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Giải:
Ta có: \(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c}\)
\(c^2=bd\Rightarrow\frac{b}{c}=\frac{c}{d}\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\)
\(\Rightarrow a=bk,b=ck,c=dk\)
Ta có:
\(\left(\frac{a+b-c}{b+c-d}\right)^3=\left(\frac{bk+ck-dk}{b+c-d}\right)^3=\left[\frac{k\left(b+c-d\right)}{b+c-d}\right]^3=k^3\) (1)
\(\left(\frac{2a+3b-4c}{2b+3c-4d}\right)^2=\left(\frac{2bk+3ck-4dk}{2b+3c-4d}\right)^3=\left[\frac{k\left(2b+3c-4d\right)}{2b+3c-4d}\right]^3=k^3\) (2)
Từ (1) và (2) suy ra \(\left(\frac{a+b-c}{b+c-d}\right)^3=\left(\frac{2a+3b-4c}{2b+3c-4d}\right)^3\) ( đpcm )
a: Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
=>a=bk; c=dk
\(\frac{ab}{cd}=\frac{bk\cdot b}{dk\cdot d}=\frac{b^2}{d^2}\)
\(\frac{a^2-b^2}{c^2-d^2}=\frac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\frac{b^2}{d^2}\)
Do đó: \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)
b: \(\frac{7a-4b}{3a+5b}=\frac{7\cdot bk-4b}{3\cdot bk+5b}=\frac{b\left(7k-4\right)}{b\left(3k+5\right)}=\frac{7k-4}{3k+5}\)
\(\frac{7c-4d}{3c+5d}=\frac{7\cdot dk-4d}{3\cdot dk+5d}=\frac{d\left(7k-4\right)}{d\left(3k+5\right)}=\frac{7k-4}{3k+5}\)
Do đó: \(\frac{7a-4b}{3a+5b}=\frac{7c-4d}{3c+5d}\)
c: \(\frac{ac}{bd}=\frac{bk\cdot dk}{bd}=k^2\)
\(\frac{a^2+c^2}{b^2+d^2}=\frac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\frac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\)
\(\frac{\left(c-a\right)^2}{\left(d-b\right)^2}=\frac{\left(dk-bk\right)^2}{\left(d-b\right)^2}=\frac{k^2\left(d-b\right)^2}{\left(d-b\right)^2}=k^2\)
Do đó; \(\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}=\frac{\left(c-a\right)^2}{\left(d-b\right)^2}\)
d: \(\frac{a^3+b^3}{c^3+d^3}=\frac{\left(bk\right)^3+b^3}{\left(dk\right)^3+d^3}=\frac{b^3\left(k^3+1\right)}{d^3\left(k^3+1\right)}=\frac{b^3}{d^3}\)
\(\frac{\left(a+b\right)^3}{\left(c+d\right)^3}=\frac{\left(bk+b\right)^3}{\left(dk+d\right)^3}=\frac{b^3\left(k+1\right)^3}{d^3\left(k+1\right)^3}=\frac{b^3}{d^3}\)
Do đó: \(\frac{a^3+b^3}{c^3+d^3}=\frac{\left(a+b\right)^3}{\left(c+d\right)^3}\)
Do đó:
Bài 1: Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)
\(\dfrac{a}{a+c}=\dfrac{ck}{ck+c}=\dfrac{ck}{c\left(k+1\right)}=\dfrac{k}{k+1}\)
\(\dfrac{b}{b+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)
Do đó: \(\dfrac{a}{a+c}=\dfrac{b}{b+d}\)
Bài 1: Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)
\(\dfrac{a}{a+c}=\dfrac{ck}{ck+c}=\dfrac{ck}{c\left(k+1\right)}=\dfrac{k}{k+1}\)
\(\dfrac{b}{b+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)
Do đó: \(\dfrac{a}{a+c}=\dfrac{b}{b+d}\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
=>a=bk; c=dk
a: \(\frac{2a+3c}{2b+3d}=\frac{2\cdot bk+3\cdot dk}{2b+3d}=\frac{k\left(2b+3d\right)}{2b+3d}=k\)
\(\frac{2a-3c}{2b-3d}=\frac{2bk-3dk}{2b-3d}=\frac{k\left(2b-3d\right)}{2b-3d}=k\)
Do đó: \(\frac{2a+3c}{2b+3d}=\frac{2a-3c}{2b-3d}\)
b: \(\frac{a^2+c^2}{b^2+d^2}=\frac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\frac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\)
\(\frac{ac}{bd}=\frac{bk\cdot dk}{bd}=k^2\)
Do đó: \(\frac{a^2+c^2}{b^2+d^2}=\frac{ac}{bd}\)
đặt a/b =c/d =k
=> a=bm , c=dm
=> 2a+3c/2b+3d =2bm+3bm/ 2b +3d = m.(2d+3d)/2d+3d =m (1)
=> 2a-3c/2d-3d=2bm-3dm /2b -3d =m.(2b-3d)/2b-3d= m (2)
Từ (1) và (2) => 2a+3c/2b+3d =2a-3c/2b-3d
câu 2 tương tự nha
Ta có: \(b^2=ac\)
=>\(\frac{a}{b}=\frac{b}{c}\) (1)
\(c^2=bd\)
=>\(\frac{b}{c}=\frac{c}{d}\) (2)
Từ (1),(2) suy ra \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\)
=>c=dk; \(b=ck=dk\cdot k=dk^2\) và \(a=bk=dk^2\cdot k=dk^3\)
\(\frac{3a^3-4b^3+5c^3}{3b^3-4c^3+5d^3}=\frac{3\cdot\left(dk^3\right)^3-4\cdot\left(dk^2\right)^3+5\cdot\left(dk\right)^3}{3\cdot\left(dk^2\right)^3-4\cdot\left(dk\right)^3+5\cdot d^3}\)
\(=k^3\cdot\frac{3\cdot d^3\cdot k^6-4\cdot d^3\cdot k^3+5d^3}{3d^3\cdot k^6-4\cdot d^3\cdot k^3+5\cdot d^3}=k^3\)
\(\frac{a}{d}=\frac{dk^3}{d}=k^3\)
Do đó: \(\frac{3a^3-4b^3+5c^3}{3b^3-4c^3+5d^3}=\frac{a}{d}\)