cho abc=2004 chứng minh
\(\dfrac{2004a}{ab+2004a+2004}\) +\(\dfrac{b}{bc+b+2004}\) +\(\dfrac{c}{ac+c+1}\) =1
Cho 3 số a, b, c thỏa mãn abc = 2004
Tính: \(M=\frac{2004a}{ab+2004a+2004}+\frac{b}{bc+b+2004}+\frac{c}{ac+c+1}\)
\(M=\frac{2004a}{ab+a^2bc+abc}+\frac{b}{bc+b+abc}+\frac{c}{ac+c+1}\)
\(M=\frac{2004a}{ab\left(1+ac+c\right)}+\frac{b}{b\left(c+1+ac\right)}+\frac{c}{ac+c+1}\)
\(M=\frac{2004ac+abc+abc^2}{abc\left(ac+c+1\right)}=\frac{a^2bc^2+abc+abc^2}{abc\left(ac+c+1\right)}=\frac{abc\left(ac+1+c\right)}{abc\left(ac+c+1\right)}=1\)
a) tính giá trị biểu thức: \(x^6-2007x^5+2007x^4-2007x^3+2007x^2-2007x+2007\)biết x=2006
b)cho \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh rằng \(\dfrac{a^{2004}-b^{2004}}{a^{2004}+b^{2004}}\)=\(\dfrac{c^{2004}-d^{2004}}{c^{2004}+d^{2004}}\)
c) tính giá trị nhỏ nhất của biểu thức \(|x-2004|+|x-1|\)
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 \(A_{Min}=2003\)
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 \(\left(x-2004\right)\left(1-x\right)\ge0\)
.....
\(cho:\frac{a^2+2004^2}{b^2+2005^2}=\frac{2004a}{2005b}\left(a,bkhac0\right).CMR:\orbr{\begin{cases}\frac{a}{2004}=\frac{b}{2005}\\\frac{a}{2004}=\frac{2005}{b}\end{cases}}\)
a,\(Cho\dfrac{a}{b}=\dfrac{c}{d}CMR,\dfrac{4a^4+5b^4}{4c^4+5d^4}=\dfrac{a^2b^2}{c^2d^2}\)
b,Cho\(\dfrac{a}{b}=\dfrac{c}{d}CMR,\dfrac{a^{2004}-b^{2004}}{a^{2004}+b^{20004}}=\dfrac{c^{2004}-d^{2004}}{c^{2004}+d^{2004}}\)
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: \(\dfrac{a^{2004}-b^{2004}}{a^{2004}+b^{2004}}=\dfrac{c^{2004}-d^{2004}}{c^{2004}+d^{2004}}\)
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....
Theo đề bài, ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a^4}{c^4}=\dfrac{b^4}{d^4}=\dfrac{4a^4}{4c^4}=\dfrac{5b^4}{5d^4}=\dfrac{4a^4+5b^4}{4c^4+5d^4}\left(1\right)\)
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}=\dfrac{a^2b^2}{b^4}=\dfrac{c^2d^2}{d^4}=\dfrac{a^2b^2}{c^2d^2}=\dfrac{b^4}{d^4}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\dfrac{4a^4+5b^4}{4c^4+5d^4}=\dfrac{a^2b^2}{c^2d^2}\)(đpcm)
b/ Theo đề bài, ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a^{2004}}{c^{2004}}=\dfrac{b^{2004}}{d^{2004}}=\dfrac{a^{2004}+b^{2004}}{c^{2004}+d^{2004}}\left(1\right)\)
\(\Rightarrow\dfrac{a^{2004}}{c^{2004}}=\dfrac{b^{2004}}{d^{2004}}=\dfrac{a^{2004}-b^{2004}}{c^{2004}-d^{2004}}\left(2\right)\)
Từ (1) và (2)\(\Rightarrow\dfrac{a^{2004}+b^{2004}}{c^{2004}+d^{2004}}=\dfrac{a^{2004}-b^{2004}}{c^{2004}-d^{2004}}=\dfrac{a^{2004}-b^{2004}}{a^{2004}+b^{2004}}=\dfrac{c^{2004}-d^{2004}}{c^{2004}+d^{2004}}\left(đpcm\right)\)
Chứng minh\(B=1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-\dfrac{1}{4^2}-...-\dfrac{1}{2004^2}>\dfrac{1}{2004}\)
Chứng minh \(S=\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-...+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+...+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004 }}< 0.2\)
a)
\(B=1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-\dfrac{1}{4^2}-...........-\dfrac{1}{2004^2}\)
\(\Leftrightarrow B=1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+..............+\dfrac{1}{2004^2}\right)\)
Đặt :
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+.............+\dfrac{1}{2004^2}\)
Ta thấy :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
..........................
