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\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x-1}{3}+\frac{x-1}{4}+\frac{x-1}{5}\)
\(\Leftrightarrow\frac{x-1}{1}+\frac{x-1}{2}-\frac{x-1}{3}-\frac{x-1}{4}-\frac{x-1}{5}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{1}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\right)=0\)
Vì \(\frac{1}{1}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\ne0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x-1}{3}+\frac{x-1}{4}+\frac{x-1}{5}\)
\(\Leftrightarrow\frac{x-1}{1}+\frac{x-1}{2}-\frac{x-1}{3}-\frac{x-1}{4}-\frac{x-1}{5}=0\)
\(\Leftrightarrow\left(x-1\right)\left(1+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\ne0\right)=0\)
\(\Leftrightarrow x=1\)
\(\frac{1}{2}+\frac{2}{3}x=\frac{1}{4}\)
\(\frac{2}{3}x=\frac{1}{4}-\frac{1}{2}\)
\(\frac{2}{3}x=-\frac{1}{4}\)
\(x=-\frac{1}{4}:\frac{2}{3}\)
\(x=-\frac{3}{8}\)
\(\frac{2}{3}x\)\(=\)\(\frac{1}{4}\)\(-\)\(\frac{1}{2}\)
\(\frac{2}{3}x\)\(=\)\(\frac{-1}{4}\)
\(x\)\(=\)\(\frac{-1}{4}\)\(:\)\(\frac{2}{3}\)
\(x\)\(=\)\(\frac{-3}{8}\)
1) \(x^2y+x\left(2y-1\right)=7\)
\(\Leftrightarrow x^2y+2xy-x=7\)
\(\Leftrightarrow xy\left(x+2\right)-x-2=7-2\)
\(\Leftrightarrow xy\left(x+2\right)-\left(x+2\right)=5\)
\(\Leftrightarrow\left(xy-1\right)\left(x+2\right)=5\)
\(\Rightarrow\)xy - 1 và x + 2 là ước của 5 là \(\pm1;\pm5\)
đến đây tự lm đc
2 ) \(B=\frac{255}{1}+\frac{254}{2}+\frac{253}{3}+....+\frac{3}{253}+\frac{2}{254}+\frac{1}{255}\)
\(=\left(\frac{254}{2}+1\right)+\left(\frac{253}{3}+1\right)+....+\left(\frac{2}{254}+1\right)+\left(\frac{1}{255}+1\right)+1\)
\(=\frac{256}{2}+\frac{256}{3}+....+\frac{256}{255}+\frac{256}{256}\)
\(=256\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{255}+\frac{1}{256}\right)=256A\)
\(\Rightarrow\frac{B}{A}=256=16^2\) Là số CP (đpcm)
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\)
Dễ thấy: \(\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\ne0\Rightarrow x+2004=0\Leftrightarrow x=-2014\)
Bài 1:
\(A=\left(\frac{-5}{11}+\frac{7}{22}-\frac{4}{33}-\frac{5}{44}\right):\left(38\frac{1}{122}-39\frac{7}{22}\right)\)
\(=\frac{-49}{132}:\left(-\frac{879}{671}\right)=\frac{2989}{105408}\)
Bài 2:
\(\frac{4}{5}-\left(\frac{-1}{8}\right)=\frac{7}{8}-x\)
<=> \(\frac{7}{8}-x=\frac{27}{40}\)
<=> \(x=\frac{7}{8}-\frac{27}{40}=\frac{1}{5}\)
Vậy...
1/2[2/1.3+2/3.5+2/5.7+.........+2/x(x+2)]=16/34
2/1.3+2/3.5+2/5.7+......+2/x(x+2)=16/34:1/2=16/17
1/1-1/3+1/3-1/5+1/5-1/7+.....+1/x-1/x+2=16/17
1-1/x+2=16/17
1/x+2=1-16/17=1/17
suy ra:x+2=17
x=17-2
x=15
\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x}{3}+\frac{x}{4}-\frac{7}{12}\)
\(\Leftrightarrow\frac{12x-12}{12}+\frac{6x-6}{12}=\frac{4x}{12}+\frac{3x}{12}-\frac{7}{12}\)
Khử mẫu : \(12x-12+6x-6=4x+3x-7\)
\(\Leftrightarrow18x-18=7x-7\Leftrightarrow11x=11\Leftrightarrow x=1\)
\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x}{3}+\frac{x}{4}-\frac{7}{12}\)
\(\Leftrightarrow\frac{12x-12}{12}+\frac{6x-6}{12}=\frac{4x}{12}+\frac{3x}{12}-\frac{7}{12}\)
\(\Leftrightarrow\frac{12x-12+6x-6}{12}=\frac{4x+3x-7}{12}\)
\(\Leftrightarrow18x-18=7x-7\)
\(\Leftrightarrow18x+7x=18+7\)
\(\Leftrightarrow25x=25\)
\(\Leftrightarrow x=1\)