xét tam giác abc và tam giác dec có ab//de

theo hệ quả của định lí Thales ta có :

\(\frac{ca}{ce}=\frac{cb}{cd}=\frac{ab}{de}\frac{5}{15}=\frac13\)

\(\frac{cb}{cd}=\frac13\rarr\frac{x}{7^{}.2}=\frac13\)

\(\rarr x=\frac{7.2\cdot1}{3}=2.4\)

\(\frac{ca}{ce}=\frac13\rarr\frac{3}{y}=\frac13\)

\(\rarr y=\frac{3\cdot3}{1}=9\)

\(\frac{x+1}{3}=\frac{2x+5}{5}\)

\(\frac{5\cdot\left(x+1\right)}{15}=\frac{3\cdot\left(2x+5\right)}{15}\)

\(5x+5=6x+15\)

\(5x-6x=15-5\)

\(-x=10\)

\(x=-10\)

a) \(P=\left(\frac{2x}{3x+1}-1\right)\div\left(1-\frac{8x^2}{9x_{}^2-1}\right)\)

\(=\frac{2x^{}-3x-1}{3x+1}\div\frac{9x^2-1-8x^2}{9x^2-1}\)

\(=-\frac{\left(x+1\right)}{3x+1}\cdot\frac{9x^2-1}{x^2-1}\)

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

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

b) thay x=2 ta được :

\(=\frac{1-3\cdot2}{2-1}\)

\(=-5\)

a) \(\frac{2y-1}{y}-\frac{2x+1}{x}\)

mẫu thức chung:\(xy\)

\(=\frac{x\cdot\left(2y-1\right)}{xy}-\frac{y\left(2x+1\right)}{xy}\)

\(=\frac{2xy-x-2xy-y}{xy}\)

\(=\frac{\left(2xy-2xy\right)-x-y}{xy}\)

\(=\frac{-x-y}{xy}\)

b) \(\frac{2x}{3}:\frac{5}{6x^2}\)

\(=\frac{2x}{3}\cdot\frac{5}{6x^2}\)

\(\frac{4x^3}{5}\)