Câu 1:
a) 2009x - 2009y = 2009·( x - y)
b) 3x - 3y + x² - 2xy + y²
= 3( x - y) + ( x- y)²
= ( 3 + x - y)·( x- y)
Câu 2:
a) 4x( x- 5) + ( x+ 1)(x -1) + x - 2020
⇒4x² - 20x + x² - 1 + x - 2020
⇒5x² - 19x - 2020
b) ( x - 3) ( x² + 3x + 9 ) + x²(x - 1) + 27
⇒x³ - 27 + x³ - x² + 27
⇒2x³ - x²
Câu 3:
a) B= $\frac{x²- 4x +4}{x(x-2)}$
⇒B = $\frac{x²- 4x +4}{x²- 2x)}$
Để tìm ĐKXD thì x² - 2x ∦ 0
x² ∦ 2x
x ∦ 2
b) B = $\frac{x²- 4x +4}{x(x-2)}$
⇒B = $\frac{(x-2)²}{x(x-2)}$
⇒B = $\frac{x-2}{x}$
Thay x = 4 ⇒B= $\frac{4 - 2}{4}$
⇒ B= $\frac{1}{2}$
Bài 4:
a)$\frac{x}{2}$ + $\frac{x+ 4}{6}$ - 2x
⇒$\frac{3x}{6}$ + $\frac{x+ 4}{6}$ -2x
⇒$\frac{3x + x + 4}{6}$ - 2x
⇒$\frac{4x +4}{6}$ - 2x
⇒$\frac{4x +4}{6}$ - $\frac{12x}{6}$
⇒$\frac{4x + 4 + 12x}{6}$
$\frac{16x + 4}{6}$
b) $\frac{1}{x+ 1}$ - $\frac{3}{x³+ 1}$ + $\frac{3}{x² - x +1}$
⇒$\frac{1}{x+ 1}$ - $\frac{3}{(x+ 1)·(x² - x + 1)}$ + $\frac{3}{x² - x +1}$ => MTC: (x + 1)·(x² -x + 1)
⇒$\frac{x²- x + 1}{(x+ 1)·(x²-x + 1)}$ - $\frac{3}{(x+ 1)·(x² - x + 1)}$ + $\frac{3x + 3}{(x + 1)·(x² - x +1)}$
⇒$\frac{x² -x + 1 - 3 + 3x + 3}{(x + 1)·(x² - x + 1)}$
⇒$\frac{x² + 2x + 1}{(x + 1)·(x² - x + 1)}$
⇒$\frac{(x + 1)²}{(x + 1)·(x² - x + 1)}$
⇒$\frac{x +1}{x² - x + 1}$
Xin câu trả lời hay nhất ·∧·
