tính nhanh
a)\(\dfrac{2006x2007+2004-2004}{2007x387+14x2007}\)
b)\(\dfrac{257x752-256}{2000+756+254}\)
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6 tháng 9 2017
a.Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) => \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
=> \(\dfrac{4\left(bk\right)^4+5b^4}{4\left(dk\right)^4+5d^4}=\dfrac{b^4\left(4k^4+5\right)}{d^4\left(4k^4+5\right)}=\dfrac{b^4}{d^4}\)(1)
\(\dfrac{a^2b^2}{c^2d^2}=\dfrac{k^2b^2b^2}{k^2d^2d^2}=\dfrac{b^4}{d^4}\)(2)
Từ (1) và (2) suy ra: \(\dfrac{4a^4+5b^4}{4c^4+5d^4}=\dfrac{a^2b^2}{c^2d^2}\)
b.Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) => \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
=> \(\dfrac{\left(bk\right)^{2004}-b^{2004}}{\left(bk\right)^{2004}+b^{2004}}=\dfrac{b^{2004}\left(k^{2004}-1\right)}{b^{2004}\left(k^{2004}+1\right)}=\dfrac{k^{2004}-1}{k^{2004}+1}\) (1)
\(\dfrac{\left(dk\right)^{2004}-d^{2004}}{\left(dk\right)^{2004}+d^{2004}}=\dfrac{d^{2004}\left(k^{2004}-1\right)}{d^{2004}\left(k^{2004}+1\right)}=\dfrac{k^{2004}-1}{k^{2004}+1}\) (2)
Từ (1) và (2) suy ra: \(\dfrac{a^{2004}-b^{2004}}{a^{2004}+b^{2004}}=\dfrac{c^{2004}-d^{2004}}{c^{2004}+d^{2004}}\)
6 tháng 9 2017
Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{4a^4+5b^4}{4c^4+5d^4}=\dfrac{4b^4k^4+5b^4}{4d^4k^4+5d^4}=\dfrac{b^4\left(4k^4+5\right)}{d^4\left(k^4+5\right)}=\dfrac{b^4}{d^4}\\\dfrac{a^2b^2}{c^2d^2}=\dfrac{bk^2b^2}{dk^2d^2}=\dfrac{k^2b^4}{k^2d^4}=\dfrac{b^4}{d^4}\end{matrix}\right.\)
Vậy.....
\(\left\{{}\begin{matrix}\dfrac{a^{2004}-b^{2004}}{a^{2004}+b^{2004}}=\dfrac{b^{2004}k^{2004}-b^{2004}}{b^{2004}k^{2004}+b^{2004}}=\dfrac{b^{2004}\left(k^{2004}-1\right)}{b^{2004}\left(k^{2004}+1\right)}=\dfrac{k^{2004}-1}{k^{2004}+1}\\\dfrac{c^{2004}-d^{2004}}{c^{2004}+d^{2004}}=\dfrac{d^{2004}k^{2004}-d^{2004}}{d^{2004}k^{2004}+d^{2004}}=\dfrac{d^{2004}\left(k^{2004}-1\right)}{d^{2004}\left(k^{2004}+1\right)}=\dfrac{k^{2004}-1}{k^{2004}+1}\end{matrix}\right.\)
Vậy....
MV
22 tháng 4 2017
\(A=\dfrac{2004^{2005}+1}{2004^{2005}-2004}>1>\dfrac{2004^{2005}}{2004^{2005}+2004}=B\)
Vậy A > B
2 tháng 5 2017
Ta có :
\(\dfrac{2004^{2005}+1}{2004^{2005}-2004}>1>\dfrac{2004^{2005}}{2004^{2005}+2004}\)
\(\Rightarrow\) \(A>1>B\)
\(\Rightarrow\) \(A>B\)
20 tháng 5 2022
a: \(A=\dfrac{\left(2004+1\right)\left(2004^2-2004+1\right)}{2004^2-2003}=2005\)
b: \(B=\dfrac{\left(2005-1\right)\left(2005^2+2005+1\right)}{2005^2+2006}=2004\)
16 tháng 3 2017
a)\(A=x^6-2007x^5+2007x^4-2007x^3+2007x^2-2007x+2007\)
Tại \(x=2006\) thì giá trị biểu thức \(A\) là:
\(A=2006^6-2007\cdot2006^5+...-2007\cdot2006+2007\)
\(=2006^6-\left(2006+1\right)\cdot2006^5+...-\left(2006+1\right)\cdot2006+2007\)
\(=2006^6-2006^6+2006^5-...-2006^2-2006+2007\)
\(=-2006+2007=1\)
b)Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Khi đó
\(VT=\dfrac{\left(bk\right)^{2004}-b^{2004}}{\left(bk\right)^{2004}+b^{2004}}=\dfrac{b^{2004}k^{2004}-b^{2004}}{b^{2004}k^{2004}+b^{2004}}=\dfrac{b^{2004}\left(k^{2004}-1\right)}{b^{2004}\left(k^{2004}+1\right)}=\dfrac{k^{2004}-1}{k^{2004}+1}\left(1\right)\)
\(VP=\dfrac{\left(dk\right)^{2004}-d^{2004}}{\left(dk\right)^{2004}+d^{2004}}=\dfrac{d^{2004}k^{2004}-d^{2004}}{d^{2004}k^{2004}+d^{2004}}=\dfrac{d^{2004}\left(k^{2004}-1\right)}{d^{2004}\left(k^{2004}+1\right)}=\dfrac{k^{2004}-1}{k^{2004}+1}\left(2\right)\)
