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\(x=9\Leftrightarrow x+1=10\\ \Leftrightarrow M=x^{2012}-\left(x+1\right)x^{2012}+...-\left(x+1\right)x^2+\left(x+1\right)x-\left(x+1\right)\\ M=x^{2012}-x^{2013}-x^{2012}+...-x^3-x^2+x^2+x-x-1\\ \Leftrightarrow M=-x^{2013}-1=-9^{2013}-1\)
Bài 4 : Tìm y , biết :
a) y ( 2y - 7 ) - 4y + 14 = 0
b) ( y + 3 )( y2 - 3y + 9 ) - y( y2 - 3 ) = 18
GIẢI HỘ LUN Ạ![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=x\left(x^2-10x+25\right)=x\left(x-5\right)^2\\ b,=y\left(x+y\right)-\left(x+y\right)=\left(y-1\right)\left(x+y\right)\\ c,=\left(x-5\right)^2\\ d,=\left(x-8\right)\left(x+8\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: x=9
nên x+1=10
Ta có: \(x^{14}-10x^{13}+10x^{12}-...+10x^2-10x+10\)
\(=x^{14}-x^{13}\left(x+1\right)+x^{12}\left(x+1\right)-...+x^2\left(x+1\right)-x\left(x+1\right)+x+1\)
\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-...+x^3+x^2-x^2-x+x+1\)
=1
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(A=x^3-30x^2-31x+1\)
\(=x^3-31x^2+x^2-31x+1\)
\(=x^2\left(x-31\right)+x\left(x-31\right)+1\)
\(=\left(x^2+x\right)\left(x-31\right)+1\)
Thay x = 31 \(\Rightarrow A=1\)
Vậy A = 1 khi x = 31
b, tách ra làm tương tự phần a
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Vì $x=9$ nên $x-9=0$
Ta có:
$F=(x^{2017}-9x^{2016})-(x^{2016}-9x^{2015})+(x^{2015}-9x^{2014})-....-(x^2-9x)+x-10$
$=x^{2016}(x-9)-x^{2015}(x-9)+x^{2014}(x-9)-....-x(x-9)+x-10$
$=x^{2016}.0-x^{2015}.0+x^{2014}.0-...-x.0+x-10$
$=x-10=9-10=-1$
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm:
Ta có: \(x=-9\Leftrightarrow-10=x-1\Rightarrow10=1-x\)nên thay vào ta tính:
\(P\left(-9\right)=1+\left(1-x\right)x+\left(1-x\right)x^2+\left(1-x\right)x^3+...+\left(1-x\right)x^{19}+\left(1-x\right)x^{20}\)
\(P\left(-9\right)=1+x-x^2+x^2-x^3+x^3-x^4+...+x^{20}-x^{21}\)
\(P\left(-9\right)=1+x-x^{21}\)
\(P\left(-9\right)=1-9+9^{21}\)
\(P\left(-9\right)=9^{21}-8\)
Vậy khi \(x=-9\)thì \(P\left(x\right)=9^{21}-8\)
Học tốt!!!!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(\dfrac{4}{x-2}-\dfrac{3}{x+2}\right):\dfrac{x+14}{x^2}\\ =\left(\dfrac{4\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\right)\cdot\dfrac{x^2}{x+14}\\ =\dfrac{4x+8-3x+6}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^2}{x+14}\\ =\dfrac{x+14}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^2}{x+14}\\=\dfrac{x^2}{x^2-4}\)
Tại `x=-3` Ta có :
\(\dfrac{x^2}{x^2-4}\\ =\dfrac{-3^2}{-3^2-4}\\ =\dfrac{9}{9-4}\\ =\dfrac{9}{5}\)