Cho dãy tỉ số bằng nhau:\(\frac{a}{2003}=\frac{b}{2005}=\frac{c}{2007}\)
Chứng minh rằng:\(\frac{\left(a-c\right)^2}{4}=\left(a-b\right)\left(b-c\right)\)
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
Cho \(\frac{a}{b}=\frac{c}{d}\). Chứng minh:
a) \(\frac{\left(a-b\right)^3}{\left(c-d\right)^3}=\frac{3a^2+2b^2}{3c^2+2d^2}\)
b)\(\frac{4a^4+5b^4}{4c^4+5d^4}=\frac{a^2b^2}{c^2d^2}\)
c)\(\left(\frac{a-b}{c-d}\right)^{2005}=\frac{2a^{2005}-b^{2005}}{2c^{2005}-d^{2005}}\)
d)\(\frac{2a^{2005}+5b^{2005}}{2c^{2005}+5d^{2005}}=\frac{\left(a+b\right)^{2005}}{\left(c+d\right)^{2005}}\)
e)\(\frac{\left(20a^{2006}+11b^{2006}\right)^{2007}}{\left(20a^{2007}-11b^{2007}\right)^{2006}}=\frac{\left(20c^{2006}+11d^{2006}\right)^{2007}}{\left(20c^{2007}-11d^{2007}\right)^{2006}}\)
f)\(\frac{\left(20a^{2007}-11c^{2007}\right)^{2006}}{\left(20a^{2006}+11c^{2006}\right)^{2007}}=\frac{\left(20b^{2007}-11d^{2007}\right)^{2006}}{\left(20b^{2006}+11d^{2006}\right)^{2007}}\)
ừ, bạn bik làm thì giúp mình nha ^^
Cho dãy tỉ số bằng nhau \(\frac{a}{m}=\frac{b}{m+n}=\frac{c}{m+2n}\)Chứng minh rằng: \(4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)
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 \(\frac{a}{2003}=\frac{b}{2004}=\frac{c}{2005}\)
Chứng minh rằng: \(4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)
Mình cũng học lớp 7 nhưng lần đầu mình thấy những loại toán này
coi \(\frac{a}{2003}=\frac{b}{2004}=\frac{c}{2005}=k\Rightarrow a=2003k;b=2004k;c=2005k\)
thay mấy cái trên vào 4(a-b)(b-c)và (c-a)2
Cho dãy tỉ số bằng nhau : \(\frac{a}{2009}=\frac{b}{2011}=\frac{c}{2013}.CMR:\frac{\left(a-c\right)^2}{4}=\left(a-b\right).\left(b-c\right)\)
Ta có : \(\frac{a}{2009}=\frac{b}{2011}=\frac{c}{2013}=\frac{a-b}{-2}=\frac{b-c}{-2}=\frac{a-c}{-4}\)
\(=>\frac{\left(a-c\right)^2}{16}=\left(\frac{a-b}{-2}\right).\left(\frac{b-c}{-2}\right)=\frac{\left(a-b\right).\left(b-c\right)}{4}\)
\(=>\frac{\left(a-c\right)^2}{4}=\left(a-b\right).\left(b-c\right)\)
Cho dãy tỉ số bằng nhau : \(\frac{a}{2009}=\frac{b}{2011}=\frac{c}{2013}.CMR:\frac{\left(a-c\right)^2}{4}=\left(a-c\right).\left(b-c\right)\)
Áp dụng t/c dãy tỉ số bằng nhau,ta có:
\(\frac{a}{2009}=\frac{b}{2011}=\frac{a-b}{2009-2011}=\frac{a-b}{-2}\)
\(\frac{b}{2011}=\frac{c}{2013}=\frac{b-c}{2011-2013}=\frac{b-c}{-2}\)
\(\frac{a}{2009}=\frac{c}{2013}=\frac{a-c}{2009-2013}=\frac{a-c}{-4}\)
=> \(\frac{a-b}{-2}=\frac{b-c}{-2}=\frac{a-c}{-4}\)
=> \(\frac{a-b}{-2}.\frac{b-c}{-2}=\left(\frac{a-c}{4}\right)^2\)
=> \(\frac{\left(a-c\right)^2}{4^2}=\frac{\left(a-b\right)\left(b-c\right)}{4}\)
=> \(\frac{\left(a-c\right)^2}{4}=\left(a-c\right)\left(b-c\right)\)
Ta có : \(\frac{a}{2009}=\frac{b}{2011}=\frac{c}{2013}=\frac{a-b}{-2}=\frac{b-c}{-2}=\frac{a-c}{-4}\)
\(=>\frac{\left(a-c\right)^2}{16}=\left(\frac{a-b}{-2}\right).\left(\frac{b-c}{-2}\right)=\frac{\left(a-b\right).\left(b-c\right)}{4}\)
\(=>\frac{\left(a-c\right)^2}{4}=\left(a-b\right).\left(b-c\right)\)
Cho dãy tỉ số bằng nhau:\(\frac{a}{n+2}=\frac{b}{n+5}=\frac{c}{n+8}\)(với \(n\in N\))
Chứng minh rằng:\(\left(a+c\right)^2=4\left(a-b\right)\left(b-c\right)\)
\(\frac{a}{n+2}=\frac{b}{n+5}=\frac{c}{n+8}=k\Leftrightarrow a=nk+2k;b=nk=5k;c=nk+8k\)
\(\left(a+c\right)^2=\left(nk+2k+nk+8k\right)^2=4k^2\left(n+5\right)^2\) ( sai nhế)
\(4\left(a-b\right)\left(b-c\right)=4\left(nk+2k-nk-5k\right)\left(nk+5k-nk-8k\right)=4\left(-3k\right)\left(-3k\right)=36k^2\)
\(\left(a-c\right)^2=\left(nk+2k-nk-8k\right)^2=4\left(-6k\right)^2=36k^2\)
=> \(\left(a-c\right)^2=4\left(a-b\right)\left(b-c\right)\)