Từ \((1) và (2)\) ta có điều phải chứng minh
c)Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(A=\left|x-2004\right|+\left|x-1\right|=\left|2004-x\right|+\left|x-1\right|\)
\(\ge\left|2004-x+x-1\right|=2003\)
Đẳng thức xảy ra khi \(1\le x\le2004\)
Vậy với \(1\le x\le2004\) thì \(A_{Min}=2003\)
16 tháng 3 2017
Ta có: \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)
Áp dụng vào bài toán \(\left|x-2004\right|+ \left|x-1\right|\ge\left|x-2004+1-x\right|=2003\)
Dấu "=" xảy ra khi \(\left(x-2004\right)\left(1-x\right)\ge0\)
.....
BM
1
IA
2 tháng 4 2022
Ta có:
N=\(\dfrac{2003+2004}{2004+2005}\)=\(\dfrac{2003}{2004+2005}\)+\(\dfrac{2004}{2004+2005}\)
Ta thấy:
\(\dfrac{2003}{2004+2005}\)<\(\dfrac{2003}{2004}\)(1)
\(\dfrac{2004}{2004+2005}\)<\(\dfrac{2004}{2005}\)(2)
Từ (1) và (2) --> M=\(\dfrac{2003}{2004}\)+\(\dfrac{2004}{2005}\)>\(\dfrac{2003}{2004+2005}\)+\(\dfrac{2004}{2004+2005}\)=N
Vậy M>N
19 tháng 6 2019
cho hỏi chút
\(\frac{a}{b}=\frac{c}{d}\)
trong đó
\(a=c\) hay \(a\ne c\)
\(b=d\) hay \(b\ne d\)
( bài có thiếu điều kiện ko vậy )
VD
1
NN
11 tháng 1 2018
Ta có:
\(B=1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-........-\dfrac{1}{2004^2}.\)
\(B=1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+........+\dfrac{1}{2004^2}\right).\)
Đặt \(M=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+........+\dfrac{1}{2004^2}.\)
Ta thấy:
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}.\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}.\)
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}.\)
..................
\(\dfrac{1}{2004^2}< \dfrac{1}{2003.2004}.\)
\(\Rightarrow M=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+........+\dfrac{1}{2004^2}.\)
\(< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+........+\dfrac{1}{2003.2004}.\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+........+\dfrac{1}{2003}-\dfrac{1}{2004}.\)
\(=\dfrac{1}{1}-\dfrac{1}{2004}.\)
\(=\dfrac{2003}{2004}.\)
\(\Rightarrow M< \dfrac{2003}{2004}.\)
\(\Rightarrow1-M>1-\dfrac{2003}{2004}.\)
\(\Rightarrow B>\dfrac{1}{2004}\) (do B = 1 - M).
\(\Rightarrowđpcm.\)
11 tháng 1 2018
\(B=1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-\dfrac{1}{4^2}-...........-\dfrac{1}{2004^2}\)
\(\Leftrightarrow B=1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...........+\dfrac{1}{2004^2}\right)\)
Đặt :
\(H=\dfrac{1}{2^2}+\dfrac{1}{3^2}+.........+\dfrac{1}{2004^2}\)
Ta có :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
.......................
\(\dfrac{1}{2004^2}< \dfrac{1}{2003.2004}\)
\(\Leftrightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+........+\dfrac{1}{2003.2004}\)
\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.......+\dfrac{1}{2003}-\dfrac{1}{2004}\)
\(\Leftrightarrow A< 1-\dfrac{1}{2004}\)
\(\Leftrightarrow A< \dfrac{2003}{2004}\)
\(\Leftrightarrow1-A< 1-\dfrac{2003}{2004}\)
\(\Leftrightarrow B< \dfrac{1}{2004}\left(đpcm\right)\)
vẽ con cá
a) \(\dfrac{2006\times2007+2004-2004}{2007\times387+14\times2007}\)
= \(\dfrac{2006\times2007}{2007\times\left(387+14\right)}\)
= \(\dfrac{2006}{401}\)