a) \(\dfrac{3}{8}x=\dfrac{9}{8}-1\)

\(\Rightarrow\dfrac{3}{8}x=\dfrac{1}{8}\)

\(\Rightarrow x=\dfrac{1}{8}:\dfrac{3}{8}=\dfrac{1}{3}\)

b) \(\dfrac{4}{5}x=\dfrac{7}{5}-\dfrac{1}{5}\)

\(\Rightarrow\dfrac{4}{5}x=\dfrac{6}{5}\)

\(\Rightarrow x=\dfrac{6}{5}:\dfrac{4}{5}=\dfrac{3}{2}\)

c) \(\dfrac{12}{7}:x+\dfrac{2}{3}=\dfrac{7}{5}\)

\(\Rightarrow\dfrac{12}{7}:x=\dfrac{7}{5}-\dfrac{2}{3}\)

\(\Rightarrow\dfrac{12}{7}:x=\dfrac{11}{15}\)

\(\Rightarrow x=\dfrac{12}{7}:\dfrac{11}{15}=\dfrac{180}{77}\)

d) \(3\left(x+7\right)-15=27\)

\(\Leftrightarrow3\left(x+7\right)=42\)

\(\Leftrightarrow x+7=14\Leftrightarrow x=7\)

a) \(\dfrac{3}{7}-\dfrac{4}{11}-\dfrac{3}{7}\)

=\(\left(\dfrac{3}{7}-\dfrac{3}{7}\right)-\dfrac{4}{11}\)

\(=-\dfrac{4}{11}\)

b) \(-\dfrac{5}{11}+\dfrac{3}{8}-\dfrac{6}{11}\)

\(=\left(-\dfrac{5}{11}-\dfrac{6}{11}\right)+\dfrac{3}{8}\)

\(=-1+\dfrac{3}{8}=-\dfrac{8}{8}+\dfrac{3}{8}=-\dfrac{5}{8}\)

c) \(\dfrac{3}{4}-\dfrac{1}{3}-\dfrac{11}{4}\)

\(=\left(\dfrac{3}{4}-\dfrac{11}{4}\right)-\dfrac{1}{3}\)

\(=-2-\dfrac{1}{3}=-\dfrac{6}{3}-\dfrac{1}{3}=-\dfrac{7}{3}\)

- Cho ăn thực phẩm phù hợp với vật nuôi.

- Vệ sinh, lau dọn sạch sẽ thân thể cũng như chuồng của vật nuôi.

- Cho khám sức khoẻ hàng tháng, tiêm phòng vắc-xin.

Có: \(A=\dfrac{11.13.15+33.39.45+55.65.75+99.117.135}{13.15.17+39.45.51+65.75.85+117.135.153}\)(sửa đề)

\(=\dfrac{11.13.15+3^3\left(11.13.15\right)+5^3\left(11.13.15\right)+9^3\left(11.13.15\right)}{13.15.17+3^3\left(13.15.17\right)+5^3\left(13.15.17\right)+9^3\left(13.15.17\right)}\)

\(=\dfrac{11.13.15\left(1+3^3+5^3+9^3\right)}{13.15.17\left(1+3^3+5^3+9^3\right)}\)

\(=\dfrac{11}{17}\)

Có: \(B=\dfrac{1111}{1717}=\dfrac{11.101}{17.101}=\dfrac{11}{17}\)

=> A = B

18x² + 21x - 4

= 18x² +24x - 3x - 4

= 6x(3x + 4) - (3x + 4)

= (6x - 1)(3x + 4)

chết làm nhầm r :(

\(S=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\)

\(=1-\dfrac{1}{101}=\dfrac{100}{101}\)