Ôn tập toán 7

NL

tìm số hữu tỉ x biết: \(\dfrac{x+2}{11}\)+\(\dfrac{x+2}{12}+\dfrac{x+2}{13}\)= \(\dfrac{x+2}{14}+\dfrac{x+2}{15}\)

H24
10 tháng 6 2017 lúc 21:43

Có:

\(\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)

\(\Leftrightarrow\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}-\dfrac{x+2}{14}-\dfrac{x+2}{15}=0\)

\(\Leftrightarrow\left(x+2\right)\left(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\right)=0\)

Dấu "=" xảy ra:

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}=0\end{matrix}\right.\)

\(\left(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\right)\ne0\)

\(\Leftrightarrow x-2=0\)

\(\Rightarrow x=0+2=2\)

Vậy \(x=2\).

Học tốt!vui

Bình luận (5)
H24
11 tháng 6 2017 lúc 14:45

\(\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)

\(\Rightarrow\left(\dfrac{1}{11}+\dfrac{1}{12}\right)\left(x+2\right)+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)

\(\Rightarrow\dfrac{23\left(x+2\right)}{132}+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)

\(\Rightarrow\left(\dfrac{23}{132}+\dfrac{1}{13}\right)\left(x+2\right)=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)

\(\Rightarrow\dfrac{431\left(x+2\right)}{1716}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)

\(\Rightarrow\dfrac{431\left(x+2\right)}{1716}=\left(\dfrac{1}{14}+\dfrac{1}{15}\right)\left(x+2\right)\)

\(\Rightarrow\dfrac{431\left(x+2\right)}{1716}=\dfrac{29\left(x+2\right)}{210}\)

\(\Rightarrow\dfrac{431\left(x+2\right)}{1716}-\dfrac{29\left(x+2\right)}{210}=0\)

\(\Rightarrow\left(\dfrac{431}{6.286}-\dfrac{29}{6.35}\right)\left(x+2\right)=0\)

\(\Rightarrow\dfrac{1}{6}\left(\dfrac{431}{286}-\dfrac{29}{35}\right)\left(x+2\right)=-2\)

Bình luận (1)

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