Câu hỏi của Thao Phuong

Question

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Nguyễn Lê Phước Thịnh · Accepted Answer

ĐKXĐ: $x\notin\left\{7;-1945\right\}$

$\dfrac{19x+8}{x-7}\cdot\dfrac{5x-9}{x+1945}+\dfrac{19x+8}{7-x}\cdot\dfrac{4x-2}{x+1945}$

$=\dfrac{19x+8}{x-7}\cdot\dfrac{5x-9}{x+1945}-\dfrac{19x+8}{x-7}\cdot\dfrac{4x-2}{x+1945}$

$=\dfrac{19x+8}{x-7}\cdot\dfrac{5x-9-4x+2}{x+1945}$

$=\dfrac{19x+8}{x+1945}\cdot\dfrac{x-7}{x-7}=\dfrac{19x+8}{x+1945}$

Câu trả lời:

ĐKXĐ: $x\notin\left\{7;-1945\right\}$

$\dfrac{19x+8}{x-7}\cdot\dfrac{5x-9}{x+1945}+\dfrac{19x+8}{7-x}\cdot\dfrac{4x-2}{x+1945}$

$=\dfrac{19x+8}{x-7}\cdot\dfrac{5x-9}{x+1945}-\dfrac{19x+8}{x-7}\cdot\dfrac{4x-2}{x+1945}$

$=\dfrac{19x+8}{x-7}\cdot\dfrac{5x-9-4x+2}{x+1945}$

$=\dfrac{19x+8}{x+1945}\cdot\dfrac{x-7}{x-7}=\dfrac{19x+8}{x+1945}$

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