Nếu \(\frac{a}{b}=\frac{c}{d}\)thì \(\frac{\left(a-b\right)^{2003}}{\left(c-d\right)^{2003}}=\frac{a^{2003}+b^{2003}}{c^{2003}+d^{2003}}\)
CMR với a>b>c
\(\frac{1}{a^{2003}}+\frac{1}{b^{2003}}+\frac{1}{c^{2003}}\)=\(\frac{1}{\left(a+b+c\right)^{2003}}\)
có\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)CMR\(\frac{\left(19a+5b+1980c\right)^{2003}}{1914^{2003}.a^{2001}.b^2}\)
Cho\(\frac{a}{2003}=\frac{b}{2005}=\frac{c}{2007}\).Chứng minh\(\frac{\left(a-c\right)^2}{4}=\left(a-b\right)\left(b-c\right)\)
Đặt \(\frac{a}{2003}=\frac{b}{2005}=\frac{c}{2007}=k\)\(\Rightarrow a=2003k;b=2005k;c=2007k\)
\(\Rightarrow VT=\frac{\left(a-c\right)^2}{4}=\frac{\left(2003k-2007k\right)^2}{4}=\frac{\left(-4k\right)^2}{4}=\frac{16k^2}{4}=4k^2\left(1\right)\)
\(VP=\left(a-b\right)\left(b-c\right)=\left(2003k-2005k\right)\left(2005k-2007k\right)\)
\(=\left(-2k\right)\cdot\left(-2k\right)=4k^2\left(2\right)\)
Từ (1) và (2) ->Đpcm
a)\(\left(2-\frac{3}{2}\right).\left(2-\frac{4}{3}\right).\left(2-\frac{5}{4}\right).\left(2-\frac{6}{4}\right)\)
b) \(\left(\frac{2003}{2004}+\frac{2004}{2003}\right):\frac{8028025}{8028024}\)
a) \(\left(2-\frac{3}{2}\right)\left(2-\frac{4}{3}\right)\left(2-\frac{5}{4}\right)\left(2-\frac{6}{4}\right)\)
\(=\frac{1}{3}\left(-\frac{4}{3}+2\right)\left(-\frac{5}{4}+2\right)\left(-\frac{6}{4}+2\right)\)
\(=\frac{1}{2}.\frac{2}{3}\left(-\frac{5}{4}+2\right)\left(-\frac{6}{4}+2\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}\left(-\frac{6}{4}+2\right)\)
\(=\frac{1.2.3\left(2-\frac{3}{2}\right)}{2.3.4}\)
\(=\frac{1.3\left(2-\frac{3}{2}\right)}{3.4}\)
\(=\frac{1.\left(2-\frac{3}{2}\right)}{4}\)
\(=\frac{2-\frac{3}{4}}{4}\)
\(=\frac{1}{2.4}\)
\(=\frac{1}{8}\)
b) \(\left(\frac{2003}{2004}+\frac{2004}{2003}\right):\frac{8028025}{8028024}\)
\(=\frac{8028024\left(\frac{2003}{2004}+\frac{2004}{2003}\right)}{8028025}\)
\(=\frac{8028024.\frac{8028025}{4014012}}{8028025}\)
\(=\frac{16056050}{8028025}\)
= 2
Giải các phương trình sau:
a) \(\frac{x-15}{2000}+\frac{x-14}{2001}+\frac{x-13}{2003}=\frac{x-12}{2003}+2\)
b) \(\left(x^2-6x+11\right)\left(y^2+2y+4\right)=2+4z-z^2\)
a. \(\frac{x-15}{2000}+\frac{x-14}{2001}+\frac{x-13}{2003}=\frac{x-12}{2003}+2\)
\(\rightarrow\frac{x}{2000}-\frac{15}{2000}+\frac{x}{2001}-\frac{14}{2001}+\frac{x}{2003}-\frac{13}{2003}=\frac{x}{2003}-\frac{12}{2003}+2\)
\(\rightarrow x.\left(\frac{1}{2000}+\frac{1}{2001}\right)=\frac{15}{2000}+\frac{14}{2001}+\frac{13}{2003}-\frac{12}{2003}+2\)
\(\rightarrow x=2015,5\)
b. \(\left(x^2-6x+11\right)\left(y^2+2y+4\right)=2+4z-z^2\)
\(\rightarrow\left\{{}\begin{matrix}x^2-6x+11=\left(x-3\right)^2+2\ge2\\y^2+2y+4=\left(y+1\right)^2+3\ge3\\2+4z-z^2=-\left(z-2\right)^2+6\le6\end{matrix}\right.\)
