Tìm GTNN của đa thức:A=x(x-6)
và GTLN của đa thức :B=-3x(x+3)-7
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Bài 5:
a) \(A=x^2-4x+9=\left(x^2-4x+4\right)+5=\left(x-2\right)^2+5\ge5\)
\(minA=5\Leftrightarrow x=2\)
b) \(B=x^2-x+1=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(minB=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}\)
c) \(C=2x^2-6x=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)
\(minC=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\)
Bài 4:
a) \(M=4x-x^2+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)
\(maxM=7\Leftrightarrow x=2\)
b) \(N=x-x^2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
\(maxN=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{2}\)
c) \(P=2x-2x^2-5=-2\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{9}{2}=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\le-\dfrac{9}{2}\)
\(maxP=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{1}{2}\)
A(x) + B(x) = x4 - 3x + 3 + x4 - x + 128
A(x) +B(x) = (x4 + x4) - (3x+x) +( 3 +128)
A(x) + B(x) = 2x4 - 4x + 131
A(x) -B(x) = x4 - 3x + 3 - (x4 - x + 128)
A(x) -B(x) = x4 - 3x + 3 - x4 + x - 128
A(x) - B(x) = (x4 - x4) - (3x - x) - ( 128 - 3)
A(x) - B(x) = 0 - 2x - 125
A(x) - B(x) = -2x - 125
A(x) = x4 + 3 - 3x
A(x) = x4 - 3x + 3
B(x) = 53 + 3 - 3x2 + x4 - 2x + 3x2 + x
B(x) = (125 + 3) - ( 3x2 - 3x2) + x4 -( 2x - x)
B(x) = 128 - 0 + x4 - x
B(x) = x4 - x + 128
b, A(2) = 24 - 3 \(\times\) 2 + 3
A(2) = 16 - 6 + 3
A(2) = 10 + 3
A(2) = 13
a) dễ tự làm
b) A(x) có bậc 6
hệ số: -1 ; 5 ; 6 ; 9 ; 4 ; 3
B(x) có bậc 6
hệ số: 2 ; -5 ; 3 ; 4 ; 7
c) bó tay
d) cx bó tay
a: \(B\left(x\right)=-\left(x^2-3x+7\right)\)
\(=-\left(x^2-3x+\dfrac{9}{4}+\dfrac{19}{4}\right)\)
\(=-\left(x-\dfrac{3}{2}\right)^2-\dfrac{19}{4}\le-\dfrac{19}{4}\)
Dấu '=' xảy ra khi x=3/2
b: Ta có: \(C\left(x\right)=-x^2+7x-20\)
\(=-\left(x^2-7x+20\right)\)
\(=-\left(x^2-7x+\dfrac{49}{4}+\dfrac{31}{4}\right)\)
\(=-\left(x-\dfrac{7}{2}\right)^2-\dfrac{31}{4}\le-\dfrac{31}{4}\)
Dấu '=' xảy ra khi x=7/2
Ta có: M=−x2−2x+5
=−(x2+2x−5)
=−(x2+2x+1)+6
=−(x+1)2+6
Vì −(x+1)2≤0∀x
⇒−(x+1)2+6≤6∀x
Dấu "=" xảy ra ⇔
Vậy
Đặt A=4x−x2+3
=−x2+4x+3=−(x2−4x−3)
=−(x2−4x+4−7)
=−[(x−2)2−7]
=−(x−2)2+7
Ta có: −(x−2)2≤0⇒−(x−2)2+7≤7
Dấu " = " khi (x−2)2=0⇔x=2
Vậy MAXA=7 khi x = 2
\(D=-3x\left(x+3\right)-7=-3x^2-9x-7=-3\left(x^2+2x.\frac{3}{2}+\frac{9}{4}-\frac{9}{4}\right)-7\)
\(=-3\left[\left(x+\frac{3}{2}\right)^2-\frac{9}{4}\right]-7=-3\left(x+\frac{3}{2}\right)^2+\frac{27}{4}-7=-3\left(x+\frac{3}{2}\right)^2-\frac{1}{4}\) < \(-\frac{1}{4}\)
Dấu "=" xảy ra <=> \(-3\left(x+\frac{3}{2}\right)^2=0< =>x=-\frac{3}{2}\)
Vậy maxD=-1/4 khi x=-3/2
a,Ta có :\(A=x\left(x-6\right)=x^2-6x\)
\(=x^2-6x+9-9\)
\(=\left(x-3\right)^2-9\)
Vì: \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow\)\(\left(x-3\right)^2-9\ge-9\forall x\)
Hay: \(A\ge-9\forall x\)
Dấu = xảy ra khi (x-3)^2=0
<=>x=3
Vậy Min A= -9 tại x=3
b,Ta có: \(B=-3x\left(x+3\right)-7\)
\(=-3x^2-9x-7\)
\(=-3\left(x^2+3x+\frac{7}{3}\right)\)
\(=-3\left[\left(x^2+3x+\frac{9}{4}\right)+\frac{1}{12}\right]\)
\(=-3\left[\left(x+\frac{3}{2}\right)^2+\frac{1}{12}\right]\)
\(=-3\left(x+\frac{3}{2}\right)^2-\frac{1}{4}\)
Vì: \(-3\left(x+\frac{3}{2}\right)^2\le0\forall x\)
\(\Rightarrow-3\left(x+\frac{3}{2}\right)^2-\frac{1}{4}\le\frac{-1}{4}\forall x\)
Hay \(B\le\frac{-1}{4}\forall x\)
Dấu = xảy ra khi \(-3\left(x+\frac{3}{2}\right)^2=0\)
\(\Rightarrow x=\frac{-3}{2}\)
Vậy Max B=-1/4 tại x=-3/2
a) \(A=x\left(x-6\right)=x^2-6x+9-9=\left(x-3\right)^2-9\ge-9\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=3\)
Vậy Min A = -9 khi x = 3
b) \(B=-3x\left(x+3\right)-7=-3x^2-9x-7=-3\left(x^2+9x+20,25\right)+53,75\)
\(=-3\left(x+4,5\right)^2+53,75\le53,75\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=-4,5\)
Vậy Max B = 53,75 khi x = -4,5