Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(A=x^2-3x-x+3+11\)
\(=\left(x^2-4x+4\right)+10\)
\(=\left(x-2\right)^2+10\ge10\forall x\in R\)
Dấu "=" xảy ra<=> \(\left(x-2\right)^2=0\Leftrightarrow x=2\)
b) \(B=5-4x^2+4x\)
\(=-\left(4x^2-4x+1\right)+6\)
\(=-\left(2x-1\right)^2+6\le6\forall x\in R\)
Dấu "=" xảy ra<=> \(-\left(2x-1\right)^2=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
c) \(C=\left(x^2-3x+1\right)\left(x^2-3x-1\right)\)
\(=\left(x^2-3x\right)^2-1\ge-1\forall x\in R\)
Dấu "=" xảy ra<=>\(\left(x^2-3x\right)^2=0\Leftrightarrow x\left(x-3\right)=0\Leftrightarrow x=0;x=3\)
a: \(A=x^2-3x+\dfrac{9}{4}+\dfrac{11}{4}=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)
Dấu '=' xảy ra khi x=3/2
b: \(B=4x^2-4x+1+x^2+4x+2\)
\(=5x^2+3>=3\)
Dấu '=' xảy ra khi x=0
d: \(D=-\left(x^2-4x+4-4\right)=-\left(x-2\right)^2+4< =4\)
Dấu '=' xảy ra khi x=2
\(A\left(x\right)=-\left(x^2-\frac{5}{3}x\right)+1=-3\left(x^2-2.x.\frac{5}{6}+\left(\frac{5}{6}\right)^2\right)+1+3.\left(\frac{5}{6}\right)^2\)
\(=-3\left(x-\frac{5}{6}\right)^2+\frac{37}{12}\le\frac{37}{12}\)
Dấu "=" xảy ra khi \(x-\frac{5}{6}=0\Leftrightarrow x=\frac{5}{6}\)
Vậy GTLN của A là 37/12.
b, c làm tương tự.
A = x2 - 3x - 5 = ( x2 - 3x + 9/4 ) - 29/4 = ( x - 3/2 )2 - 29/4 ≥ -29/4 ∀ x
Dấu "=" xảy ra khi x = 3/2
=> MinA = -29/4 <=> x = 3/2
B = 5x - x2 - 2021 = -( x2 - 5x + 25/4 ) - 8059/4 = -( x - 5/2 )2 - 8059/4 ≤ -8059/4 ∀ x
Dấu "=" xảy ra khi x = 5/2
=> MaxB = -8059/4 <=> x = 5/2
C = 4x2 - 4x - 11 = ( 4x2 - 4x + 1 ) - 12 = ( 2x - 1 )2 - 12 ≥ -12 ∀ x
Dấu "=" xảy ra khi x = 1/2
=> MinC = -12 <=> x = 1/2
D = 3x - x2 - 15 = -( x2 - 3x + 9/4 ) - 51/4 = -( x - 3/2 )2 - 51/4 ≤ -51/4 ∀ x
Dấu "=" xảy ra khi x = 3/2
=> MaxD = -51/4 <=> x = 3/2
a) \(A= 2x^2- 3x +1\)
\(=2\left(x^2-\dfrac{3}{2}x+\dfrac{1}{2}\right)\)
\(=2\left(x^2-2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{1}{16}\right)\)
\(=2\left(x-\dfrac{3}{4}\right)^2-\dfrac{1}{8}\ge-\dfrac{1}{8}\)
Vậy Amin = \(-\dfrac{1}{8}\) khi \(x=\dfrac{3}{4}\)
b) \(B= 4x^2 +7x + 13\)
\(=\left(2x\right)^2+2\cdot2x\cdot\dfrac{7}{4}+\dfrac{49}{16}+\dfrac{159}{16}\)
\(=\left(2x+\dfrac{7}{4}\right)^2+\dfrac{159}{16}\ge\dfrac{159}{16}\)
Vậy Bmin = \(\dfrac{159}{16}\) khi \(x=-\dfrac{7}{8}\)
c) \(C= 5-8x+x^2\)
\(=x^2-2\cdot x\cdot4+16+9\)
\(=\left(x-4\right)^2+9\ge9\)
Vậy Cmin = 9 khi x = 4
d) \(D = (x-1)(x+2)(x+3)(x+6)\)
\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(=\left(x^2+5x\right)^2-36\ge-36\)
Vậy Dmin = - 36 khi \(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
\(A=x^2-4x^2+2-1=\left(x-2\right)^2-1\)
suy ra Amin=-1
\(B=4x^2+4x+11=4\left(x^2+x+\frac{11}{4}\right)=4\left(x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{10}{4}\right)=4\left(x+\frac{1}{2}\right)^2+10\) Suy ra Bmin = 10
Theo mình nghĩ thì phải là giá trị lớn nhất
A=-(x^2-4x+5)
A=-[(x-2)^2+1]
Mà (x-2)^2+1>=1
Nên A<=-1
B=-(x^2+6x-1)
B=-[(x+3)^2-10]
nên B<=10
C=-(x^2+3x+2)
C=-(x^2+3x+9/4-1/4)
C=-[(x+3/2)^2-1/4]
Nên C<=1/4
D=-(2x^2-3x+1)
D=-2(x^2-3x/2+1/2)
D=-2(x^2-3x/2+9/16-1/16)
D=-2[(x-3/2)^2-1/16]
Nên D<=1/8
Chúc bạn học tốt!
a)C=(x2-3x+1)2>=0
c ) \(C=\left(x^2-3x+1\right)\left(x^2-3x+1\right)=\left(x^2-3x+1\right)^2\ge0\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x^2-3x+1=0\)
\(\Leftrightarrow x^2-3x+\dfrac{9}{4}-\dfrac{5}{4}=0\)
\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=\dfrac{5}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{\sqrt{5}}{2}\\x-\dfrac{3}{2}=\dfrac{-\sqrt{5}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{5}+3}{2}\\x=\dfrac{3-\sqrt{5}}{2}\end{matrix}\right.\)
Vậy Min C là : \(0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{5}+3}{2}\\x=\dfrac{3-\sqrt{5}}{2}\end{matrix}\right.\)
d ) \(D=\left(x^2-4x+1\right)\left(x^2-4x+5\right)\)
\(=\left(x^2-4x+3-2\right)\left(x^2-4x+3+2\right)\)
\(=\left(x^2-4x+3\right)^2-4\ge-4\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x^2-4x+3=0\)
\(\Leftrightarrow x^2-3x-x+3=0\)
\(\Leftrightarrow x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
Vậy Min D là : \(-4\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)