Question

[le tuan my tam]

So sánh

\(A=\dfrac{2003.2004-1}{2003.2004};B=\dfrac{2004.2005-1}{2004.2005}\)

Tính :

(3x - 2^4) .7^5 = 2.7^6 . 1/2009^0

Answer 1

[Trần Hùng Luyện]

a) A=\(\dfrac{2003.2004-1}{2003.2004}=\dfrac{2003.2004}{2003.2004}-\dfrac{1}{2004}=1-\dfrac{1}{2003.2004}\)

B = \(\dfrac{2004.2005-1}{2004.2005}=\dfrac{2004.2005}{2004.2005}-\dfrac{1}{2004.2005}=1-\dfrac{1}{2004.2005}\)

Vì \(\dfrac{1}{2003.2004}>\dfrac{1}{2004.2005}\)

\(\Rightarrow1-\dfrac{1}{2003.2004}< 1-\dfrac{1}{2004.2005}\)

Vậy A < B

b) \(\left(3X-2^4\right).7^5=2.7^6.\dfrac{1}{2009^0}\)

\(\left(3X-2^4\right).7^5=2.7^6.1\)

\(\left(3X-2^4\right).7^5=2.7^6\)

\(\left(3X-2^4\right).=2.7^6:7^5\)

\(3X-2^4=2.7\)

\(3X-16=14\)

\(3X=16+14=30\)

\(X=30:3=10\)

Vậy X = 10

Answer 2

[Nguyễn Thanh Hằng]

1/ \(A=\dfrac{2003.2004-1}{2003.2004}=\dfrac{2003.2004}{2003.2004}-\dfrac{1}{2003.2004}=1-\dfrac{1}{2003.2004}\)

\(B=\dfrac{2004.2005-1}{2004.2005}=\dfrac{2004.2005}{2004.2005}-\dfrac{1}{2004.2005}=1-\dfrac{1}{2004.2005}\)

Vì \(1-\dfrac{1}{2003.2004}< 1-\dfrac{1}{2004.2005}\Leftrightarrow A< B\)

2/ \(\left(3x-2^4\right).7^5=2.7^6.\dfrac{1}{2009^0}\)

\(\Leftrightarrow\left(3x-2^4\right).7^5=2.7^6.1\)

\(\Leftrightarrow3x-2^4=2.7^6:7^5\)

\(\Leftrightarrow3x-2^4=2.7\)

\(\Leftrightarrow3x-16=14\)

\(\Leftrightarrow3x=30\)

\(\Leftrightarrow x=10\left(tm\right)\)

Vậy ..