\(\frac{x+3}{x-2}=\frac{x+2}{x-1}\)(ĐKXĐ: x khác 2 và 1)

\(\Leftrightarrow\frac{\left(x+3\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}=\frac{\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-1\right)}\)

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

\(\Leftrightarrow x^2-x+3x-3=x^2-2x+2x-4\)

\(\Leftrightarrow x^2-x+3x-x^2-2x+2x=3-4\)

\(\Leftrightarrow2x=-1\Leftrightarrow x=-\frac{1}{2}\)

\(\frac{x+3}{x-2}=\frac{x+2}{x-1}\)

<=> ( x + 3 ) ( x - 1 ) = ( x + 2 ) ( x - 2 )

<=> x² + 2x - 3 = x² - 4

<=> x² + 2x - 3 - x² + 4 = 0

<=> 2x + 1 = 0

<=> x = - 1/2

a, Sửa \(2x=3y;5y=4z\)và \(x-y+z=2\)

ta có : \(\frac{x}{3}=\frac{y}{2};\frac{y}{4}=\frac{z}{5}\Leftrightarrow\frac{x}{12}=\frac{y}{8};\frac{y}{8}=\frac{z}{10}\)

\(\Leftrightarrow\frac{x}{12}=\frac{y}{8}=\frac{z}{10}\)

ADTC dãy tỉ số bằng nhau ta có : 

\(\frac{x}{12}=\frac{y}{8}=\frac{z}{10}=\frac{x-y+z}{12-8+10}=\frac{2}{14}=\frac{1}{7}\)

\(x=\frac{12}{7};y=\frac{8}{7};z=\frac{10}{7}\)

b, Ta có : \(3x=4y=5z\Leftrightarrow\frac{x}{20}=\frac{y}{15}=\frac{z}{12}\)

ADTC dãy tỉ số bằng nhau ta có : 

\(\frac{x}{20}=\frac{y}{15}=\frac{z}{12}=\frac{x+y+z}{20+15+12}=\frac{47}{47}=1\)

\(x=20;y=15;z=12\)

Từ \(\frac{2019a+2020c}{2019a-2021c}=\frac{2019b+2020d}{2019b-2021d}\)

<=> \(\frac{2019a-2021c+4041c}{2019a-2021c}=\frac{2019b-2021d+4041d}{2019b-2021d}\)

<=> \(1+\frac{4041c}{2019a-2021c}=1+\frac{4041d}{2019b-2021d}\)

<=> \(\frac{4041c}{2019a-2021c}=\frac{4041d}{2019b-2021d}\)

<=> 4041c( 2019b - 2021d ) = 4041d( 2019a - 2021c )

<=> c( 2019b - 2021d ) = d( 2019a - 2021c )

<=> 2019bc - 2021dc = 2019ad - 2021cd

<=> 2019bc - 2021dc - 2019ad + 2021cd = 0

<=> 2019( bc - ad ) = 0

<=> bc - ad = 0

<=> bc = ad

<=> a/b = c/d

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

Ta có : \(\left(\frac{a+b}{c+d}\right)^{2020}=\left(\frac{kb+b}{kd+d}\right)^{2020}=\left[\frac{b\left(k+1\right)}{d\left(k+1\right)}\right]^{2020}=\left(\frac{b}{d}\right)^{2020}=\frac{b^{2020}}{d^{2020}}\)(1)

\(\)\(\frac{a^{2020}+b^{2020}}{c^{2020}+d^{2020}}=\frac{\left(kb\right)^{2020}+b^{2020}}{\left(kd\right)^{2020}+d^{2020}}=\frac{k^{2020}b^{2020}+b^{2020}}{k^{2020}d^{2020}+d^{2020}}=\frac{b^{2020}\left(k^{2020}+1\right)}{d^{2020}\left(k^{2020}+1\right)}=\frac{b^{2020}}{d^{2020}}\)(2)

Từ (1) và (2) ta có đpcm