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x = 2013 => x + 1 = 2014
Ta có:\(B=x^{2013}-2014x^{2012}+2014x^{2011}-2014x^{2010}+...+2014x-1\)
\(=x^{2013}-\left(x+1\right)x^{2012}+\left(x+1\right)x^{2011}-\left(x+1\right)x^{2010}+...+\left(x+1\right)x-1\)
\(=x^{2013}-x^{2013}-x^{2012}+x^{2012}+x^{2011}-x^{2011}-x^{2010}+...+x^2+x-1\)
\(=x-1\)
\(=2013-1\)
\(=2012\)
\(X=2013\Rightarrow2014=X+1\Rightarrow B=X^{2013}-\left(X+1\right)\times X^{2012}+...+\left(X+1\right)\times X-1\)\(X-1\)
\(\Rightarrow B=X^{2013}-X^{2013}-X^{2012}+...+X^2+X-1\)
\(\Rightarrow B=X-1\)\(=2013-1=2012\)
Ta thấy 2014=2013+1=x+1
B=x2013-2014x2012+2014x2011-2014x2011-2014x2010+.....-2014x2+2014x
B=x2013-(2013+1).x2012+(2013+1).x2011-(2013+1).x2011-(2013+1).x2010+....-(2013+1).x2+(2013+1).x
B=x2013-(x+1).x2012+(x+1).x2011-(x+1).x2011-(x+1).x2010+......-(x+1).x2+(x+1).x
B=x2013-x2013-x2012+x2012+x2011-x2012-x2011-x2011-x2010+....-x3-x2+x2+x
B=.....................(tự triệt tiêu tiếp)
Bài 1:
|\(x\)| = 1 ⇒ \(x\) \(\in\) {-\(\dfrac{1}{3}\); \(\dfrac{1}{3}\)}
A(-1) = 2(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)) + 5
A(-1) = \(\dfrac{2}{9}\) + 1 + 5
A (-1) = \(\dfrac{56}{9}\)
A(1) = 2.(\(\dfrac{1}{3}\) )2- \(\dfrac{1}{3}\).3 + 5
A(1) = \(\dfrac{2}{9}\) - 1 + 5
A(1) = \(\dfrac{38}{9}\)
|y| = 1 ⇒ y \(\in\) {-1; 1}
⇒ (\(x;y\)) = (-\(\dfrac{1}{3}\); -1); (-\(\dfrac{1}{3}\); 1); (\(\dfrac{1}{3};-1\)); (\(\dfrac{1}{3};1\))
B(-\(\dfrac{1}{3}\);-1) = 2.(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)).(-1) + (-1)2
B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) - 1 + 1
B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\)
B(-\(\dfrac{1}{3}\); 1) = 2.(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)).1 + 12
B(-\(\dfrac{1}{3};1\)) = \(\dfrac{2}{9}\) + 1 + 1
B(-\(\dfrac{1}{3}\); 1) = \(\dfrac{20}{9}\)
B(\(\dfrac{1}{3};-1\)) = 2.(\(\dfrac{1}{3}\))2 - 3.(\(\dfrac{1}{3}\)).(-1) + (-1)2
B(\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) + 1 + 1
B(\(\dfrac{1}{3}\); -1) = \(\dfrac{20}{9}\)
B(\(\dfrac{1}{3}\); 1) = 2.(\(\dfrac{1}{3}\))2 - 3.(\(\dfrac{1}{3}\)).1 + (1)2
B(\(\dfrac{1}{3}\); 1) = \(\dfrac{2}{9}\) - 1 + 1
B(\(\dfrac{1}{3}\);1) = \(\dfrac{2}{9}\)
1. Để F nguyên dương
=> 3x - 2 chia hết cho x + 3
3x + 9 - 11 chia hết cho x + 3
3(x + 3) - 11 chia hết cho x + 3
=> 11 chia hết cho x + 3
=> x + 3 thuộc Ư(11) = {1 ; -1; 11; -11}
Tự lập bảng xét 4 giá trị của ước , x lớn hơn 0 thì đáp ứng nhu cầu đề bài !
2. Cậu vẽ hình đi , tớ hình yếu lắm
Xét đề bài , ta thấy :
\(\widehat{xOy'}+\widehat{y'Ox'}=90^0\) \(\Rightarrow\widehat{x'Oy}=90^0-\widehat{y'Ox'}\)
\(\widehat{yOx'}+\widehat{x'Oy'}=90^0\) \(\Rightarrow\widehat{yOx'}=90^0-\widehat{x'Oy'}\)
=> \(\dfrac{\widehat{x'Oy}}{\widehat{y'Ox}}=\dfrac{1}{1}=1\)
3. Ta có :
|3 - 2014x| \(\ge0\)
=> 8 - |3 - 2014x| \(\le8\)
=> MaxA = 8
<=> |3 - 2014x| = 0
<=> x = \(\dfrac{3}{2014}\)
4. \(E=2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2}}}}=2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}=2+\dfrac{1}{2+\dfrac{1}{1+\dfrac{2}{3}}}\)
\(E=2+\dfrac{1}{2+\dfrac{1}{\dfrac{5}{3}}}=2+\dfrac{1}{2+\dfrac{3}{5}}=2+\dfrac{1}{\dfrac{13}{5}}=2+\dfrac{5}{13}=\dfrac{31}{13}\)
Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{x}{y}=\frac{y}{z}=\frac{z}{x}=\frac{x+y+z}{y+z+x}=1\)
=> x = y = z
Ta có: \(A=\frac{2013x^2+y^2+z^2}{x^2+2013y^2+z^2}=\frac{2013x^2+x^2+x^2}{x^2+2013x^2+x^2}=\frac{2015x^2}{2015x^2}=1\)
Câu 7:
x=2014 nên x-1=2013
\(A=x^{2014}-x^{2013}\left(x-1\right)-x^{2012}\left(x-1\right)-...-x\left(x-1\right)+1\)
\(=x^{2014}-x^{2014}+x^{2013}-x^{2013}+x^{2012}-...-x^2+x+1\)
=x+1
=2014+1=2015
Vì \(\dfrac{x}{y}=\dfrac{2}{3}->x=2,y=3\)
A =\(\dfrac{2014.2+2014.3}{2014.2-2014.3}=\dfrac{4028+6042}{4028-6042}=\dfrac{10070}{-2014}=-5\)