Question

[vietdat vietdat]

Bài1

a)a⁴+b⁴≥a³b+ab³ ∀ a,b ∈ R

b)(x-3)(x-4)(x-5)(x-6)+3 >0 ∀ x

mn giải giup mk vs mk đang cần gấp thanks mn nha

Answer 1

[Akai Haruma]

Lời giải:

a) Xét hiệu:

\(a^4+b^4-(a^3b+ab^3)\)

\(=(a^4-a^3b)-(ab^3-b^4)\)

\(=a^3(a-b)-b^3(a-b)=(a-b)(a^3-b^3)=(a-b)(a-b)(a^2+ab+b^2)\)

\(=(a-b)^2(a^2+ab+b^2)\)

Ta thấy: \((a-b)^2\geq 0, \forall a,b\in\mathbb{R}\)

\(a^2+ab+b^2=(a+\frac{b}{2})^2+\frac{3b^2}{4}\geq 0, \forall a,b\in\mathbb{R}\)

\(\Rightarrow a^4+b^4-(a^3b+ab^3)=(a-b)^2(a^2+ab+b^2)\geq 0, \forall a,b\in\mathbb{R}\)

\(\Rightarrow a^4+b^4\geq ab^3+a^3b\) với mọi $a,b\in\mathbb{R}$

Ta có đpcm.

Dấu "=" xảy ra khi $a=b$

b)

\((x-3)(x-4)(x-5)(x-6)+3\)

\(=[(x-3)(x-6)][(x-4)(x-5)]+3\)

\(=(x^2-9x+18)(x^2-9x+20)+3\)

\(=a(a+2)+3\) (đặt \(x^2-9x+18=a)\)

\(=a^2+2a+3=(a+1)^2+2\geq 2>0, \forall a\in\mathbb{R}\)

hay \((x-3)(x-4)(x-5)(x-6)+3>0, \forall x\in\mathbb{R}\) (đpcm)

Answer 2

[nguyen the vuong]

a) Xét hiệu:

a4+b4−(a3b+ab3)a4+b4−(a3b+ab3)

=(a4−a3b)−(ab3−b4)=(a4−a3b)−(ab3−b4)

=a3(a−b)−b3(a−b)=(a−b)(a3−b3)=(a−b)(a−b)(a2+ab+b2)=a3(a−b)−b3(a−b)=(a−b)(a3−b3)=(a−b)(a−b)(a2+ab+b2)

=(a−b)2(a2+ab+b2)=(a−b)2(a2+ab+b2)

Ta thấy: (a−b)2≥0,∀a,b∈R(a−b)2≥0,∀a,b∈R

a2+ab+b2=(a+b2)2+3b24≥0,∀a,b∈Ra2+ab+b2=(a+b2)2+3b24≥0,∀a,b∈R

⇒a4+b4−(a3b+ab3)=(a−b)2(a2+ab+b2)≥0,∀a,b∈R⇒a4+b4−(a3b+ab3)=(a−b)2(a2+ab+b2)≥0,∀a,b∈R

⇒a4+b4≥ab3+a3b⇒a4+b4≥ab3+a3b với mọi a,b∈Ra,b∈R

Ta có đpcm.

Dấu "=" xảy ra khi a=ba=b

b)

(x−3)(x−4)(x−5)(x−6)+3(x−3)(x−4)(x−5)(x−6)+3

=[(x−3)(x−6)][(x−4)(x−5)]+3=[(x−3)(x−6)][(x−4)(x−5)]+3

=(x2−9x+18)(x2−9x+20)+3=(x2−9x+18)(x2−9x+20)+3

=a(a+2)+3=a(a+2)+3 (đặt x2−9x+18=a)x2−9x+18=a)

=a2+2a+3=(a+1)2+2≥2>0,∀a∈R=a2+2a+3=(a+1)2+2≥2>0,∀a∈R

hay (x−3)(x−4)(x−5)(x−6)+3>0,∀x∈R(x−3)(x−4)(x−5)(x−6)+3>0,∀x∈R (đpcm)

a) Xét hiệu:

a4+b4−(a3b+ab3)a4+b4−(a3b+ab3)

=(a4−a3b)−(ab3−b4)=(a4−a3b)−(ab3−b4)

=a3(a−b)−b3(a−b)=(a−b)(a3−b3)=(a−b)(a−b)(a2+ab+b2)=a3(a−b)−b3(a−b)=(a−b)(a3−b3)=(a−b)(a−b)(a2+ab+b2)

=(a−b)2(a2+ab+b2)=(a−b)2(a2+ab+b2)

Ta thấy: (a−b)2≥0,∀a,b∈R(a−b)2≥0,∀a,b∈R

a2+ab+b2=(a+b2)2+3b24≥0,∀a,b∈Ra2+ab+b2=(a+b2)2+3b24≥0,∀a,b∈R

⇒a4+b4−(a3b+ab3)=(a−b)2(a2+ab+b2)≥0,∀a,b∈R⇒a4+b4−(a3b+ab3)=(a−b)2(a2+ab+b2)≥0,∀a,b∈R

⇒a4+b4≥ab3+a3b⇒a4+b4≥ab3+a3b với mọi a,b∈Ra,b∈R

Ta có đpcm.

Dấu "=" xảy ra khi a=ba=b

b)

(x−3)(x−4)(x−5)(x−6)+3(x−3)(x−4)(x−5)(x−6)+3

=[(x−3)(x−6)][(x−4)(x−5)]+3=[(x−3)(x−6)][(x−4)(x−5)]+3

=(x2−9x+18)(x2−9x+20)+3=(x2−9x+18)(x2−9x+20)+3

=a(a+2)+3=a(a+2)+3 (đặt x2−9x+18=a)x2−9x+18=a)

=a2+2a+3=(a+1)2+2≥2>0,∀a∈R=a2+2a+3=(a+1)2+2≥2>0,∀a∈R

hay (x−3)(x−4)(x−5)(x−6)+3>0,∀x∈R(x−3)(x−4)(x−5)(x−6)+3>0,∀x∈R (đpcm)v