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

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

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

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

=> \(4041c\left(2019b-2021d\right)=4041d\left(2019a-2021c\right)\)

=> \(c\left(2019b-2021d\right)=d\left(2019a-2021c\right)\)( rút 4041 ở cả hai vế )

=> \(2019bc-2021cd=2019ad-2021cd\)

=> \(2019ad-2021cd-2019bc+2021cd=0\)

=> \(2019\left(ad-bc\right)=0\)

=> \(ad-bc=0\)

=> \(ad=bc\)

=> \(\frac{a}{b}=\frac{c}{d}\)

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

+) \(\left(\frac{a+b}{c+d}\right)^{2020}=\left(\frac{kb+b}{kd+d}\right)^{2020}=\left[\frac{b\left(k+1\right)}{d\left(k+1\right)}\right]^{2020}=\left(\frac{b}{d}\right)^{2020}=\frac{b^{2020}}{d^{2020}}\)(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 điều phải chứng minh

PT <=> \(2\sqrt{x}-x-1=0\)

\(\Leftrightarrow2\sqrt{x}=x+1\)

Bình phương 2 vế ta được : \(4x=x^2+2x+1\)

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

Vậy x = 1

\(9\left(3x+1\right)=\left(3x+1\right)\Leftrightarrow9\left(3x+1\right)-\left(3x+1\right)=0\)

\(\Leftrightarrow8\left(3x+1\right)=0\Leftrightarrow3x+1=0\Leftrightarrow x=-\frac{1}{3}\)

\(\Rightarrow\frac{x+15}{x}=\frac{4}{3}\Rightarrow3.\left(x+15\right)=4.x\Rightarrow3x+45=4x\Rightarrow4x-3x=0-45\Rightarrow x=-45\)

\(\frac{1}{6}-\left(\frac{2}{3}\right)^2+\frac{5}{18}\)

\(=\frac{1}{6}-\frac{4}{9}+\frac{5}{18}\)

\(=\frac{3}{18}-\frac{8}{18}+\frac{5}{18}\)

\(=\frac{3-8+5}{18}\)

\(=0\)

e có nhầm đề không nhỉ, chỗ a+b+c không có mũ 2 hả e.

nếu không có, ta lấy 1 trường hợp cụ thể của a=b=c=1 thì đã thấy đề có vấn đề r e