2,
a,P=[TEX]x^2-2.x+5=x^2-2.x+1+4=(x-1)^2+4[/TEX]
Vì [TEX](x-1)^2 \ \geq \ 0[/TEX] nên [TEX]P \ \geq \ 4[/TEX]
Vậy min [TEX]P=4 \ \Leftrightarrow \ x=1[/TEX]
b,[TEX] Q=2x^2-6x=2.(x^2-3x)=2[(x^2-2.x.\frac{3}{2}+\frac{9}{4})-\frac{9}{4}][/TEX]
[TEX]=2.(x-\frac{3}{2})^2-\frac{9}{2}[/TEX]
Vì [TEX]2.(x-\frac{3}{2})^2 \ \geq \ 0 \Rightarrow Q \ \geq \ \frac{-9}{2}[/TEX]
Vậy min [TEX] Q=\frac{-9}{2} \ \Leftrightarrow \ x=\frac{3}{2}[/TEX]
c, [TEX]M=x^2 + y^2-x + 6y + 10=(x^2-x+\frac{1}{4})+(y^2+6y+9)+\frac{3}{4}[/TEX]
[TEX]=(x-\frac{1}{2})^2+(y+3)^2+\frac{3}{4}[/TEX]
Vì [TEX](x - \frac{1}{2})^2 \geq 0 ; (y+3)^2 \geq 0 \Rightarrow (x - \frac{1}{2})^2+(y+3)^2 \geq 0[/TEX]
[TEX]\Rightarrow \ M \geq \frac{3}{4}[/TEX]
Vậy min [TEX]M=\frac{3}{4} \ \Leftrightarrow \ x=\frac{1}{2} ; y=-3[/TEX]
a, [TEX]A = 4x - x^2 + 3 = - ( x^2 - 4x + 4 ) + 7 = - ( x - 2 )^2 + 4 \leq 4[/TEX]
Vậy max[TEX] A = 4 \Leftrightarrow x = 2 [/TEX]
b,[TEX] B = x - x^2 = - ( x^2 - x) = - ( x^2 - 2.\frac{1}{2}.x + \frac{1}{4} ) + \frac{1}{4 }[/TEX]
[TEX]= ( x - \frac{1}{2} )^2 + \frac{1}{4 } \leq \frac{1}{4}[/TEX]
Vậy max[TEX] B = \frac{1}{4} \Leftrightarrow x = \frac{1}{2}[/TEX]
c,[TEX] N = 2x - 2x^2 - 5 = -2.( x^2 - x + \frac{5}{2} ) = -2.( x^2 - x + \frac{1}{4} + \frac{9}{4} )[/TEX]
[TEX]= -2. (x - \frac{1}{2} )^2 - \frac{9}{2} \leq \frac{-9}{2}[/TEX]
Vậy max [TEX]N = \frac{-9}{2} \Leftrightarrow x = \frac{1}{2}[/TEX]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn