a, \(2x\left(x-2y\right)-2y\left(y-2x\right)=2x^2-4xy-2y^2+4xy=2x^2-2y^2\)

b, \(\left(x-2\right)\left(3x^2+4x-5\right)=3x^3+4x^2-5x-6x^2-8x+10=3x^3-2x^2-13x+10\)

c, \(\left(x+2y\right)\left(x^2-2xy+4y^2\right)=x^3-2x^2y+4xy^2+2x^2y-4xy^2+8y^3=x^3+8y^3\)

Đặt A = \(\frac{n+1}{n+2}\)

=> \(\frac{1}{A}=\frac{n+2}{n+1}\)

=> \(\frac{1}{A}-1=\frac{n+2-n-1}{n+1}=\frac{1}{n+1}\)

Đặt B = \(\frac{n+3}{n+4}\)

=> \(\frac{1}{B}=\frac{n+4}{n+3}\)

=> \(\frac{1}{B}-1=\frac{n+4-n-3}{n+3}=\frac{1}{n+3}\)

Vì \(\frac{1}{n+1}>\frac{1}{n+3}\Rightarrow\frac{1}{A}-1>\frac{1}{B}-1\Rightarrow\frac{1}{A}>\frac{1}{B}\Rightarrow A< B\)

Vậy \(\frac{n+1}{n+2}< \frac{n+3}{n+4}\)

Đặt \(A=\frac{n+1}{n+2}\)

\(\Rightarrow\frac{1}{A}=\frac{n+2}{n+1}\)

\(\Rightarrow\frac{1}{A}-1=\frac{n+2-n+1}{n+1}=\frac{1}{n+1}\)

Đặt \(B=\frac{n+3}{n+4}\)

\(\Rightarrow\frac{1}{B}=\frac{n+4}{n+3}\)

\(\Rightarrow\frac{1}{B}-1=\frac{n+4-n-3}{n+3}=\frac{1}{n+3}\)

Vì \(\frac{1}{n+1}>\frac{1}{n+3}\Rightarrow\frac{1}{A}-1>\frac{1}{B}-1\Rightarrow\frac{1}{A}>\frac{1}{B}\Rightarrow A< B\)

Vậy \(\frac{n+1}{n+2}< \frac{n+3}{n+4}\)

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

 Sửa đề \(VT=\frac{ac}{bd}=\frac{bkdk}{bd}=k^2\)(1)

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

Từ (1) ( 2) => VT = VP (ĐPCM)

Bài 1:

\(3\frac{1}{5}-5\frac{1}{4}+2\)

\(=\frac{16}{5}-\frac{21}{4}+2\)

\(=\left(-\frac{41}{20}\right)+2\)

\(=-\frac{1}{20}.\)

\(3\frac{1}{5}-5\frac{1}{4}+2=3+\frac{1}{5}-5-\frac{1}{4}+2=\left(3-5+2\right)+\left(\frac{1}{5}-\frac{1}{4}\right)=0+\frac{-1}{20}=\frac{-1}{20}\)