Câu hỏi của Tô Hiểu Nguyệt

Question

$\dfrac{x+1}{60}$+$\dfrac{x+2}{59}$=$\dfrac{x+3}{58}$+$\dfrac{x+4}{57}$

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Nhã Doanh · Accepted Answer

$\dfrac{x+1}{60}+\dfrac{x+2}{59}=\dfrac{x+3}{58}+\dfrac{x+4}{57}$

$\Leftrightarrow\dfrac{x+1}{60}+1+\dfrac{x+2}{59}+1=\dfrac{x+3}{58}+1+\dfrac{x+4}{57}+1$

$\Leftrightarrow\dfrac{x+1+60}{60}+\dfrac{x+2+59}{59}=\dfrac{x+3+58}{58}+\dfrac{x+4+57}{57}$

$\Leftrightarrow\dfrac{x+61}{60}+\dfrac{x+61}{59}-\dfrac{x+61}{58}-\dfrac{x+61}{57}=0$

$\Leftrightarrow\left(x+61\right)\left(\dfrac{1}{60}+\dfrac{1}{59}-\dfrac{1}{58}-\dfrac{1}{57}\right)=0$

$\Leftrightarrow x+61=0$

$\Leftrightarrow x=-61$Câu trả lời: $\dfrac{x+1}{60}+\dfrac{x+2}{59}=\dfrac{x+3}{58}+\dfrac{x+4}{57}$

$\Leftrightarrow\dfrac{x+1}{60}+1+\dfrac{x+2}{59}+1=\dfrac{x+3}{58}+1+\dfrac{x+4}{57}+1$

$\Leftrightarrow\dfrac{x+1+60}{60}+\dfrac{x+2+59}{59}=\dfrac{x+3+58}{58}+\dfrac{x+4+57}{57}$

$\Leftrightarrow\dfrac{x+61}{60}+\dfrac{x+61}{59}-\dfrac{x+61}{58}-\dfrac{x+61}{57}=0$

$\Leftrightarrow\left(x+61\right)\left(\dfrac{1}{60}+\dfrac{1}{59}-\dfrac{1}{58}-\dfrac{1}{57}\right)=0$

$\Leftrightarrow x+61=0$

$\Leftrightarrow x=-61$

Bài 4: Phương trình tích