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a: \(A=x^2-4x+4-3=\left(x-2\right)^2-3>=-3\)
Dấu = xảy ra khi x=2
b: \(x^2+4x-10=x^2+4x+4-14=\left(x+2\right)^2-14>=-14\)
\(\Leftrightarrow\dfrac{4}{x^2+4x-10}< =-\dfrac{4}{14}\)
=>B>=2/7
Dấu = xảy ra khi x=-2
c: \(x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}\)
=>2/x^2-x+1<=2:3/4=8/3
=>C>=-8/3
Dấu = xảy ra khi x=1/2
d: x^2-6x+12=(x-3)^2+3>=3
=>6/x^2-6x+12<=2
=>D>=-2
Dấu = xảy ra khi x=3
\(a,14x^2+3x+9=14\left(x^2+\dfrac{14}{3}x+\dfrac{49}{9}\right)-\dfrac{605}{9}\ge\dfrac{-605}{9}\)(câu a âm mà)
Câu b cũng thế !
\(x^2+8x+16=\left(x+4\right)^2\ge0\)
Vậy ....
làm nốt
d) (2x-1)(3x+2)(3-x)
=(6x2+x-2)(3-x)
=-6x3+17x2+5x-6
e) (x+3)(x2+3x-5)
=x3+6x2+4x-15
f) (xy-2)(x3-2x-6)
=x4y-2x3-2x2y-6xy+4x+12
g) (5x3-x2+2x-3)(4x2-x+2)
=20x5-9x4+19x3-16x2+7x-6
Bài 1:
a) (x-2)(x2+3x+4)
=x(5x+4)-2(5x+4)
= 5x2+4x-10x-8
=5x2-6x-8
p =1 => x = 27
p = 2 => x= 125
p = 3 => x = 343
.................
a)= -(x2 -2x +1) +1 +4
GTLN = 5
b)= -( x2 -4x +4) +4
GTLN = 4
c) = -4( x2 - x/4 + 1/16) +1/4 -5
GTLN = -19/5
\(\left(4x+1\right)\left(-4x+1\right)-16x\left(-5\right)=17\)
\(\Leftrightarrow\left(1+4x\right)\left(1-4x\right)+80x-17=0\)
\(\Leftrightarrow1-16x^2+80x-17=0\)
\(\Leftrightarrow-16x^2-16+80x=-16\left(x^2-5x+1\right)=0\Leftrightarrow-16\left[\left(x^2-5x+\dfrac{25}{4}\right)-\dfrac{21}{4}\right]=0\Leftrightarrow-16\left(x-\dfrac{5}{2}\right)^2+84=0\Rightarrow\left(x-\dfrac{5}{2}\right)^2=\dfrac{21}{4}\) \(\Rightarrow\left[{}\begin{matrix}x-\dfrac{5}{2}=\sqrt{\dfrac{21}{4}}\\x-\dfrac{5}{2}=-\sqrt{\dfrac{21}{4}}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{21}{4}}+\dfrac{5}{2}\\x=-\sqrt{\dfrac{21}{4}}+\dfrac{5}{2}\end{matrix}\right.\)