Mình nghĩ đề là (x^3-3x^2y+3xy^2-y^3):(x-y) 

 Cách 2: x^3 - 3x^2y + 3xy^2 - y^3 = (x-y)^3 là hằng đẳng thức luôn bạn nhé

Đề được viết lại: (x-y)^3:(x-y) = (x-y)^2=x^2-2xy+y^2

\(\dfrac{8}{3}+\dfrac{10}{7}=\dfrac{56}{21}+\dfrac{30}{21}\\ =\dfrac{56+30}{21}=\dfrac{86}{21}\)

\(\dfrac{4}{5}+\dfrac{3}{4}=\dfrac{16}{20}+\dfrac{15}{20}\\ =\dfrac{16+15}{20}=\dfrac{31}{20}\)

Vì tam giác ABC vuông tại A

Nên: \(\widehat{B}+\widehat{C}=90^o\\ \Rightarrow0^o< \widehat{C}< 90^o\)

\(\Rightarrow0< \sin C< 1\) 

Ta có: \(\sin^2C+\cos^2C=1\Rightarrow\sin^2C=1-\left(\dfrac{4}{5}\right)^2=\dfrac{9}{25}\\ \Rightarrow\sin C=\dfrac{3}{5}\)

Lại có: \(\tan C=\dfrac{\sin C}{\cos C}=\dfrac{\dfrac{3}{5}}{\dfrac{4}{5}}=\dfrac{3}{4}\\ \cot C=\dfrac{1}{\tan C}=\dfrac{4}{3}\)

\(\dfrac{2}{3}-\left(-\dfrac{1}{4}\right)+\dfrac{3}{5}-\dfrac{7}{45}-\left(-\dfrac{5}{9}\right)+\dfrac{1}{12}+\dfrac{1}{35}\\ =\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{7}{45}+\dfrac{5}{9}+\dfrac{1}{12}+\dfrac{1}{35}\\ =\left(\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{1}{12}\right)+\left(\dfrac{3}{5}+\dfrac{5}{9}-\dfrac{7}{45}\right)+\dfrac{1}{35}\\ =\left(\dfrac{8}{12}+\dfrac{3}{12}+\dfrac{1}{12}\right)+\left(\dfrac{27}{45}+\dfrac{25}{45}-\dfrac{7}{45}\right)+\dfrac{1}{35}\\ =\dfrac{12}{12}+\dfrac{45}{45}+\dfrac{1}{35}\\ =1+1+\dfrac{1}{35}\\ =2+\dfrac{1}{35}\\ =\dfrac{70}{35}+\dfrac{1}{35}=\dfrac{71}{35}\)

\(4^{x+2}.3^x=16.12^5\\ \Rightarrow4^{x+2}.3^x=4^2.4^5.3^5\\ \Rightarrow4^{x+2}.3^x=4^7.3^5\\ \Rightarrow\dfrac{4^{x+2}}{4^7}.\dfrac{3^x}{3^5}=1\\ \Rightarrow4^{x-5}.3^{x-5}=1\\ \Rightarrow12^{x-5}=1\\ \Rightarrow x-5=0\\ \Rightarrow x=5\)

Số số hạng tổng trên:

  (62-5):3+1=20 (số hạng)

Tổng dãy trên:

  (62+5)x20:2=670

\(x-\left(\dfrac{5}{4}-\dfrac{7}{5}\right)=\dfrac{9}{20}\\ \Rightarrow x=\dfrac{9}{20}+\dfrac{5}{4}-\dfrac{7}{5}\\ \Rightarrow x=\dfrac{9}{20}+\dfrac{25}{20}-\dfrac{28}{20}\\ \Rightarrow x=\dfrac{6}{20}=\dfrac{3}{10}\)

\(\left(x+\dfrac{3}{5}\right)-\dfrac{1}{2}=\dfrac{1}{3}\\ \Rightarrow x+\dfrac{3}{5}-\dfrac{1}{2}=\dfrac{1}{3}\\ \Rightarrow x=\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{1}{2}\\ \Rightarrow x=\dfrac{10}{30}-\dfrac{18}{30}+\dfrac{15}{30}\\ \Rightarrow x=\dfrac{7}{30}\)