\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
Đặt x + 1 ra ngoài ta được:
\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
MÀ \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\)
=> x + 1 = 0
=> x = 0 - 1 = -1
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
Tìm x nha!
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
a) \(-\frac{2}{5}+\frac{2}{3}.x+\frac{1}{6}.x=-\frac{4}{5}\)
b)\(\frac{3}{2}.x-\frac{2}{5}-\frac{2}{3}.x=-\frac{4}{15}\)
c)\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
Giải giúp mình với, được câu nào thì làm câu nấy ạ! Mình cần gấp
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
Các bạn vui lòng giải đầy đủ giúp mình.Thanks trước!
a,\(x+\frac{1}{10}+x+\frac{1}{11}+x+\frac{1}{12}=x+\frac{1}{13}+x+\frac{1}{14}\)
\(\frac{x+4}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
Tìm x:
\(\frac{X+1}{10}+\frac{X+1}{11}+\frac{X+1}{12}=\frac{X+1}{13}+\frac{X+1}{14}\)
Tìm x , biết : \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
Tìm x
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
