Toán Lớp 8: Tìm GTLN của A=(2x+1)^2-(3x-2)^2+x-11
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Toán Lớp 8: Tìm GTLN của A=(2x+1)^2-(3x-2)^2+x-11
Comments ( 2 )
Giải đáp:
Lời giải và giải thích chi tiết:
A=(2x+1)^2 -(3x-2)^2 +x-11
=(2x+1-3x+2)(2x+1+3x-2)+x-11
=(-x+3)(5x-1)+x-11
=-5x^2 +15x+x-3+x-11
=-5x^2 +(15x+x+x)+(-3-11)
=-5x^2 +17x-14
=-5(x^2 -[17]/5 x +[289]/[100])+9/[20]
=-5(x -[17]/[10] )^2 +9/[20]
Vì -5(x -[17]/[10] )^2 ≤0 ∀x
=>-5(x-[17]/[10] )^2 +9/[20] ≤ 9/[20] ∀x
=>Max_A =9/[20] <=>x-[17]/[10] =0
<=>x=[17]/[10]
Giải đáp:
A=(2x+1)^2−(3x−2)^2+x−11
-> A = [(2x)^2 + 2.2x.1 + 1^2] – [(3x)^2 – 2.3x.2 + 2^2] + x – 11
-> A = (4x^2 + 4x +1) – (9x^2 – 12x + 4) + x – 11
-> A = 4x^2 + 4x +1 – 9x^2 + 12x – 4 + x – 11
-> A = (4x^2 – 9x^2) + (4x + 12x + x) + (1 – 4 – 11)
-> A = -5x^2 + 17x – 14
-> A = -5.(x^2 – 17/5x + 289/100) + 9/20
-> A = -5.(x- 17/10)^2 + 9/20
Nhận xét
(x – 17/10)^2 >= 0 ∀ x
-> -5.(x – 17/10)^2 ≤ 0 ∀x
-> -5.(x – 17/10)^2 + 9/20 ≤ 9/20
Dấu “=” xảy ra ↔ x = 17/10
Vậy A_(max) = 9/20 ↔ x = 17/10
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