1, Cho a, b, c, d là 4 số khác 0 thỏa mãn \(b^2\)=ac và \(c^2\)=bd
Chứng minh rằng: \(\dfrac{2016a^3+2017b^3+2018c^3}{2016b^3+2017c^3+2018d^3}\)=\(\dfrac{a}{d}\)
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26 tháng 10 2021
Chứng minh (2016a-2017b)/(2017c+2018d)=(2016c-2017d)/(2017a+2018b) - Nguyễn Minh Hải
5 tháng 11 2021
\(\left\{{}\begin{matrix}b^2=ac\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\\c^2=bd\Rightarrow\dfrac{b}{c}=\dfrac{c}{d}\end{matrix}\right.\)\(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
Áp dụng t/c dtsbn:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a^3}{b^3}\left(1\right)\)
Và \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\Rightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\left(đpcm\right)\)
25 tháng 11 2021
\(b^2=ac\Rightarrow\dfrac{a}{b}=\dfrac{b}{c};c^2=bd\Rightarrow\dfrac{b}{c}=\dfrac{c}{d}\\ \Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\\ \Rightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{c^3+b^3+d^3}\left(1\right)\\ \text{Đặt }\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=k\\ \Rightarrow a=bk;b=ck;c=dk\\ \Rightarrow a=bk=ck^2=dk^3\\ \Rightarrow\dfrac{a}{d}=k^3\\ \text{Mà }\dfrac{a}{b}=k\Rightarrow\dfrac{a^3}{b^3}=k^3\\ \Rightarrow\dfrac{a}{d}=\dfrac{a^3}{b^3}\left(2\right)\\ \left(1\right)\left(2\right)\RightarrowĐpcm\)
GH
7 tháng 10 2017
b) Ta có: [tex]\frac{a^{2} + c^{2}}{b^{2} + a^{2}}[/tex]= [tex]\frac{bc + c^{2}}{b^{2} + bc}= \frac{c(b +c)}{b(b + c)}= \frac{c}{b}[/tex] (đpcm)
A
12 tháng 7 2017
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)
\(VT=\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{a\left(7a+5c\right)}{a\left(7a-5c\right)}=\dfrac{7ck+5c}{7ck-5c}=\dfrac{c\left(7k+5\right)}{c\left(7k-5\right)}=\dfrac{7k+5}{7k-5}\left(1\right)\)
\(VP=\dfrac{7b^2+5bd}{7b^2-5bd}=\dfrac{b\left(7b+5d\right)}{b\left(7b-5d\right)}=\dfrac{7dk+5d}{7dk-5d}=\dfrac{d\left(7k+5\right)}{d\left(7k-5\right)}=\dfrac{7k+5}{7k-5}\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\)
\(\Rightarrow\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7b^2+5bd}{7b^2-5bd}\left(đpcm\right)\)
Vậy \(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7b^2+5bd}{7b^2-5bd}\)
11 tháng 8 2017
Đặt:
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2bk+5b}{3bk-4b}=\dfrac{b\left(2k+5\right)}{b\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\)
\(\Rightarrow\dfrac{2c+5d}{3c-4d}=\dfrac{2dk+5d}{3dk-4d}=\dfrac{d\left(2k+5\right)}{d\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\)
\(\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
\(\dfrac{2016a-2017b}{2017c+2018d}=\dfrac{2016bk-2017b}{2017dk+2018d}=\dfrac{b\left(2016k-2017\right)}{d\left(2017k+2018\right)}\)
\(\dfrac{2016c-2017d}{2017a+2018b}=\dfrac{2016dk-2017d}{2017bk+2018b}=\dfrac{d\left(2016k-2017\right)}{b\left(2017k+2018\right)}\)
\(\Rightarrow\dfrac{2016a-2017b}{2017c+2018d}=\dfrac{2016c-2017d}{2017a+2018b}\)
\(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7bk^2+5bdk^2}{7bk^2-5bdk^2}=\dfrac{k^2\left(7b+5bd\right)}{k^2\left(7b-5bd\right)}=\dfrac{7b+5bd}{7b-5bd}\)
\(\dfrac{7b^2+5ab}{7b^2-5ab}=\dfrac{7b^2+5kb^2}{7b^2-5kb^2}=\dfrac{b^2\left(7+5k\right)}{b^2\left(7-5k\right)}=\dfrac{7+5k}{7-5k}\)
Hình như sai sai
Làm gì mà căng!!!
Ta có: \(\left\{{}\begin{matrix}b^2=ac\\c^2=bd\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{b}{c}\\\dfrac{b}{c}=\dfrac{c}{d}\end{matrix}\right.\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
Đặt: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=t\)
ta có: \(\dfrac{2016a^3}{2016b^3}=\dfrac{2017b^3}{2017c^3}=\dfrac{2018c^3}{2018d^3}=t^3\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(t^3=\dfrac{2016a^3+2017b^3+2018c^3}{2016b^3+2017c^3+2018d^3}\)
Mặt khác: \(\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=t.t.t=t^3=\dfrac{a}{d}\)
Ta có đpcm
a^3/d^3 chứ,mang tính chất tương ứng mà:v