Câu hỏi của Trịnh Đức Thịnh

Question

Tính giá trị của biểu thức : $A=\dfrac{45}{19}-\left\{\dfrac{1}{2}+\left[\dfrac{1}{3}+\left(\dfrac{1}{4}\right)^{-1}\right]^{-1}\right\}^{-1}$

O=C=O · Accepted Answer

$A=\dfrac{45}{19}-\left\{\dfrac{1}{2}+\left[\dfrac{1}{3}+\left(\dfrac{1}{4}\right)^{-1}\right]^{-1}\right\}^{-1}$

$A=\dfrac{45}{19}-\left\{\dfrac{1}{2}+\left[\dfrac{1}{3}+4\right]^{-1}\right\}^{-1}$

$A=\dfrac{45}{19}-\left\{\dfrac{1}{2}+\left(\dfrac{13}{3}\right)^{-1}\right\}^{-1}$

$A=\dfrac{45}{19}-\left\{\dfrac{1}{2}+\dfrac{3}{13}\right\}^{-1}$

$A=\dfrac{45}{19}-\left(\dfrac{19}{26}\right)^{-1}$

$A=\dfrac{45}{19}-\dfrac{26}{19}$

$A=\dfrac{19}{19}=1.$Câu trả lời: $A=\dfrac{45}{19}-\left\{\dfrac{1}{2}+\left[\dfrac{1}{3}+\left(\dfrac{1}{4}\right)^{-1}\right]^{-1}\right\}^{-1}$

$A=\dfrac{45}{19}-\left\{\dfrac{1}{2}+\left[\dfrac{1}{3}+4\right]^{-1}\right\}^{-1}$

$A=\dfrac{45}{19}-\left\{\dfrac{1}{2}+\left(\dfrac{13}{3}\right)^{-1}\right\}^{-1}$

$A=\dfrac{45}{19}-\left\{\dfrac{1}{2}+\dfrac{3}{13}\right\}^{-1}$

$A=\dfrac{45}{19}-\left(\dfrac{19}{26}\right)^{-1}$

$A=\dfrac{45}{19}-\dfrac{26}{19}$

$A=\dfrac{19}{19}=1.$

Chương I : Số hữu tỉ. Số thực