Question

[Thầy Cao Đô]

Câu 1. (2 điểm): Tính (thu gọn).

a) $(6 x-7)(7 x-1)$;

b) $(4 x-1)^2+(2 x-5)(2 x+5)$;

c) $\dfrac{x+5}{x}+\dfrac{x}{x-5}+\dfrac{25}{x^2-5 x}$.

Answer 1

[GV Nguyễn Trần Thành Đạt]

\(a,\\ \left(6x-7\right).\left(7x-1\right)=6x.7x-7x.7-6x.1-7.\left(-1\right)\\ =42x^2-49x-6x+7=42x^2-55x+7\\ b,\\ \left(4x-1\right)^2+\left(2x-5\right).\left(2x+5\right)=16x^2-8x+1+4x^2-25\\ =20x^2-8x-24\)

\(c,\\ \dfrac{x+5}{x}+\dfrac{x}{x-5}+\dfrac{25}{x^2-5x}\\ =\dfrac{\left(x-5\right).\left(x+5\right)}{x.\left(x-5\right)}+\dfrac{x.x}{x.\left(x-5\right)}+\dfrac{25}{x.\left(x-5\right)}\\ =\dfrac{x^2-25+x^2+25}{x.\left(x-5\right)}=\dfrac{2x^2}{x.\left(x-5\right)}=\dfrac{2x}{\left(x-5\right)}\left(ĐK:x\ne0;x\ne5\right)\)

Answer 2

[Đặng Thảo Nguyên]