a) \(A=\frac{2}{x-3}+\frac{x-3}{x+3}-\frac{x-15}{9-x^2}\left(x\ne\pm3\right)\)

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

\(=\frac{2}{x-3}+\frac{x-3}{x+3}+\frac{x-15}{\left(x-3\right).\left(x+3\right)}\)

\(=\frac{2.\left(x+3\right)}{\left(x-3\right).\left(x+3\right)}+\frac{\left(x-3\right).\left(x-3\right)}{\left(x-3\right).\left(x+3\right)}+\frac{x-15}{\left(x-3\right).\left(x+3\right)}\)

\(=\frac{2x+6+\left(x-3\right)^2+x-15}{\left(x-3\right).\left(x+3\right)}\)

\(=\frac{2x+6+x^2-6x+9+x-15}{\left(x-3\right).\left(x+3\right)}\)

\(=\frac{x^2+\left(2x-6x+x\right)+\left(6+9-15\right)}{\left(x-3\right).\left(x+3\right)}\)

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

\(=\frac{x}{x+3}\)

b) Thay vào ta được:

\(\frac{1}{2}:\left(\frac{1}{2}+3\right)=\frac{1}{7}\)

Vậy biểu thức \(A=\frac{1}{7}\) khi \(x=\frac{1}{2}\)

[1495247x25978];5x[123987x456123=

Ta có: 1(x−2)2−1x+2−14−x21(x−2)2−1x+2−14−x2

=x+2−(x−2)2+x−2(x−2)2⋅(x+2)=x+2−(x−2)2+x−2(x−2)2⋅(x+2)

=2x−x2+4x−4(x−2)2⋅(x+2)=2x−x2+4x−4(x−2)2⋅(x+2)

em ko bt lớp 8

em ko bt lớp 8

a) (3x³ - 5x² + 9x - 15) : (3x - 5)

= [x²(3x - 5) + 3(3x - 5)] : (3x - 5)

= [(3x - 5)(x² + 3)] : (3x - 5)

= x² + 3

a) (3x³ - 5x² + 9x - 15) : (3x - 5)

= [x²(3x - 5) + 3(3x - 5)] : (3x - 5)

= [(3x - 5)(x² + 3)] : (3x - 5)

= x² + 3