\(\dfrac{1}{2004^2}< \dfrac{1}{2003.2004}\)
\(\Leftrightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+..............+\dfrac{1}{2003.2004}\)
\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+..........+\dfrac{1}{2003}-\dfrac{1}{2004}\)
\(\Leftrightarrow A< 1-\dfrac{1}{2004}\)
\(\Leftrightarrow A< \dfrac{2003}{2004}\)
\(\Leftrightarrow1-A< 1-\dfrac{2003}{2004}\)
\(\Leftrightarrow B< \dfrac{1}{2004}\left(đpcm\right)\)
b) \(S=\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-........+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+.......+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\)
\(\Leftrightarrow2^2S=2^2\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+.....+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+....+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)\)
\(\Leftrightarrow4S=1-\dfrac{1}{2^2}+.......+\dfrac{1}{2^{4n}}-\dfrac{1}{2^{4n+2}}+.......+\dfrac{1}{2^{2000}}-\dfrac{1}{2^{2002}}\)
\(\Leftrightarrow4S+S=\left(1-\dfrac{1}{2^2}+.....+\dfrac{1}{2^{2000}}-\dfrac{1}{2^{2002}}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+.......+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)\)\(\Leftrightarrow5S=1-\dfrac{1}{2^{2004}}< 1\)
\(\Leftrightarrow S< \dfrac{1}{5}=0,2\)
\(\Leftrightarrow S< 0,2\left(đpcm\right)\)
Chứng minh rằng: \(B=1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-\dfrac{1}{4^2}-...-\dfrac{1}{2004^2}>\dfrac{1}{2004}\)
Help me!
Tao có: \(B=1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-\dfrac{1}{4^2}-...-\dfrac{1}{2004^2}\)
\(B>1-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2003\cdot2004}\right)\)
\(B>1-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2003}-\dfrac{1}{2004}\right)\)
\(B>1-\left(1-\dfrac{1}{2004}\right)=1-1+\dfrac{1}{2004}=\dfrac{1}{2004}\left(đpcm\right)\)
Chứng tỏ B = \(1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-.......-\dfrac{1}{2004^2}>\dfrac{1}{2004}\)
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).
\(\Rightarrowđpcm.\)
\(B=1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-\dfrac{1}{4^2}-...........-\dfrac{1}{2004^2}\)
\(\Leftrightarrow B=1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...........+\dfrac{1}{2004^2}\right)\)
Đặt :
\(H=\dfrac{1}{2^2}+\dfrac{1}{3^2}+.........+\dfrac{1}{2004^2}\)
Ta có :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
.......................
\(\dfrac{1}{2004^2}< \dfrac{1}{2003.2004}\)
\(\Leftrightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+........+\dfrac{1}{2003.2004}\)
\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.......+\dfrac{1}{2003}-\dfrac{1}{2004}\)
\(\Leftrightarrow A< 1-\dfrac{1}{2004}\)
\(\Leftrightarrow A< \dfrac{2003}{2004}\)
\(\Leftrightarrow1-A< 1-\dfrac{2003}{2004}\)
\(\Leftrightarrow B< \dfrac{1}{2004}\left(đpcm\right)\)
1. Cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\). Chứng minh rằng \(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)
2. Cho \(\dfrac{a}{2003}=\dfrac{b}{2004}=\dfrac{c}{2005}\). Chứng minh rằng \(4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)
Bài 1:
Áp dụng t.c của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\\ =\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a^3}{b^3}=\dfrac{a.b.c}{b.c.d}=\dfrac{a}{d}\left(dpcm\right)\)
Cho \(\dfrac{a}{b} = \dfrac{c}{d}\) . Chứng minh :
a, \(\dfrac{a^{2005}}{b^{2005}} = \dfrac{(a-c)^{2005}}{(b-d)^{2005}}\)
b, \(\dfrac{(a^2+b^2)^3}{(c^2+d^2)^3}\) =\(\dfrac{a^3+b^3)^2}{(c^3+d^3)^2}\)
c, \((\dfrac{a-b}{c-d})^{2005}\) = \(\dfrac{2.a^{2005}-b^{2005}}{2.c^{2005}-d^{2005}}\)
d, \(\dfrac{(a^2-b^2)^5}{(c^2-d^2)^5} = \) \(\dfrac{a^{10}+b^{10}}{c^{10}+d^{10}}\)
e, \(\dfrac{2.a^{2005}+5.b^{2005}}{2.c^{2005}+5.d^{2005}}\) = \(\dfrac{(a+b)^{2005}}{(c+d)^{2005}}\)
f, \(\dfrac{(a^{2004}+b^{2004})^{2005}}{(c^{2004}+d^{2004})^{2005}}\) = \(\dfrac{(a^{2005} -b^{2005})^{2004}}{(c^{2005}-d^{2005})^{2004}}\)
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 )