Toán Lớp 9: Rút gọn biểu thức :
$A$ = ( $\frac{12}{\sqrt{x}-1}$ +$\frac{12}{1+\sqrt{x}}$ ):$\frac{2017}{x-1}$ với ($x$ $\geq$ 0; x$\neq$ 1)
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Toán Lớp 9: Rút gọn biểu thức :
$A$ = ( $\frac{12}{\sqrt{x}-1}$ +$\frac{12}{1+\sqrt{x}}$ ):$\frac{2017}{x-1}$ với ($x$ $\geq$ 0; x$\neq$ 1)
Comments ( 2 )
A=(12/(\sqrt{x}-1)+12/(1+\sqrt{x})):2017/(x-1)
ĐKXĐ:x>0n; x\ne 1
<=>A=(12/(\sqrt{x}-1)+12/(\sqrt{x}+1)):2017/((\sqrt{x})²-1²)
<=>A=(12/(\sqrt{x}-1)+12/(\sqrt{x}+1)):2017/((\sqrt{x}-1)(\sqrt{x}+1))
<=>A=((12(\sqrt{x}+1))/((\sqrt{x}-1)(\sqrt{x}+1))+(12(\sqrt{x}-1))/((\sqrt{x}-1)(\sqrt{x}+1)) ):2017/((\sqrt{x}-1)(\sqrt{x}+1))
<=>A=((12\sqrt{x}+12)/((\sqrt{x}-1)(\sqrt{x}+1))+(12\sqrt{x}-12)/((\sqrt{x}-1)(\sqrt{x}+1))):2017/((\sqrt{x}-1)(\sqrt{x}+1))
<=>A=(12\sqrt{x}+12+12\sqrt{x}-12)/((\sqrt{x}-1)(\sqrt{x}+1)):2017/((\sqrt{x}-1)(\sqrt{x}+1))
<=>A=((12+12)\sqrt{x}+(12-12))/((\sqrt{x}-1)(\sqrt{x}+1)). ((\sqrt{x}-1)(\sqrt{x}+1))/2017
<=>A=(24\sqrt{x})/((\sqrt{x}-1)(\sqrt{x}+1)).((\sqrt{x}-1)(\sqrt{x}+1))/2017
<=>A=(24\sqrt{x}(\sqrt{x}-1)(\sqrt{x}+1))/((\sqrt{x}-1)(\sqrt{x}+1).2017)
<=>A=(24\sqrt{x})/2017 ( giản ước (\sqrt{x}-1)(\sqrt{x}+1) )
Vậy A=(24\sqrt{x})/2017
$\textit{Giải đáp + Lời giải và giải thích chi tiết:}$
A=(12/(\sqrt{x}-1)+12/(1+\sqrt{x})):2017/(x-1) (x \ge 0,x \ne 1)
=[(12(\sqrt{x}+1))/((\sqrt{x}-1)(\sqrt{x}+1))+(12(\sqrt{x}-1))/((\sqrt{x}-1)(\sqrt{x}+1))].(x-1)/2017
=(12\sqrt{x}+12+12\sqrt{x}-12)/((\sqrt{x})^2-1^2).(x-1)/2017
=(24\sqrt{x})/(x-1).(x-1)/2017
=(24\sqrt{x})/2017