x+420-1007=2583
X + 420 - 1007 = 2583
X+420-1007=2583
X + 420 = 2583 + 1007
X + 420 = 3590
X = 3590 - 420
X= 3170
\(x+420−1007=2583\)
\(x−587=2583\)
\(x=2583+587\)
\(x=3170\)
x + 420 = 2583 + 1007
x + 420 = 3590
x = 3590 - 420
x = 3170
\(\left\{{}\begin{matrix}x+y=1500\\x+\dfrac{75}{100}x+y+\dfrac{68}{100}y=2583\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x+y=1500\\x+\dfrac{75}{100}x+y+\dfrac{68}{100}y=2583\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y=1500\\\left(x+y\right)+\dfrac{3}{4}x+\dfrac{17}{25}y=2583\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y=1500\\1500+\dfrac{3}{4}x+\dfrac{17}{25}y=2583\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y=1500\\\dfrac{3}{4}x+\dfrac{17}{25}y=1083\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=900\\y=600\end{matrix}\right.\)
Cho x,y,z thỏa mãn:
\(x^{2014}+y^{2014}+x^{2014}=x^{1007}y^{1007}+y^{1007}.z^{1007}+z^{1007}.x^{1007}\)
Tính giá trị của biểu thức \(P=\left(x-y\right)^{2014}+\left(y-z\right)^{2014}+\left(x-z\right)^{2014}\)
(173+159+2583)*(235237*189*189189*235)
Tìm x, y thoả mãn sao cho x+1007=y-1007 và x.y=5
\(x+1007=y-1007\)
\(\Rightarrow x=y-2014\)(1)
Lại có
\(xy=5\Rightarrow x=\frac{5}{y}\)(2)
Từ (1) và (2) \(\Rightarrow y-2014=\frac{5}{y}\)
\(\Rightarrow\frac{y^2-2014y}{y}=\frac{5}{y}\)
\(\Rightarrow y\left(y-2014\right)=5\)
Lập bảng tìm y sau đó thay vô tìm x
giải pt sau: \(\frac{x-1009}{1010}+\frac{x-1007}{1012}=\frac{x-1010}{1009}+\frac{x-1012}{1007}\)
\(\frac{x-1009}{1010}+\frac{x-1007}{1012}=\frac{x-1010}{1009}+\frac{x-1012}{1007}\)
\(\Rightarrow(\frac{x-1009}{1010}-1)+\left(\frac{x-1007}{1012}-1\right)=\left(\frac{x-1010}{1009}-1\right)+\left(\frac{x-1012}{1007}-1\right)\)
\(\Rightarrow\frac{x-2019}{1010}+\frac{x-2019}{1012}-\frac{x-2019}{1009}-\frac{x-2019}{1007}\)
\(\Rightarrow\left(x-2019\right)\left(\frac{1}{1010}+\frac{1}{1012}-\frac{1}{1009}-\frac{1}{1007}\right)=0\)
Ta có
\(\frac{1}{1010}+\frac{1}{1012}-\frac{1}{1009}-\frac{1}{1007}\ne0\Rightarrow x-2019=0\Rightarrow x=2019\)
\(\frac{x-1009}{1010}+\frac{x-1007}{1012}=\frac{x-1010}{1009}+\frac{x-1012}{1007}\)
\(\frac{x-1009}{1010}-1+\frac{x-1007}{1012}-1=\frac{x-1010}{1009}-1+\frac{x-1012}{1007}\)\(\frac{x-2019}{1010}+\frac{x-2019}{1012}-\frac{x-2019}{1009}-\frac{x-2019}{1007}=0\)
\(\left(x-2019\right)\left(\frac{1}{1010}+\frac{1}{1012}-\frac{1}{1009}-\frac{1}{1007}\right)=0\)
1/1010 + 1/1012 - 1/1009 - 1/1007 khác 0
=> x - 2019 =0 => x = 2019
Tìm x:
giải pt sau: \(\frac{x-1009}{1010}+\frac{x-1007}{1012}=\frac{x-1010}{1009}+\frac{x-1012}{1007}\)