Ta có:

\(\frac{u+2}{u-2}=\frac{v+3}{v-3}\)

<=> \(\left(u+2\right)\left(v-3\right)=\left(u-2\right)\left(v+3\right)\)

<=> \(uv+2v-3u-6=uv-2v+3u-6\)

<=> \(2v-3u=3u-2v\)

<=> \(2v+2v=3u+3u\)

<=> \(4v=6u\)

<=> \(2v=3u\)

<=> \(\frac{u}{2}=\frac{v}{3}\)

Ta có:

\(\frac{u+2}{u-2}=\frac{v+3}{v-3}\)

\(\Leftrightarrow\left(u+2\right)\left(v-3\right)=\left(u-2\right)\left(v+3\right)\)

mình có sửa lại đề 1 chút!

đặt \(T=\sqrt{\frac{u-8\sqrt[6]{u^3v^2}+4\sqrt[3]{v^2}}{\sqrt{u}-2\sqrt[3]{v}+2\sqrt[12]{u^3v^2}}+3\sqrt[3]{v}}+\sqrt[6]{v}=1\)

đặt \(u=a^4;v=b^6\)(a,b>0) ta có

\(T=\frac{u-8\sqrt[6]{u^3v^2}+4\sqrt[3]{v^2}}{\sqrt{u}-2\sqrt[3]{v}+2\sqrt[12]{u^3v^2}}+3\sqrt[3]{v}=\frac{a^4-8a^2b^2+4b^2}{a^2-2b^2+2ab}+3b^2\)

vậy \(T=\frac{a^4-8a^2b^2+4b^4}{a^2-2b^2+2ab}+3b^2=\frac{a^4-5a^2b^2-2b^4+6ab^3}{a^2-2b^2+2ab}=a^2-2ab+b^2\)

từ đó suy ra \(\sqrt{\frac{u-8\sqrt[6]{u^3v^2}+4\sqrt[3]{v^2}}{\sqrt{u}-2\sqrt[3]{v}+2\sqrt[12]{u^3v^2}}+3\sqrt[3]{v}}+\sqrt[6]{v}=\left|\sqrt[4]{u}-\sqrt[6]{v}\right|+\sqrt[6]{v}\)

vì \(u^3\ge v^2\)nên \(\left|\sqrt[4]{u}-\sqrt[6]{v}\right|+\sqrt[6]{v}=\sqrt[4]{u}\)

\(\sqrt{\frac{u-8\sqrt[6]{u^3v^2}+4\sqrt[3]{v^2}}{\sqrt{u}-2\sqrt[3]{v}+2\sqrt[12]{u^3v^2}}+3\sqrt[3]{v}}+\sqrt[6]{v}=1\)

với u=1 ta có \(T=\sqrt{\frac{1-8\sqrt[6]{v^2}+4\sqrt[3]{v^2}}{1-2\sqrt[3]{v}+2\sqrt[6]{v^2}}+3\sqrt[3]{v}}+\sqrt[6]{v}\)

nếu \(1-2\sqrt[3]{v}+2\sqrt[6]{v}=0\)thì \(\sqrt[3]{v}=\frac{3+1}{2}>0\)

do \(v^2>1=u^3\), mâu thuẫn suy ra \(1-2\sqrt[3]{v}+2\sqrt[6]{v}\ne0\)

tóm lại với \(u^3\ge v^2\)và u,v\(\inℚ^+\)để \(\sqrt{\frac{u-8\sqrt[6]{u^3v^2}+4\sqrt[3]{v^2}}{\sqrt{u}-2\sqrt[3]{v}+2\sqrt[12]{u^3v^2}}+3\sqrt[3]{v}}+\sqrt[6]{v}=1\)cần và đủ là u=1 và v<1, v\(\inℚ^+\)được lấy tùy ý

Ta có :

<=> u³ - 3u - 2 \(\le\) v³ - 3v + 2 <=> ( u + 1 )²( u - 2 ) \(\le\) ( v - 1 )²( v + 2 )

Đặt x = u + 1 , y = v -1 thì :

BĐT <=> x³ - 3x² \(\le\) y³ + 3y² <=> x³ - y^{3 \(\le\)} 3(x² + y²)

Ta có : x - y = ( u - v ) + 2 \(\le\)2

=> ( x - y ) ( x² + xy + y² ) \(\le\)2( x² + xy + y²) = 2(x² + y²) + 2xy \(\le\) 2(x² + y²) + ( x² + y² ) = 3(x² + y² ) => x³ - y³ \(\le\) 3(x² +y² ) ( đpcm)

Dấu bằng xảy ra khi <=> x = y = 0 <=> u = -1 ; v = 1

lô

xem lại đề bạn ơi. nếu( u+2v+1)+(2u-2v+2)=3u+3 và chưa chắc cái này đã lẻ

đề đúng mà

\(B=\dfrac{2u+\sqrt{uv}-3v}{2u-5\sqrt{uv}+3v}\)

\(=\dfrac{2u+3\sqrt{uv}-2\sqrt{uv}-3v}{2u-2\sqrt{uv}-3\sqrt{uv}+3v}\)

\(=\dfrac{\sqrt{u}.\left(2\sqrt{u}+3\sqrt{v}\right)-\sqrt{v}.\left(2\sqrt{u}+3\sqrt{v}\right)}{2\sqrt{u}.\left(\sqrt{u}-\sqrt{v}\right)-3\sqrt{v}.\left(\sqrt{u}-\sqrt{v}\right)}\)

\(=\dfrac{\left(2\sqrt{u}+3\sqrt{v}\right)\left(\sqrt{u}-\sqrt{v}\right)}{\left(\sqrt{u}-\sqrt{v}\right)\left(2\sqrt{u}-3\sqrt{v}\right)}\)

\(=\dfrac{2\sqrt{u}+3\sqrt{v}}{2\sqrt{u}-3\sqrt{v}}\\ =\dfrac{4u+12\sqrt{uv}+9v}{4u-9v}\)

a) Đặt A = u² + v² - 2u + 3v + 15

= (u² - 2u + 1) + (v² + 3v + 9/4) + 47/4

= (u - 1)² + (v + 3/2)² + 47/4 \(\ge\frac{47}{4}\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}u-1=0\\v+\frac{3}{2}=0\end{cases}}\Rightarrow\hept{\begin{cases}u=1\\v=-\frac{3}{2}\end{cases}}\)

Vậy Min A = 47/4 <=> u = 1 ; y = -3/2

(u-1)^2 + (v+3/2)^2 + 11,75 \(\ge\)11,75

''='' <=> u = 1, v = -3/2

=> Min = 11,75 <=> u = 1, v = -3/2

Đặt \(A=u^2+v^2-2u+3v+15\)

\(=\left(u^2-2u+1\right)+\left(v^2+3v+\frac{9}{16}\right)+\frac{215}{16}\)

\(=\left(u-1\right)^2+\left(v+\frac{3}{4}\right)^2+\frac{215}{16}\ge\frac{215}{16}\)

Dấu "=" xảy ra khi u = 1, v = -3/4

Vậy Amin = 215/16 khi u = 1, v = -3/4

Sửa lại

\(=\left(u^2-2u+1\right)+\left(v^2+3v+\frac{9}{4}\right)+\frac{47}{4}=\left(u-1\right)^2+\left(v+\frac{3}{2}\right)^2+\frac{47}{4}\ge\frac{47}{4}\)

Dấu "=" xảy ra khi u=1,v=-3/2

Vậy Amin=47/4  khi u=1,v=-3/2