tinh gia tri cua bieu thuc 02/09/2021 Bởi Aubrey tinh gia tri cua bieu thuc A=(1\7+1\23-1\1009):(1\7+1\23-1\1009+1\7.1\23.1\1009)+1:(30.1009-160)
Đáp án: Giải thích các bước giải: $A= \dfrac{(\dfrac{1}{23}+ \dfrac{1}{7}- \dfrac{1}{1009}. 23. 7. 1009)}{(\dfrac{1}{23}+ \dfrac{1}{7}- \dfrac{1}{1009}+ \dfrac{1}{23}. \dfrac{1}{7}. \dfrac{1}{1009})}= \dfrac{1}{(23+ 7). 1009- 161+ 1}$ $= \dfrac{7. 1009- 23. 1009- 23. 7}{7. 1009- 23. 1009- 23. 7+ 1}= \dfrac{1}{7. 1009- 23. 1009- 23. 7+ 1}= 1$ $⇒ A= 1$ Bình luận
$A=\left ( \dfrac{1}{7}+\dfrac{1}{23}-\dfrac{1}{1009} \right ):\left ( \dfrac{1}{7}+\dfrac{1}{23}-\dfrac{1}{1009}+\dfrac{1}{7}.\dfrac{1}{23}.\dfrac{1}{1009} \right )+1:\left ( 30.1009-160 \right )\\A=\dfrac{30109}{162449}:\left ( \dfrac{1}{7}+\dfrac{1}{23}-\dfrac{1}{1009}+\dfrac{1}{162449} \right )+1:(30270-160)\\A=\dfrac{30109}{162449}:\dfrac{30110}{162449}+1:30110\\A=\dfrac{30109}{162449}.\dfrac{162449}{30110}+\dfrac{1}{30110}\\A=30109.\dfrac{1}{30110}+\dfrac{1}{30110}\\A=\dfrac{30109}{30110}+\dfrac{1}{30110}\\A=1$Vậy $A=1$ Bình luận
Đáp án:
Giải thích các bước giải:
$A= \dfrac{(\dfrac{1}{23}+ \dfrac{1}{7}- \dfrac{1}{1009}. 23. 7. 1009)}{(\dfrac{1}{23}+ \dfrac{1}{7}- \dfrac{1}{1009}+ \dfrac{1}{23}. \dfrac{1}{7}. \dfrac{1}{1009})}= \dfrac{1}{(23+ 7). 1009- 161+ 1}$
$= \dfrac{7. 1009- 23. 1009- 23. 7}{7. 1009- 23. 1009- 23. 7+ 1}= \dfrac{1}{7. 1009- 23. 1009- 23. 7+ 1}= 1$
$⇒ A= 1$
$A=\left ( \dfrac{1}{7}+\dfrac{1}{23}-\dfrac{1}{1009} \right ):\left ( \dfrac{1}{7}+\dfrac{1}{23}-\dfrac{1}{1009}+\dfrac{1}{7}.\dfrac{1}{23}.\dfrac{1}{1009} \right )+1:\left ( 30.1009-160 \right )\\A=\dfrac{30109}{162449}:\left ( \dfrac{1}{7}+\dfrac{1}{23}-\dfrac{1}{1009}+\dfrac{1}{162449} \right )+1:(30270-160)\\A=\dfrac{30109}{162449}:\dfrac{30110}{162449}+1:30110\\A=\dfrac{30109}{162449}.\dfrac{162449}{30110}+\dfrac{1}{30110}\\A=30109.\dfrac{1}{30110}+\dfrac{1}{30110}\\A=\dfrac{30109}{30110}+\dfrac{1}{30110}\\A=1$
Vậy $A=1$