Đáp án:
Giải thích các bước giải:
1A. a) $2x-6\neq 0\Leftrightarrow x\neq \frac{6}{2}\Leftrightarrow x\neq 3$
b) $x^2-4x+4\neq 0\Leftrightarrow (x-2)^2\neq 0\Leftrightarrow x\neq 2$
c) $27x^3+27x^2+9x+1\neq 0\Leftrightarrow (3x)^3+3.(3x)^2.1+3.3x.1^2+1\neq 0\Leftrightarrow (3x
+1)^3\neq 0\Leftrightarrow 3x+1\neq 0\Leftrightarrow x\neq \frac{-1}{3}$
d) $4x-3x^2\neq 0\Leftrightarrow x(4-3x)\neq 0\Leftrightarrow x\neq 0,4-3x\neq 0\Leftrightarrow x\neq 0,x\neq \frac{4}{3}$
1B.
a) ĐK: $(x+y)(3x-3y)\neq 0\Leftrightarrow x\neq -y,x\neq y$
Chứng minh bằng hằng số
$\frac{x^2-y^2}{(x+y)(3x-3y)}=\frac{(x+y)(x-y)}{(x+y)3(x-y)}=\frac{1}{3}$
b)ĐK: $ 25kx+15x+9y+15ky\neq 0\Leftrightarrow 5x(5k+3)+3y(3+5k)\neq 0\Leftrightarrow (3+5k)(5x+3y)\neq 0\Leftrightarrow 5x+3y\neq 0\Leftrightarrow x\neq \frac{-3y}{5}$
Chứng minh bằng hằng số
$\frac{5kx-5x-3y+3ky}{25kx+15x+9y+15ky}=\frac{5x(k-1)+3y(k-1)}{5x(5k+3)+3y(3+5k)}=\frac{(k-1)(5x+3y)}{3+5k)(5x+3y)}=\frac{k-1}{3+5k}$
