kẻ đường cao BH

BC²=BH²+HC²(pytago)

BH=AB.sin60; HC=AC-AH=AC-ABcos60 thay vào trên

BC²=(AB.sin60)²+(AC-ABcos60)²=AB².sin²60+AC²-2AB.ACcos60+AB².cos²60=AB²+AC²-2AB.AC.\(\frac{1}{2}\)=AB²+AC²-AB.AC

kẻ BH _|_ AC (H thuộc AC)

xét tam giác ABH có : góc A + góc ABH + góc AHB = 180 (ĐL)

Có : góc A = 60 (gt)

góc AHB = 90 do BH _|_ AC (Cách vẽ)

=> góc ABH = 180 - 90 - 60 = 30 

xét tam giác ABH vuông tại H có góc ABH = 30 

=> AH = 1/2.AB (đl)

=> AB = 2AH     (1)

xét tam giác ABH vuông tại H 

=> AB^2 = AH^2 + BH^2 (Đl PTG)

=> BH^2 = AB^2 - AH^2    (2) 

xét tam giác BHC vuông tại H : 

=> BC^2 = HC^2 + BH^2 (đl PTG)

=> BC^2 = BH^2 + (AC - AH)^2 

=> BC^2 = BH^2 + AC^2 - 2AH.AC + AH^2 

thay (1)(2) vào ta được : 

BC^2 = (AB^2 - AH^2) + AC^2 - AB.AC + AH^2

=> BC^2 = AB^2 - AH^2+ AC^2 - AB.AC + AH^2

=> BC^2 = AB^2 + AC^2 - AB.AC

cotg²a-cotg²a.cos²a=  cotg²a(1-cos²a)    =\(\frac{cos^2a}{sin^2a}sin^2a=cos^2a\)

\(DK:x,y>0\)

HPT\(\Leftrightarrow\hept{\begin{cases}3x^2y=2y^2+1\\3xy^2=2x^2+1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}3x^2y=2y^2+1\left(1\right)\\3xy\left(x-y\right)-2\left(x^2-y^2\right)=0\left(2\right)\end{cases}}\)

Xet PT(2)

\(\Leftrightarrow\left(x-y\right)\left(3xy-2x-2y\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=y\\y=\frac{2x}{3x-2}\end{cases}}\)

Xet \(x=y\)

Thay vao PT(1) ta duoc:

\(3x^3=2x^2+1\)

\(\Leftrightarrow3x^3-2x^2-1=0\)

\(\Leftrightarrow3x^2\left(x-1\right)+\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x^2+x+1\right)=0\)

Vi \(3x^2+x+1>0\left(\forall x>0\right)\)

\(\Rightarrow x=1\left(n\right)\)

\(\Rightarrow y=1\left(n\right)\)

Xet \(y=\frac{2x}{3x-2}\left(DK:x>\frac{3}{2}\right)\)

Cái này thay vào roi giải(dài lắm) 

Áp dụng BĐT AM - GM

\(A=\left(a+1\right)\left(1+\frac{1}{b}\right)+\left(b+1\right)\left(1+\frac{1}{a}\right)\)

\(=\frac{a}{b}+\frac{b}{a}+a+\frac{1}{a}+b+\frac{1}{b}+2\)

\(=\frac{a}{b}+\frac{b}{a}+\left(a+\frac{1}{2a}\right)+\left(b+\frac{1}{2b}\right)+\frac{1}{2a}+\frac{1}{2b}+2\)

\(\ge2\sqrt{\frac{a}{b}.\frac{b}{a}}+2\sqrt{a.\frac{1}{2a}}+2\sqrt{b.\frac{1}{2b}}+2\sqrt{\frac{1}{2a}.\frac{1}{2b}}+2\)

\(=4+2\sqrt{2}+\frac{1}{\sqrt{ab}}\ge4+2\sqrt{2}+\frac{1}{\frac{\sqrt{2\left(a^2+b^2\right)}}{2}}\)

\(=4+3\sqrt{2}\)

Dấu " = " xảy ra khi \(a=b=\frac{1}{\sqrt{2}}\)

Ta co:\(1=a^2+b^2\ge\frac{\left(a+b\right)^2}{2}\Rightarrow a+b\le\sqrt{2}\)

Ta lai co:

\(A=\frac{a}{b}+\frac{b}{a}+\frac{1}{a}+\frac{1}{b}+a+b+2\)

\(=\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{1}{a}+2a\right)+\left(\frac{1}{b}+2b\right)-\left(a+b\right)+2\)

\(\ge2+2\sqrt{2}+2\sqrt{2}-\sqrt{2}+2=4+3\sqrt{2}\)

Dau '=' xay ra khi \(a=b=\frac{1}{\sqrt{2}}\)

Vay \(A_{min}=4+3\sqrt{2}\)khi \(a=b=\frac{1}{\sqrt{2}}\)