\(=\left[\dfrac{a^6b^3}{c^3d^6}\cdot\dfrac{ac^4}{b^2d^3}\right]:\left[\dfrac{a^8b^8}{c^4d^{12}}\cdot\dfrac{c^3}{b^9d^3}\right]\)

\(=\dfrac{a^7b^3c^4}{c^3d^9b^2}:\dfrac{a^8}{bcd^{15}}\)

\(=\dfrac{a^7bc}{d^9}\cdot\dfrac{bcd^{15}}{a^8}=\dfrac{d^6\cdot b^2\cdot c^2}{a}\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\). Ta có:

\(\frac{\left(a-b\right)^3}{\left(c-d\right)^3}=\frac{\left(bk-b\right)^3}{\left(dk-d\right)^3}=\frac{b^3\left(k-1\right)^3}{d^3\left(k-1\right)^3}=\frac{b^3}{d^3}\)

\(\frac{3a^2+2b^2}{3c^2+2d^2}=\frac{3\left(bk\right)^2+2b^2}{3\left(dk\right)^2+2d^2}=\frac{3b^2k^2+2b^2}{3d^2k^2+2d^2}=\frac{b^2\left(3k^2+2\right)}{d^2\left(3k^2+2\right)}=\frac{b^2}{d^2}\)

Đến đây nhìn có vẻ đề sai

\(\frac{a}{b}=\frac{c}{d}=k\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)ta có:

\(\frac{\left(a-b\right)^3}{\left(c-d\right)^3}=\frac{\left(bk-b\right)^3}{\left(dk-d\right)^3}=\frac{\left[b\left(k-1\right)\right]^3}{\left[d\left(k-1\right)\right]^3}=\frac{b^3}{d^3}\)

\(\frac{2b^2+3a^2}{2d^2+3c^2}=\frac{4.b^2+9.k^2.b^2}{4.d^2+9.d^2.k^2}=\frac{b^2\left(4+k^2.9\right)}{d^2\left(4+9.k^2\right)}=\frac{b^2}{d^2}\)

\(Taco:\frac{b^3}{d^3}=\frac{b^2}{d^2}\Leftrightarrow b=d\)

có thể chứng minh mà ko phải đặt k ko

1. Đặt \(\frac{a}{b}=\frac{c}{d}=k=>a=bk,c=dk\)

Thay vào 2 vế là sẽ CM được

1. Đặt \(\frac{a}{b}=\frac{c}{d}=k>a=bk.c=dk\)

Thay vào 2 vế để chứng minh

1 ) 

Ta có : 

\(\frac{a}{b}=\frac{c}{d}\)

\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)

\(\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)(  Áp dụng t/c DTSBN ) 

\(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\left(1\right)\)

Lại có : \(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{3a^2}{3c^2}=\frac{2b^2}{2d^2}=\frac{3a^2+2b^2}{3c^2+2d^2}\) (  Áp dụng t/c DTSBN )  \(\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\)

\(\Rightarrow\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{3a^2+2b^2}{3c^2+2d^2}\left(đpcm\right)\)

2 ) 

Ta có : 

\(x+y+2xy=83\)

\(\Rightarrow2\left(x+y+2xy\right)=166\)
\(\Rightarrow2x+2y+4xy+1=167\)

\(\Rightarrow2x\left(2y+1\right)+\left(2y+1\right)=167\)

\(\Rightarrow\left(2x+1\right)\left(2y+1\right)=167\)

Do \(x;y\in Z\)
\(\Leftrightarrow2x+1;2y+1\in Z\)
\(\Leftrightarrow2x+1;2y+1\in\left\{\pm1;\pm167\right\}\)

Ta có bảng sau : 

Vậy \(\left(x;y\right)\in\left\{\left(0;83\right),\left(83;0\right),\left(-1;-84\right),\left(-84;-1\right)\right\}\)
 

\(\Sigma_{sym}a^4b^4\ge\frac{\left(\Sigma_{sym}a^2b^2\right)^2}{3}\ge\frac{\left(\Sigma_{sym}ab\right)^4}{27}\ge\frac{a^2b^2c^2\left(a+b+c\right)^2}{3}=3a^4b^4c^4\)

\(\Sigma\frac{a^5}{bc^2}\ge\frac{\left(a^3+b^3+c^3\right)^2}{abc\left(a+b+c\right)}\ge\frac{\left(a^2+b^2+c^2\right)^4}{abc\left(a+b+c\right)^3}\ge\frac{\left(a+b+c\right)^6\left(a^2+b^2+c^2\right)}{27abc\left(a+b+c\right)^3}\)

\(\ge\frac{\left(3\sqrt[3]{abc}\right)^3\left(a^2+b^2+c^2\right)}{27abc}=a^2+b^2+c^2\)

\(\frac{a^3}{\left(b+2c\right)^2}+\frac{b+2c}{27}+\frac{b+2c}{27}\ge\frac{a}{3}\)\(\Leftrightarrow\)\(\frac{a^3}{\left(b+2c\right)^2}\ge\frac{1}{3}a-\frac{2}{27}b-\frac{4}{27}c\)

tương tự rồi cộng lại