1. Phân tích ĐTTNT
a, x^3 – x^2 -x -2
b, x^2 + 2xy + y^2 – x-y-12
c,(x^2+x+1)(x^2+x+2)-12
2.tìm GTLN của P
P= 4^2- 6xy +9y^2 -16x +12x +2020
1. Phân tích ĐTTNT
a, x^3 – x^2 -x -2
b, x^2 + 2xy + y^2 – x-y-12
c,(x^2+x+1)(x^2+x+2)-12
2.tìm GTLN của P
P= 4^2- 6xy +9y^2 -16x +12x +2020
Bài 1:
a) $x^3 – x^2 – x – 2$
$= x^3 – 2x^2 + x^2 – 2x + x – 2$
$= x^2(x- 2) + x(x – 2) + x – 2$
$= (x -2)(x^2 + x + 1)$
b) $x^2 + 2xy + y^2 – x – y – 12$
$= (x+y)^2 – (x + y) – 12$
Đặt $t = x + y$ ta được:
$t^2 – t – 12$
$= t^2 – 4t + 3t – 12$
$= t(t -4) + 3(t – 4)$
$= (t – 4)(t + 3)$
$= (x + y – 4)(x + y + 3)$
c) $(x^2 + x + 1)(x^2 + x + 2) – 12$
Đặt $t = x^2 + x + 1$ ta được:
$t(t + 1) – 12$
$= t^2 + t – 12$
$= t^2 + 4t – 3t – 12$
$= (t+4)(t – 3)$
$= (x^2 + x + 5)(x^2 + x – 2)$
Bài 2:
$P = 4x^2 – 6xy + 9y^2- 16x + 12y + 2020$
$= \dfrac{1}{4}(16x^2 + 9y^2 + 64 – 24xy – 64x + 48y) + \dfrac{27}{4}y^2 + 2004$
$= \dfrac{1}{4}(4x -3y – 8)^2 +\dfrac{27}{4}y^2 + 2004$
Ta có:
$\begin{cases}(4x – 3y – 8)^2 \geq 0, \, \forall x,y\\y^2 \geq 0, \, \forall y\end{cases}$
Do đó:
$\dfrac{1}{4}(4x -3y – 8)^2 +\dfrac{27}{4}y^2 + 2004 \geq 2004$
Dấu = xảy ra $\Leftrightarrow \begin{cases}4x – 3y – 8 = 0\\y = 0\end{cases}\Leftrightarrow \begin{cases}x = \\y = 0\end{cases}$
Vậy $\min P = 2004 \Leftrightarrow (x;y) = (2;0)$