Tìm GTNN của biểu thức A=a^4-2a^3+3a^2-4a+5

\(A=a^4-2a^3+3a^2-4a+5\)

\(A=\left(a^4-2a^3+a^2\right)+\left(2a^2-4a+2\right)+3\)

\(A=\left(a^2-a\right)^2+\left(\sqrt{2}a-\sqrt{2}\right)^2+3\)

Do \(\left(a^2-a\right)^2+\left(\sqrt{2}a-\sqrt{2}\right)^2\ge0\forall a\)

Nên \(\left(a^2-a\right)^2+\left(\sqrt{2}a-\sqrt{2}\right)^2+3\ge3\forall a\)

Dấy "=" xả ra khi a = 1

Vậy Min A = 3 khi a = 1

1. Tìm GTNN

a) E = x² - 2x + y² + 4y + 8

b) F = x² - 4x + y² - 8y + 6

2. Phân tích đa thức thành nhân tử

a) 3x² - 3y² - 2 (x-y)²

b) x³ - 4x² - 9x + 36

c) 3x² - 6xy + 3y² - 12z²

d) 5x² - 10xy + 5y² - 20x²

Câu 1:

\(\text{ a) }E=x^2-2x+y^2+4y+8\\ E=x^2-2x+y^2+4y+1+4+3\\ E=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+3\\ E=\left(x-1\right)^2+\left(y+2\right)^2+3\\ Do\text{ }\left(x-1\right)^2\ge0\forall x\\ \left(y+2\right)^2\ge0\forall y\\ \Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x;y\\ \Rightarrow E=\left(x-1\right)^2+\left(y+2\right)^2+3\ge3\forall x;y\\ \text{ Dấu “=” xảy ra khi : }\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\\ \text{ Vậy }E_{\left(Min\right)}=3\text{ }khi\text{ }\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

\(\text{b) }F=x^2-4x+y^2-8y+6\\ F=x^2-4x+y^2-8y+4+16-14\\ F=\left(x^2-4x+4\right)+\left(y^2-8y+16\right)-14\\ F=\left(x-2\right)^2+\left(y-4\right)^2-14\\ Do\text{ }\left(x-2\right)^2\ge0\forall x\\ \left(y-4\right)^2\ge0\forall y\\ \left(x-2\right)^2+\left(y-4\right)^2\ge0\forall x;y\\ F=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\forall x;y\\ \text{Dấu “=” xảy ra khi : }\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left(y-4\right)^2=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\\ \text{Vậy }F_{\left(Min\right)}=-14\text{ }khi\text{ }\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)

Bai 1: tim x

a) 5x(x-3)² - 5(x-1)³ + 15(x+2)(x-2)= 5

b) (x+2) (3-4x) = x² + 4x + 4

c) x² (x² + 1) - x² -1= 0

a/ \(5x\left(x-3\right)^2-5\left(x-1\right)^3+15\left(x+2\right)\left(x-2\right)=5\)

\(\Leftrightarrow5x\left(x^2-6x+9\right)-5\left(x^3-3x^2+3x-1\right)+15\left(x^2-4\right)=5\)

\(\Leftrightarrow5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-60-5=0\)

\(\Leftrightarrow30x-60=0\)

\(\Leftrightarrow30x=60\)

\(\Leftrightarrow x=2\)

vậy x=2

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

\(\Leftrightarrow3x-4x^2+6-8x=x^2+4x+4\)

\(\Leftrightarrow x^2+4x^2+4x+18x-3x+4-6=0\)

\(\Leftrightarrow5x^2+9x-2=0\)

\(\Leftrightarrow\left(5x-1\right)\left(x+2\right)=0\)

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

vậy \(x=\dfrac{1}{5}\) hoặc \(x=-2\)

c/ \(x^2\left(x^2+1\right)-x^2-1=0\)

\(\Leftrightarrow x^2\left(x^2+1\right)-\left(x^2+1\right)=0\)

\(\Leftrightarrow\left(x^2+1\right)\left(x^2-1\right)=0\)

vì x²+1 >0 nên x² - 1 = 0 \(\Rightarrow x^2=1\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

vậy \(x=1\) hoặc \(x=-1\)