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\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}=\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\left(1\right)\\ \text{Đặt }\dfrac{a}{b}=\dfrac{b}{c}=k\Leftrightarrow a=bk;b=ck\Leftrightarrow a=ck^2\\ \Leftrightarrow\dfrac{a^2}{bc}=\dfrac{c^2k^4}{c^2k}=k^3=\left(\dfrac{a}{b}\right)^3\left(2\right)\\ \left(1\right)\left(2\right)\Leftrightarrow\left(\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\right)^3=\dfrac{a^2}{bc}\)
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{a}=\dfrac{a+b+c+d}{a+b+c+d}=1\\ \Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=d\\d=a\end{matrix}\right.\Rightarrow a=b=c=d\\ \Rightarrow VT=\left(\dfrac{2019a+2020a-2021a}{2019a+2020a-2021a}\right)^3=1^3=1=\dfrac{a^2}{a\cdot a}=VP\)
Sửa đề: \(\dfrac{2018a-2019b}{2019a+2020b}=\dfrac{2018c-2019d}{2019c+2020d}\)
\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2020a}{2020b}=\dfrac{2020c}{2020d}=\dfrac{2019a}{2019c}=\dfrac{2019b}{2019d}=\dfrac{2018a}{2018c}=\dfrac{2018b}{2018d}=\dfrac{2018a-2019b}{2018c-2019d}=\dfrac{2019a+2020b}{2019c+2020d}\\ \Leftrightarrow\dfrac{2018a-2019b}{2019a+2020b}=\dfrac{2018c-2019d}{2019c+2020d}\)
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
\(\Leftrightarrow\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}\)
\(\Leftrightarrow\dfrac{2020a^2}{2020c^2}=\dfrac{2021b^2}{2021d^2}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{2020a^2}{2020c^2}=\dfrac{2021b^2}{2021d^2}=\dfrac{2020a^2+2021b^2}{2020c^2+2021d^2}=\dfrac{2020a^2-2021b^2}{2020c^2-2021d^2}\)
Ta có: \(\dfrac{2020a^2+2021b^2}{2020c^2+2021d^2}=\dfrac{2020a^2-2021b^2}{2020c^2-2021d^2}\)(cmt)
nên \(\dfrac{2020a^2+2021b^2}{2020a^2-2021b^2}=\dfrac{2020c^2+2021d^2}{2020c^2-2021d^2}\)(đpcm)
Ta có \(\frac{a}{b}=\frac{c}{d}=>\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}=>\frac{a}{a-b}=\frac{c}{c-d} \)
Bài 1:
Áp dụng t.c của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\\ =\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a^3}{b^3}=\dfrac{a.b.c}{b.c.d}=\dfrac{a}{d}\left(dpcm\right)\)
ta có: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}\Rightarrow\dfrac{a^4}{b^4}=\dfrac{b^4}{c^4}=\dfrac{c^4}{d^4}=\dfrac{d^4}{e^4}\)
\(\dfrac{2a^4+3b^4+4c^4+5d^4}{2b^4+3c^4+4d^4+5e^4}=\dfrac{2a^4}{2b^4}=\dfrac{3b^4}{3c^4}=\dfrac{4c^4}{4d^4}=\dfrac{4d^4}{4e^4}\\ =\dfrac{a^4}{b^4}=\dfrac{b^4}{c^4}=\dfrac{c^4}{d^4}=\dfrac{d^4}{e^4}\\ \dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}\)
Bài 1:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
Khi đó: \(\left\{\begin{matrix} \frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b(2k+5)}{b(3k-4)}=\frac{2k+5}{3k-4}\\ \frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{d(2k+5)}{d(3k-4)}=\frac{2k+5}{3k-4}\end{matrix}\right.\)
\(\Rightarrow \frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)
Ta có đpcm.
Bài 2:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
Khi đó: \(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{(bk)^2+b^2}{(dk)^2+d^2}=\frac{b^2(k^2+1)}{d^2(k^2+1)}=\frac{b^2}{d^2}\)
Do đó: \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}(=\frac{b^2}{d^2})\) . Ta có đpcm.
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