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a) \(\frac{x^2-9}{x+3}-ĐKXĐ:x+3\ne0\Leftrightarrow x\ne-3\)

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

Cho tam giác ABC có AB = 3cm, AC = 4cm, BC = 5cm. a) Chứng minh tam giác ABC vuông tại A. b) Vẽ tia phân giác BD (D thuộc AC). Vẽ tia phân giác BD (D thuộc AC), từ D vẽ DE vuông góc với BC (E thuộc BD). AD cắt AB tại F, ED cắt AB tại F. Chứng minh DA = DE và DF > DE  Phần c

nhầm nha

\(a^3-7a=6\)

\(=a^3+a^2-a^2-a-6a-6\)

\(=a^2\left(a+1\right)-a\left(a+1\right)-6\left(a+1\right)\)

\(=\left(a+1\right)\left(a^2-a-6\right)\)

\(=\left(a+1\right)\left(a^2-3a+2a-6\right)\)

\(=\left(a+1\right).\left[a\left(a-3\right)+2\left(a-3\right)\right]\)

\(=\left(a+1\right)\left(a-3\right)\left(a+2\right)\)

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

\(=\left[\frac{x+2}{\left(x+1\right)^2}-\frac{x-2}{\left(x-1\right)\left(x+1\right)}\right].\frac{\left(x^2-1\right)\left(x+1\right)}{2x^2+x}\)

\(=\frac{\left(x+2\right)\left(x-1\right)-\left(x-2\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)^2}.\frac{\left(x-1\right)\left(x+1\right)^2}{2x^2+x}\)

\(=\frac{\left(x^2+x-2\right)-\left(x^2-x-2\right)}{2x^2+x}\)

\(=\frac{2x}{2x^2+x}=\frac{2}{2x+1}\)

\(\frac{\left(x+y\right)^2}{x}.\left(\frac{x}{\left(x+y\right)^2}-\frac{x}{x^2-y^2}\right)-\frac{5x-3y}{y-x}\left(đk:x\text{≠}0-y;y\right).\)

\(=\frac{\left(x+y\right)^2}{x}.\left(\frac{x}{\left(x+y\right)^2}-\frac{x}{\left(x-y\right)\left(x+y\right)}\right)-\frac{5x-3y}{y-x}\)

\(=\frac{\left(x+y\right)^2}{x}.\frac{x\left(x-y\right)-x\left(x+y\right)}{\left(x+y\right)^2\left(x-y\right)}+\frac{5x-3y}{x-y}\)

\(=\frac{1}{x}.\frac{x^2-xy-x^2-xy}{\left(x+y\right)^2\left(x-y\right)}+\frac{5x-3y}{x-y}\)

\(=\frac{1}{x}.\frac{-2xy}{x-y}+\frac{5x-3y}{x-y}\)

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

\(=\frac{-2xy+5x-3y}{x-y}\)

\(=\frac{5\left(x-y\right)}{x-y}\)

\(=5\)

Ta có đpcm

\(\frac{\left(x+y\right)^2}{x}.\left[\frac{x}{\left(x+y\right)^2}-\frac{x}{x^2-y^2}\right]-\frac{5x-3y}{y-x}\)

\(=\frac{\left(x+y\right)^2}{x}.\left[\frac{x}{\left(x+y\right)^2}-\frac{x}{\left(x-y\right)\left(x+y\right)}\right]-\frac{5x-3y}{y-x}\)

\(=\frac{\left(x+y\right)^2}{x}.\left[\frac{x\left(x-y\right)}{\left(x+y\right)^2\left(x-y\right)}-\frac{x\left(x+y\right)}{\left(x-y\right)\left(x+y\right)^2}\right]-\frac{5x-3y}{y-x}\)

\(=\frac{\left(x+y\right)^2}{x}.\left[\frac{x^2-xy-x^2-xy}{\left(x+y\right)^2\left(x-y\right)}\right]-\frac{5x-3y}{y-x}\)

\(=\frac{\left(x+y\right)^2}{x}.\frac{-2xy}{\left(x+y\right)^2\left(x-y\right)}-\frac{5x-3y}{y-x}\)

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

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

\(=\frac{5x-5y}{x-y}\)

\(=\frac{5\left(x-y\right)}{x-y}\)

\(=5\)

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