Question

[Natsu Dragneel]

O.1       Tìm giá trị nhỏ nhất của các biểu thức sau:

a)  A = – 2x² + 5x – 17           b)  B =  – x² + 4x – 5     

c)  C = – 4x² – 4x – 2             d)  D = – 6 – 8x – 16x²  

Answer 1

[l҉o҉n҉g҉ d҉z҉]

Max chứ không phải Min bạn nhé !

A = -2x² + 5x - 17

A = -2( x² - 5/2x + 25/16 ) - 111/8

A = -2( x - 5/4 )² - 111/8

\(-2\left(x-\frac{5}{4}\right)^2\le0\forall x\Rightarrow-2\left(x-\frac{5}{4}\right)^2-\frac{111}{8}\le-\frac{111}{8}\)

Dấu " = " xảy ra <=> x - 5/4 = 0 => x = 5/4

=> MaxA = -111/8 <=> x = 5/4

B = -x² + 4x - 5

B = -x² + 4x - 4 - 1

B = -( x² - 4x + 4 ) - 1

B = -( x - 2 )² - 1

\(-\left(x-2\right)^2\le0\forall x\Rightarrow-\left(x-2\right)^2-1\le-1\)

Dấu " = " xảy ra <=> x - 2 = 0 => x = 2

=> MaxB = -1 <=> x = 2

C = -4x² - 4x - 2

C = -( 4x² + 4x + 1 ) - 1

C = -( 2x + 1 )² - 1

\(-\left(2x+1\right)^2\le0\forall x\Rightarrow-\left(2x+1\right)^2-1\le-1\)

Dấu " = " xảy ra <=> 2x + 1 = 0 => x = -1/2

=> MaxC = -1 <=> x = -1/2

D = -6 - 8x - 16x²

D = -16( x² + 1/2x + 1/16 ) - 5

D = -16( x + 1/4 )² - 5

\(-16\left(x+\frac{1}{4}\right)^2\le0\forall x\Rightarrow-16\left(x+\frac{1}{4}\right)^2-5\le-5\)

Dấu " = " xảy ra <=> x + 1/4 = 0 => x = -1/4

=> MaxD = -5 <=> x = -1/4

Answer 2

[Phan Nghĩa]

\(A=-2x^2+5x-17=-2\left(x^2-\frac{5}{2}+\frac{5^2}{4^2}\right)-\frac{111}{8}\)

\(=-2\left(x-\frac{5}{4}\right)^2-\frac{111}{8}\le-\frac{111}{8}\)

Dấu = xảy ra \(< =>-2\left(x-\frac{5}{4}\right)^2=0\Leftrightarrow x-\frac{5}{4}=0\Leftrightarrow x=\frac{5}{4}\)

Vậy \(Max_A=-\frac{111}{8}\)khi \(x=\frac{5}{4}\)

\(B=-x^2+4x-5=-\left(x^2-4x+4\right)-1\)

\(=-\left(x-2\right)^2-1\le-1\)

Dấu = xảy ra \(< =>-\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy \(Max_B=-1\)khi \(x=2\)

\(C=-4x^2-4x-2=-\left(4x^2+4x+2\right)\)

\(=-\left(4x^2+4x+1\right)-1=-\left(2x+1\right)^2-1\le-1\)

Dấu = xảy ra \(< =>-\left(2x+1\right)^2=0\Leftrightarrow2x+1=0\Leftrightarrow x=-\frac{1}{2}\)

Vậy \(Max_C=-1\)khi \(x=-\frac{1}{2}\)

\(D=-6-8x-16x^2=-\left(16x^2+8x+6\right)\)

\(=-\left[\left(4x\right)^2+2.4x+1\right]-5=-\left(4x+1\right)^2-5\le-5\)

Dấu = xảy ra \(< =>-\left(4x+1\right)^2=0\Leftrightarrow4x+1=0\Leftrightarrow x=-\frac{1}{4}\)

Vậy \(Max_D=-5\)khi \(x=-\frac{1}{4}\)