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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.
Ta có:
\(b^2=ac\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\left(1\right)\)
\(c^2=bd\Rightarrow\dfrac{b}{c}=\dfrac{c}{d}\left(2\right)\)
Từ (1) và (2), suy ra: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
\(\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}\)
Vậy \(\dfrac{a}{d}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\)(đpcm)
~ Học tốt!~
\(\dfrac{x^2-yz}{a}=\dfrac{y^2-zx}{b}=\dfrac{z^2-xy}{c}\)
\(\Rightarrow\dfrac{a}{x^2-yz}=\dfrac{b}{y^2-zx}=\dfrac{c}{z^2-xy}\)
\(\Rightarrow\left[\dfrac{a}{x^2-yz}\right]^2=\left[\dfrac{b}{y^2-zx}\right]^2=\left[\dfrac{c}{z^2-xy}\right]^2\)
Ta có
\(\left[\dfrac{a}{x^2-yz}\right]^2=\dfrac{b}{y^2-zx}.\dfrac{c}{z^2-xy}=\)
\(=\dfrac{a^2-bc}{x^4-2x^2yz+y^2z^2-y^2z^2+xy^3+xz^3-x^2yz}=\)
\(=\dfrac{a^2-bc}{x\left(x^3+y^3+z^3-3xyz\right)}=\dfrac{a^2-bc}{x}.\dfrac{1}{\left(x^3+y^3+z^3-3xyz\right)}\)
Tương tự
\(\left[\dfrac{b}{y^2-zx}\right]^2=\dfrac{b^2-ca}{y}.\dfrac{1}{x^3+y^3+z^3-3xyx}\)
\(\left[\dfrac{c}{z^2-xy}\right]^2=\dfrac{c^2-ab}{z}.\dfrac{1}{x^3+y^3+z^3-3xyz}\)
\(\Rightarrow\dfrac{a^2-bc}{x}=\dfrac{b^2-ca}{y}=\dfrac{c^2-ab}{z}\)