Violympic toán 8

\(Q=4x^2+y^2+4x-10y+2016\)

\(Q=\left(2x\right)^2+2.2x+1+y^2-2.y.5+25+1990\)

\(Q=\left(2x+1\right)^2+\left(y-5\right)^2+1990\)

Vì \(\left(2x+1\right)^2\ge0\) với mọi x

\(\left(y-5\right)^2\ge0\) với mọi y

\(\Rightarrow\left(2x+1\right)^2+\left(y-5\right)^2\ge0\) với mọi x và y

\(\Rightarrow\left(2x+1\right)^2+\left(y-5\right)^2+1990\ge1990\)

\(\Rightarrow Qmin=1990\Leftrightarrow\left\{{}\begin{matrix}2x+1=0\\y-5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=5\end{matrix}\right.\)

\(A=2\left(a^3+b^3\right)-3\left(a^2+b^2\right)\)

\(A=2\left(a+b\right)\left(a^2-ab+b^2\right)-3\left(a^2+b^2\right)\)

\(A=2\left(a+b\right)\left[\left(a+b\right)^2-3ab\right]-3\left[\left(a+b\right)^2-2ab\right]\)

Thay a + b = 1

\(A=2\left(1-3ab\right)-3\left(1-2ab\right)\)

\(A=2-6ab-3+6ab\)

\(A=-1\)

\(=2013^4-\left(2012\cdot2014\right)\left(2013^2+1\right)\\ =2013^4-\left(2013^2-1\right)\left(2013^2+1\right)\\ =2013^4-\left(2013^4-1\right)\\ =1\)

Mysterious Person

\(P=\left(x+y\right)^3-3\left(x+y\right)^2+3\left(x+y\right)+2017\)

\(=\left(x+y-1\right)^3+2018\)

\(=100^3+2018\)

Ta có : P = \(\dfrac{1}{1975}\left(\dfrac{2}{1945}-1\right)-\dfrac{1}{1945}\left(1-\dfrac{2}{1975}\right)+\dfrac{1974}{1975}.\dfrac{1946}{1945}\)

\(-\dfrac{3}{1975.1945}\)

= \(\dfrac{2}{1975.1945}-\dfrac{1}{1975}-\dfrac{1}{1945}+\dfrac{2}{1975.1945}+\dfrac{1974}{1975}.\dfrac{1946}{1945}\)

\(-\dfrac{3}{1975.1945}\)

= \(\dfrac{2+2+1974.1946-3-1975-1945}{1975.1945}\)

= \(\dfrac{2+2+1974.1946-3-1975-1945}{1975.1945}\)

= \(\dfrac{1973}{1975}\)

E hèm

\(\left(x-1\right)^3+4\left(x+1\right)\left(1-x\right)+3\left(x-1\right)\left(x^2+x+1\right)\)

\(=x^3-3x^2+3x-1+4\left(1-x^2\right)+3\left(x^3-1\right)\)

\(=x^3-3x^2+3x-1+4-4x^2+3x^3-3\)

\(=\left(x^3+3x^3\right)-\left(3x^2+4x^2\right)+3x+\left(4-1-3\right)\)

\(=4x^3-7x^2+3x\)

:D

x=7 nen x+1=8

\(A=x^{15}-x^{14}\left(x+1\right)+x^{13}\left(x+1\right)-...+x^3\left(x+1\right)-x^2\left(x+1\right)+x\left(x+1\right)-5\)

\(=x^{15}-x^{15}-x^{14}+x^{14}+...+x^4+x^3-x^3-x^2+x^2+x-5\)

=x-5

=2

Ta có : \(Q=-2x^2-6x+5\)

\(\Leftrightarrow Q=-2\left(x^2+3x-\dfrac{5}{2}\right)\)

\(\Leftrightarrow Q=-2\left(x^2+2x.\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{19}{4}\right)\)

\(\Leftrightarrow Q=-2\left[\left(x+\dfrac{3}{2}\right)^2-\dfrac{19}{4}\right]\)

\(\Leftrightarrow Q=-2\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{2}\le\dfrac{19}{2}\forall x\)

Dấu " = " xảy ra

\(\Leftrightarrow x+\dfrac{3}{2}=0\Leftrightarrow x=-\dfrac{3}{2}\)

Vậy GTNN của Q là : \(\dfrac{19}{2}\Leftrightarrow x=-\dfrac{3}{2}\)

:D

\(\dfrac{19}{2}\)