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a, => (x-10/30 - 3) + (x-14/43 - 2) + (x-5/95 - 1) + x-100/8 = 0 ( vì x-148/8 = x-100/8 + 48/8 = x-100/8 + 6 )
=> x-100/30 + x-100/43 + x-100/95 + x-100/8 = 0
=> (x-100).(1/30 + 1/43 + 1/95 + 1/8) = 0
=> x-100 = 0 ( vì 1/30+1/43+1/95+1/8 > 0 )
=> x = 100
Vậy x = 100
Tk mk nha
#)Giải :
a) x + 2x + 3x + ... + 100x = - 213
=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213
=> 100x + 5049 = - 213
<=> 100x = - 5262
<=> x = - 52,62
#)Giải :
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)
\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{2}{3}\)
c) \(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
\(\Leftrightarrow\left(\frac{x-1}{2009}-1\right)+\left(\frac{x-2}{2008}-1\right)=\left(\frac{x-3}{2007}-1\right)+\left(\frac{x-4}{2006}-1\right)\)
\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right).\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)
\(\Leftrightarrow x-2010=0\)
\(\Leftrightarrow x=0+2010\)
\(\Rightarrow x=2010\)
Vậy \(x=2010.\)
Mình chỉ làm câu c) thôi nhé.
Chúc bạn học tốt!
a) \(-\frac{3}{x}=\frac{15}{7}\)
=> -3.7 = 15x
=> 15x = -21
=> x = -21:15
=> x = -1,4
Vậy x = -1,4
b) \(\frac{x+3}{4}=\frac{5}{20}\)
\(\Rightarrow\frac{x+3}{4}=\frac{1}{4}\)
=> x + 3 = 1
=> x = 1 - 3
=> x = -2
Vậy x = -2
d) \(\frac{x-1}{3}=\frac{x+1}{5}\)
=> 5(x - 1) = 3(x + 1)
=> 5x - 5 = 3x + 3
=> 5x - 3x = 5 + 3
=> 2x = 8
=> x = 8:2
=> x = 4
Vậy x = 4
\(a,\frac{-3}{x}=\frac{15}{7}\)
=> -21 = 15x
=> \(x=-\frac{21}{15}=-\frac{7}{5}\)
b,
\(\frac{x+3}{4}=\frac{5}{20}\)
=> \(\frac{5(x+3)}{20}=\frac{5}{20}\)
=> 5\((x+3)\)= 5
=> x + 3 = 1
=> x = -2
\(c,\frac{1,2}{30}=\frac{3x+4}{50}\)
=> \(\frac{\frac{12}{10}}{30}=\frac{3x+4}{50}\)
=> \(\frac{\frac{6}{5}}{30}=\frac{3x+4}{50}\)
=> \(\frac{2}{50}=\frac{3x+4}{50}\)
=> 3x + 4 = 2
=> 3x = -2
=> x = -2/3
\(d,\frac{x-1}{3}=\frac{x+1}{5}\)
=> 5[x - 1] = 3[x + 1]
=> 5x - 5 = 3x + 3
=> 5x - 5 - 3x = 3
=> 5x - 3x - 5 = 3
=> 2x = 8
=> x = 4
Mấy bài còn lại tương tự nhé cậu
a ) Ta có : \(\frac{x+11}{10}+\frac{x+21}{20}+\frac{x+31}{30}=\frac{x+41}{40}+\frac{x+101}{5}\)
\(\Leftrightarrow\left(\frac{x+11}{10}-1\right)+\left(\frac{x+21}{10}-1\right)+\left(\frac{x+31}{30}-1\right)=\left(\frac{x+41}{40}-1\right)+\left(\frac{x+101}{50}-2\right)\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}=\frac{x+1}{40}+\frac{x+1}{50}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}-\frac{x+1}{40}-\frac{x+1}{50}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)\ne0\)
Nên x + 1 = 0
=> x = -1
còn b vs c thì sao ạ