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a.\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Mà: \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\Rightarrow x+1=0\Rightarrow x=-1\)
b.
\(\frac{x+4}{1990}+\frac{x+3}{1991}=\frac{x+2}{1992}+\frac{x+1}{1993}\Rightarrow2+\frac{x+4}{1990}+\frac{x+3}{1991}=2+\frac{x+2}{1992}+\frac{x+1}{1993}\)
\(\Rightarrow\left(1+\frac{x+4}{1990}\right)+\left(1+\frac{x+3}{1991}\right)=\left(1+\frac{x+2}{1992}\right)+\left(1+\frac{x+1}{1993}\right)\)
\(\Rightarrow\frac{x+1994}{1990}+\frac{x+1994}{1991}=\frac{x+1994}{1992}+\frac{x+1994}{1993}\)
\(\Rightarrow\frac{x+1994}{1990}+\frac{x+1994}{1991}-\frac{x+1994}{1992}-\frac{x+1994}{1993}=0\)
\(\Rightarrow\left(x+1994\right)\left(\frac{1}{1990}+\frac{1}{1991}-\frac{1}{1992}-\frac{1}{1993}\right)=0\)
\(\frac{1}{1990}+\frac{1}{1991}-\frac{1}{1992}-\frac{1}{1993}\ne0\Rightarrow x+1994=0\Rightarrow x=-1994\)
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Có: \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\)
\(\Rightarrow x+1=0\)
\(\Rightarrow x=-1\)
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}+\frac{x+1}{14}=0\)
\(\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\)\(\ne\)0 nên x + 1 = 0 \(\Rightarrow\)x = -1
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Mà \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\)
\(\Rightarrow x+1=0\)
\(\Rightarrow x=-1\)
Vậy \(x=-1.\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\).Do\(\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\ne0\)
\(\Leftrightarrow x=-1\)
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\ne0\)
\(\Rightarrow x+1=0\)
\(\Leftrightarrow x=0-1\)
\(\Leftrightarrow x=-1\)
Vậy \(x=-1\)
x+1/10 + x+1/11 + x+1/12 = x+1/13 + x+1/14 = 0
<=> x+1/10 + x+1/11 + x+1/12 - x+1/13 - x+1/14 = 0
<=> (x+1)(1/10 + 1/11 + 1/12 - 1/13 - 1/14) = 0
Mà (1/10 + 1/11+ 1/12 - 1/13 - 1/14) khác 0
=> x+1 = 0
<=> x = 0-1
<=> x = -1
Vậy x = -1
<=> \(\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Do \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\ne0\)
=> x+1=0 => x=-1
Đáp số: x=-1
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Dễ thấy \(\frac{1}{10}>\frac{1}{11}>\frac{1}{12}>\frac{1}{13}>\frac{1}{14}\)
nên \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\ne0\)
Do đó: x + 1 = 0 => x = -1
Vậy x = -1
#)Giải :
a) x + 2x + 3x + ... + 100x = - 213
=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213
=> 100x + 5049 = - 213
<=> 100x = - 5262
<=> x = - 52,62
#)Giải :
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)
\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{2}{3}\)
a) \(\frac{x+1}{3}=\frac{x-2}{4}\)
=> (x+1).4 = (x - 2) . 3
=> 4x + 4 = 3x - 6
=> 4x - 3x = - 6 - 4
=> x = - 10
b) \(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
\(\Rightarrow\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)
\(\Rightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}=\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}\)
\(\Rightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}-\frac{x+1}{12}\) = 0
\(\Rightarrow\left(x+1\right).\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)\)
Vì \(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\ne0\) nên x + 1 =0
=> x = -1
c) Xem lại đề