Câu 23 : Trang 46 sgk toán 8 tập 1
Làm các phép tính sau.
a) \( \frac{y}{2x^{2}-xy}+\frac{4x}{y^{2}-2xy}\);
b) \( \frac{1}{x+2}+\frac{3}{x^{2}-4}+\frac{x-14}{(x^{2}+4x+4)(x-2)}\);
c) \( \frac{1}{x+2}+\frac{1}{(x+2)(4x+7)}\);
d) \( \frac{1}{x+3}+\frac{1}{(x+3)(x+2)}+\frac{1}{(x+2)(4x+7)}\)
Bài Làm:
Hướng dẫn: Đưa các phân thức về dạng cùng mẫu thức, sau đó cộng tử số với nhau.
a) \( \frac{y}{2x^{2}-xy}+\frac{4x}{y^{2}-2xy}\) \( =\frac{y}{x(2x-y)}+\frac{4x}{y(y-2x)}\)
\( =\frac{y}{x(2x-y)}+\frac{-4x}{y(2x-y)}=\frac{y^{2}}{xy(2x-y)}+\frac{-4x^{2}}{xy(2x-y)}\)
\( = \frac{y^{2}-4x^{2}}{xy(2x-y)}=\frac{(y-2x)(y+2x)}{xy(2x-y)}=\frac{-(2x-y)(y+2x)}{xy(2x-y)}\)
\( =\frac{-(2x+y)}{xy}\)
b) \( \frac{1}{x+2}+\frac{3}{x^{2}-4}+\frac{x-14}{(x^{2}+4x+4)(x-2)}\)
\( =\frac{1}{x+2}+\frac{3}{(x-2)(x+2)}+\frac{x-14}{(x+2)^{2}(x-2)}\)
\( =\frac{(x+2)(x-2)}{(x+2)^{2}(x-2)}+\frac{3(x+2)}{(x-2)(x+2)^{2}}+\frac{x-14}{(x+2)^{2}(x-2)}\)
\( =\frac{x^{2}-4+3x+6+x-14}{(x+2)^{2}(x-2)}= \frac{x^{2}+4x-12}{(x+2)^{2}(x-2)}\)
\( =\frac{x^{2}-2x+6x-12}{(x+2)^{2}(x-2)}= \frac{x(x-2)+6(x-2)}{(x+2)^{2}(x-2)}\)
\( = \frac{(x-2)(x+6)}{(x+2)^{2}(x-2)}=\frac{x+6}{(x+2)^{2}}\)
c) \( \frac{1}{x+2}+\frac{1}{(x+2)(4x+7)}\)
\( =\frac{4x+7}{(x+2)(4x+7)}+\frac{1}{(x+2)(4x+7)}\)
\( =\frac{4x+8}{(x+2)(4x+7)}=\frac{4(x+2)}{(x+2)(4x+7)}=\frac{4}{4x+7}\)
d) \( \frac{1}{x+3}+\frac{1}{(x+3)(x+2)}+\frac{1}{(x+2)(4x+7)}\)
\( =\frac{x+2}{(x+3)(x+2)}+\frac{1}{(x+3)(x+2)}+\frac{1}{(x+2)(4x+7)}\)
\( =\frac{x+3}{(x+3)(x+2)}+\frac{1}{(x+2)(4x+7)}\) \( =\frac{1}{x+2}+\frac{1}{(x+2)(4x+7)}\)
\( =\frac{4x+7}{(x+2)(4x+7)}+\frac{1}{(x+2)(4x+7)}=\frac{4x+8}{(x+2)(4x+7)}\)
\( =\frac{4(x+2)}{(x+2)(4x+7)}+\frac{4}{4x+7}\)