\(\rightarrow\left(x^2-6x+11\right)\left(y^2+2y+4\right)\ge6\)
\(\rightarrow\left(x^2-6x+11\right)\left(y^2+2y+4\right)=2+4z-z^2\)
\(\rightarrow\left\{{}\begin{matrix}x=3\\y=-1\\z=2\end{matrix}\right.\)
a. Tìm GTNN của các biểu thức sau
A=|x-2013|+|2014-x|
B=|x-123|+|x-456|
C=|x-1|+|x-2|+|x-3|
D=|x-1|+|x-2|+|x-3|+|x-4
b. Tìm GTLN của biểu thức
A=\(\frac{2003}{\left|x\right|+2004}\)
B=\(\frac{\left|x\right|+2003}{\left|x\right|+2002}\)
a)
\(A=\left|x-2013\right|+\left|2014-x\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(A=\left|x-2013\right|+\left|2014-x\right|\ge\left|x-2013+2014-x\right|\)
\(\Rightarrow A\ge\left|1\right|\)
\(\Rightarrow A\ge1.\)
Dấu '' = '' xảy ra khi:
\(\left\{{}\begin{matrix}x-2013\ge0\\2014-x\le0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge2013\\x\le2014\end{matrix}\right.\Rightarrow2013\le x\le2014.\)
Vậy \(MIN_A=1\) khi \(2013\le x\le2014.\)
Chúc bạn học tốt!
cho \(\frac{a}{2003}=\frac{b}{2004}=\frac{c}{2005}\) chứng minh rằng : \(4\left(a-b\right).\left(b-c\right)=\left(c-a\right)^2\)
Áp dụng tính chất của dãy tỉ số bằng nhau :
\(\frac{a}{2003}=\frac{b}{2004}=\frac{c}{2005}=\frac{a-b}{2003-2004}=\frac{b-c}{2004-2005}=\frac{c-a}{2005-2003}\)
\(\Leftrightarrow\frac{a-b}{-1}=\frac{b-c}{-1}=\frac{c-a}{2}\)
\(\Rightarrow\left(\frac{a-b}{-1}\right)\left(\frac{b-c}{-1}\right)=\left(\frac{c-a}{2}\right)^2\)
\(\Rightarrow\left(a-b\right)\left(b-c\right)=\frac{\left(c-a\right)^2}{4}\)
\(\Rightarrow4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)
Vậy ...
Cho \(\frac{a}{2003}\)=\(\frac{b}{2004}=\frac{c}{2005}\). Chứng minh rằng :\(4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)
Đặt: \(\frac{a}{2003}=\frac{b}{2004}=\frac{c}{2005}=b\Rightarrow\hept{\begin{cases}a=2003b\\b=2004b\\c=2005b\end{cases}}\)
\(\Rightarrow4\left(a-b\right)\left(b-c\right)=4\left(2003b-2004b\right)\left(2004b-2005b\right)=4.-b.-b=4b^2\)
\(\Rightarrow\left(c-a\right)^2=\left(2005b-2003b\right)^2=2k^2=4k^2\)
\(\Rightarrow4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\left(đpcm\right)\)
Đặt a/2003=b/2004=c/2005=k
Suy ra a=2003k, b=2004k, c=2005k (*)
Thay (*) vào 4(a-b)(b-c) ta được:
4(a-b)(b-c)=4(2003k-2004k) (2004k-2005k)
=4k(2003-2004).k(2004-2005)=4k2 .-1.-1
=4.k2 (1)
Thay (*) vào (c-a)2 ta được:
(c-a)2 =(2005k-2003k)2
= k2 (2005-2003)2
=k2 .4 (2)
Từ (1) và (2)
Suy ra ĐPCM
nha
cho \(\frac{a}{2003}=\frac{b}{2004}=\frac{c}{2005}\)CMR 4(a-b)(b-c)=\(\left(c-a\right)^2\)
Đặt \(\frac{a}{2003}\) = \(\frac{b}{2004}\) = \(\frac{c}{2005}\) = k
=> a = 2003k; b = 2004k và c = 2005k
Xét hiệu:
4(a - b)(b - c) - (c - a)2
= 4(2003k - 2004k)(2004k - 2005k) - (2005k - 2003k)2
= 4(-k)(-k) - (2k)2
= 4k2 - 22.k2
= 4k2 - 4k2 = 0
Do đó 4(a - b)(b - c) = (c - a)2.
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