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\(\frac{x+1}{2015}+\frac{x+2}{2014}=\frac{x+3}{2013}+\frac{x+4}{2012}\)
\(\Leftrightarrow\frac{x+1}{2015}+1+\frac{x+2}{2014}+1=\frac{x+3}{2013}+1+\frac{x+4}{2012}+1\)
\(\Leftrightarrow\frac{x+2016}{2015}+\frac{x+2016}{2014}=\frac{x+2016}{2013}+\frac{x+2016}{2012}\)
\(\Leftrightarrow\frac{x+2016}{2015}+\frac{x+2016}{2014}-\frac{x+2016}{2013}-\frac{x+2016}{2012}=0\)
\(\Leftrightarrow\left(x+2016\right)\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)
\(\Leftrightarrow x+2016=0\).Do \(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\ne0\)
\(\Leftrightarrow x=-2016\)
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\(\frac{x+1}{2015}+\frac{x+2}{2014}=\frac{x+3}{2013}+\frac{x+4}{2012}\)
\(\Leftrightarrow\)\(\frac{x+1}{2015}+1+\frac{x+2}{2014}+1=\frac{x+3}{2013}+1+\frac{x+4}{2012}+1\)
\(\Leftrightarrow\)\(\frac{x+1}{2015}+\frac{2015}{2015}+\frac{x+20}{2014}+\frac{2014}{2014}=\frac{x+3}{2013}+\frac{2013}{2013}+\frac{x+4}{2012}+\frac{2012}{2012}\)
\(\Leftrightarrow\frac{x+2016}{2015}+\frac{x+2016}{2014}=\frac{x+2016}{2013}+\frac{x+2016}{2012}\)
\(\Leftrightarrow\)\(\frac{x+2016}{2015}+\frac{x+2016}{2014}-\frac{x+2016}{2013}-\frac{x+2016}{2012}=0\)
\(\Leftrightarrow\left(x+2016\right)\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)
Suy ra \(x+2016=0\) \(\Leftrightarrow x=-2016\)
Vậy \(x\in\left\{-2016\right\}\)
bạn cộng thêm 1 vào mỗi phân thức đó
sau đó sẽ có phân tử chung là x+2016
kết quả là x=-2016
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a) \(x^2-5x+4=0\)
\(\Leftrightarrow\)\(x^2-x-4x+4=0\)
\(\Leftrightarrow\)\(x\left(x-1\right)-4\left(x-1\right)=0\)
\(\Leftrightarrow\)\(\left(x-1\right)\left(x-4\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-1=0\\x-4=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=1\\x=4\end{cases}}\)
Vậy tổng các giá trị nguyên của x thỏa mãn là:
\(1+4=5\)
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Ta có : \(\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x+2=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-2\end{array}\right.\)
Vậy \(x\in\left\{1;-2\right\}\)
Đây giống bài lớp 6 hơn
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bt sau khi nhân ra sẽ bằng x^3 - 4x^2 + 4x + x^2 - 4x + 4 + 4x ^2 - x^3 = 13 <=> x ^ 2 + 4 = 13 <=> x ^2 = 9 <=> x thuộc {-3; 3}
vậy x thuộc {-3; 3}