Lời giải:

$a^2-2ab-3b^2\geq 0$

$\Leftrightarrow (a^2+ab)-(3ab+3b^2)\geq 0$

$\Leftrightarrow a(a+b)-3b(a+b)\geq 0$

$\Leftrightarrow (a+b)(a-3b)\geq 0$

$\Leftrightarrow a-3b\geq 0$ (do $a+b>0$ với mọi $a,b>0$)

$\Leftrightarrow a\geq 3b$

Xét hiệu:

$P-\frac{37}{3}=\frac{4a^2+b^2}{ab}-\frac{37}{3}$

$=\frac{12a^2+3b^2-37ab}{3ab}=\frac{(a-3b)(12a-b)}{3ab}\geq 0$ do $a\geq 3b>0$

$\Rightarrow P\geq \frac{37}{3}$

Vậy $P_{\min}=\frac{37}{3}$

C

c

Bài 1:

\(\left(a+b+c\right)^2=a^2+b^2+c^2\)

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2\)

\(\Leftrightarrow ab+bc+ac=0\Leftrightarrow bc=-ab-ac\)

\(\dfrac{a^2}{a^2+2bc}=\dfrac{a^2}{a^2+bc-ab-ac}=\dfrac{a^2}{\left(a-c\right)\left(a-b\right)}\)

CMTT: \(\left\{{}\begin{matrix}\dfrac{b^2}{b^2+2ca}=\dfrac{b^2}{\left(b-c\right)\left(b-a\right)}\\\dfrac{c^2}{c^2+2ab}=\dfrac{c^2}{\left(b-c\right)\left(a-c\right)}\end{matrix}\right.\)

\(M=\dfrac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=\dfrac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)

Bài 2:

\(a^3+b^3+c^3-3abc=\left(a^3+3a^2b+3ab^2+b^3\right)+c^3-3abc-3a^2b-3ab^2\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)(do \(a+b+c=0\))

\(\Rightarrow A=\dfrac{0}{\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3}=0\)

Chọn D

a) \(\sqrt{\dfrac{3+\sqrt{5}}{2x^2}}-\sqrt{\dfrac{3-\sqrt{5}}{2}}\)

= \(\sqrt{\dfrac{6+2\sqrt{5}}{4x^2}}-\sqrt{\dfrac{6-2\sqrt{5}}{4}}=\sqrt{\dfrac{5+2\sqrt{5}+1}{4x^2}}-\sqrt{\dfrac{5-2\sqrt{5}+1}{4}}\) = \(\sqrt{\dfrac{\left(\sqrt{5}+1\right)^2}{\left(2x\right)^2}}-\sqrt{\dfrac{\left(\sqrt{5}-1\right)^2}{2^2}}=\dfrac{\left|\sqrt{5}+1\right|}{\left|2x\right|}-\dfrac{\left|\sqrt{5}-1\right|}{2}=\dfrac{\sqrt{5}+1}{2x}-\dfrac{\sqrt{5}-1}{2}\)

Thay x = 1 vào biểu thức \(\dfrac{\sqrt{5}+1}{2x}-\dfrac{\sqrt{5}-1}{2}\) ta được :

\(\dfrac{\sqrt{5}+1}{2}-\dfrac{\sqrt{5}-1}{2}=\dfrac{\sqrt{5}+1-\sqrt{5}+1}{2}=1\)

Vậy tại x =1 thì giá trị của biểu thức \(\sqrt{\dfrac{3+\sqrt{5}}{2x^2}}-\sqrt{\dfrac{3-\sqrt{5}}{2}}\) là bằng 1

b) \(\dfrac{\sqrt{a^3+4a^2+4a}}{\sqrt{a\left(a^2-2ab+b^2\right)}}-\dfrac{\sqrt{b^3-4b^2+4b}}{\sqrt{b\left(a^2-2ab+b^2\right)}}+ab\)

= \(\sqrt{\dfrac{a\left(a^2+4a+4\right)}{a\left(a^2-2ab+b^2\right)}}-\sqrt{\dfrac{b\left(b^2-4b+4\right)}{b\left(a^2-2ab+b^2\right)}}+ab\)

= \(\dfrac{\sqrt{\left(a+2\right)^2}}{\sqrt{\left(a-b\right)^2}}-\dfrac{\sqrt{\left(b-2\right)^2}}{\sqrt{\left(a-b\right)^2}}+ab=\dfrac{a+2}{a-b}-\dfrac{b-2}{a-b}+ab\) = a - b + ab

Thay a = 4 và b = 3 vào biểu thức a - b +ab ta được :

4 - 3 + 4.3 = 13

Vậy tại a = 4 ; b = 3 thì giá trị của biểu thức \(\dfrac{\sqrt{a^3+4a^2+4a}}{\sqrt{a\left(a^2-2ab+b^2\right)}}-\dfrac{\sqrt{b^3-4b^2+4b}}{\sqrt{b\left(a^2-2ab+b^2\right)}}+ab\) là bằng 13

c) \(ab^2.\sqrt{\dfrac{4}{a^2b^4}}+ab=ab^2.\dfrac{2}{ab^2}+ab=2+ab\)

Thay a = 1 và b = -2 vào BT : 2 + ab ta được :

2 + 1.(-2) = 2 + (-2) = 0

Vậy tại a = 1 ; b = -2 thì giá trị của biểu thức \(ab^2.\sqrt{\dfrac{4}{a^2b^4}}+ab\) là bằng 0

d) \(\dfrac{a+b}{b^2}.\sqrt{\dfrac{a^2b^2}{a^2+2ab+b^2}}\) = \(\dfrac{a+b}{b^2}.\dfrac{\sqrt{a^2b^2}}{\sqrt{a^2+2ab+b^2}}=\dfrac{a+b}{b^2}.\dfrac{ab}{a+b}=\dfrac{ab}{b^2}\)

Thay a = 1 ; b =2 vào BT : \(\dfrac{ab}{b^2}\) ta được : \(\dfrac{1.2}{2^2}=\dfrac{1}{2}\)

Vậy tại a =1 ; b =2 GT của BT : \(\dfrac{a+b}{b^2}.\sqrt{\dfrac{a^2b^2}{a^2+2ab+b^2}}\) là \(\dfrac{1}{2}\)

\(\left(a+b+c\right)^2=a^2+b^2+c^2\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2\)

\(\Leftrightarrow2\left(ab+bc+ac\right)=0\Leftrightarrow ab+bc+ac=0\Leftrightarrow bc=-ab-ac\)

\(\dfrac{a^2}{a^2+2bc}=\dfrac{a^2}{a^2+bc-ac-ab}=\dfrac{a^2}{\left(a-c\right)\left(a-b\right)}\)

CMTT: \(\left\{{}\begin{matrix}\dfrac{b^2}{b^2+2ca}=\dfrac{b^2}{\left(b-a\right)\left(b-c\right)}\\\dfrac{c^2}{c^2+2ab}=\dfrac{c^2}{\left(c-a\right)\left(c-b\right)}=\dfrac{c^2}{\left(a-c\right)\left(b-c\right)}\end{matrix}\right.\)

\(\Rightarrow A=\dfrac{a^2}{\left(a-c\right)\left(a-b\right)}+\dfrac{b^2}{\left(b-a\right)\left(b-c\right)}+\dfrac{c^2}{\left(a-c\right)\left(b-c\right)}=\dfrac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\dfrac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=1\)

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

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

(a+3)^3=(a+b)^2*(a+b)

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

=a^3+a^2b+2a^2b+2ab^2+b^2a+b^3

=a^3+3a^2b+3ab^2+b^3