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\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Leftrightarrow x+1.\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\div\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Ta có :
Vậy x = -1
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\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
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
a) \(\frac{-2^2}{3}.x=\frac{-2^5}{3}\)
\(x=\frac{-2^5}{3}:\frac{-2^2}{3}\)
\(x=\frac{-2^5}{3}.\frac{-3}{2^2}\)
\(x=\frac{2^3}{1}\)
\(x=2^3\)
\(x=8\)
Vậy x = 8
b)\(\frac{-1^3}{3}.x=\frac{1}{81}\)
\(x=\frac{1}{81}:\frac{-1^3}{3}\)
\(x=\frac{1}{81}.\frac{-3}{1^3}\)
\(x=\frac{1}{3^4}.\frac{-3}{1^3}\)
\(x=\frac{1}{3^3}.\frac{-1}{1^2}\)
\(x=\frac{-1}{3^3}\)
\(x=\frac{-1}{27}\)
Vậy \(x=\frac{-1}{27}\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\) khác 0
=> x+1 = 0
=> x= 0-1
=> x= -1
Vậy x = -1
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Có: \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\)
\(\Rightarrow x+1=0\)
\(\Rightarrow x=-1\)