac{13}{12}\)
2(x+y+z)=\(\frac{13}{12}\)
=>x+y+z=\(\frac{13}{24}\)
z=(x+y+z)-(x+y)
y=y+z-z
x=x+Y-y