Question

Bài 1: Tìm Min A = x² - 3x + 9

Min B = 9x² - 6x + 2

Max C = -x² + 2x + 4

Max D = -x² + 4x

Bài 2: Cho x + y = 2

x² + y² = 10. Tính x³ + y³.

Answer 1

Câu 1:

\(A=x^2-3x+9\\ =x^2-3x+\dfrac{9}{4}+\dfrac{27}{4}\\ =\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{27}{4}\\ =\left(x-\dfrac{3}{2}\right)^2+\dfrac{27}{4}\\ Do\text{ }\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\\ \Rightarrow A=\left(x-\dfrac{3}{2}\right)^2+\dfrac{27}{4}\ge0\forall x\\ \text{Dấu “=” xảy ra khi: }\\ \left(x-\dfrac{3}{2}\right)^2=0\\ \Leftrightarrow x-\dfrac{3}{2}=0\\ \Leftrightarrow x=\dfrac{3}{2}\\ Vậy\text{ }A_{\left(Min\right)}=\dfrac{27}{4}\text{ }khi\text{ }x=\dfrac{3}{2}\)

\(B=9x^2-6x+2\\ =9x^2-6x+1+1\\ =\left(9x^2-6x+1\right)+1\\ =\left(3x-1\right)^2+1\\ Do\text{ }\left(3x-1\right)^2\ge0\forall x\\ \Rightarrow B=\left(3x-1\right)^2+1\ge1\forall x\\ \text{Dấu “=” xảy ra khi: }\\ \left(3x-1\right)^2=0\\ \Leftrightarrow3x-1=0\\ \Leftrightarrow3x=1\\ \Leftrightarrow x=\dfrac{1}{3}\\ Vậy\text{ }B_{\left(Min\right)}=1\text{ }khi\text{ }x=\dfrac{1}{3}\)

\(C=-x^2+2x+4\\ =-x^2+2x-1+5\\ =-\left(x^2-2x+1\right)+5\\ =-\left(x-1\right)^2+5\\ Do\text{ }\left(x-1\right)^2\ge0\forall x\\ \Rightarrow-\left(x-1\right)^2\le0\forall x\\ \Rightarrow C=-\left(x-1\right)^2+5\le5\forall x\\ \text{ Dấu “=” xảy ra khi: }\\ \left(x-1\right)^2=0\\ \Leftrightarrow x-1=0\\ \Leftrightarrow x=1\\ \text{Vậy }C_{\left(Max\right)}=5\text{ }khi\text{ }x=1\)

\(D=-x^2+4x\\ =-x^2+4x-4+4\\ =-\left(x^2-4x+4\right)+4\\ =-\left(x-2\right)^2+4\\ \\ Do\text{ }\left(x-2\right)^2\ge0\forall x\\ \Rightarrow-\left(x-2\right)^2\le0\forall x\\ \Rightarrow C=-\left(x-2\right)^2+4\le4\forall x\\ \text{ Dấu “=” xảy ra khi: }\\ \left(x-2\right)^2=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\\ \text{Vậy }C_{\left(Max\right)}=4\text{ }khi\text{ }x=2\)

Answer 2

Câu 2:

\(\text{Ta có : }x+y=2\\ \Rightarrow\left(x+y\right)^2=2^2\\ \Rightarrow x^2+2xy+y^2=4\\ Thay\text{ }x^2+y^2=10\text{ }vào\\ \Rightarrow2xy+10=4\\ \Rightarrow2xy=-6\\ \Rightarrow xy=-3\\ \text{Ta lại có : }x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\\ Thay\text{ }x^2+y^2=10;x+y=2;xy=-3\text{ }ta\text{ }được:\\ x^3+y^3=2\cdot\left(10+3\right)=26\)

Vậy \(x^3+y^3=26\text{ }tại\text{ }x+y=2;x^2+y^2=10\)