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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.
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=b.k\\c=d.k\end{matrix}\right.\)
\((\dfrac{a+b}{c+d})^3=\left(\dfrac{bk+b}{dk+d}\right)^3=\left(\dfrac{b.\left(k+1\right)}{d.\left(k+1\right)}\right)^3=\left(\dfrac{b}{d}\right)^3\left(1\right)\)
\(\dfrac{a^3-b^3}{c^3-d^3}=\dfrac{\left(bk\right)^3-b^3}{\left(dk\right)^3-d^3}=\dfrac{b^3.\left(k^3-1\right)}{d^3.\left(k^3-1\right)}=\dfrac{b^3}{d^3}=\left(\dfrac{b}{d}\right)^3\)(2)
Từ (1) và (2) \(\Rightarrow\)\(\left(\dfrac{a+b}{c+d}\right)^3=\dfrac{a^3-b^3}{c^3-d^3}\)
Giải:
Từ \(\left\{{}\begin{matrix}b^2=ac\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\\c^2=bd\Rightarrow\dfrac{b}{c}=\dfrac{c}{d}\end{matrix}\right.\) \(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
Theo tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b-c}{b+c-d}\)
\(\Rightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{\left(a+b-c\right)^3}{\left(b+c-d\right)^3}\left(1\right)\)
Mà \(\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3-c^3}{b^3+c^3-d^3}\left(2\right)\)
Kết hợp \(\left(1\right)\) và \(\left(2\right)\) suy ra:
\(\dfrac{a^3+b^3-c^3}{b^3+c^3-d^3}=\dfrac{\left(a+b-c\right)^3}{\left(b+c-d\right)^3}\) (Đpcm)
\(\left\{{}\begin{matrix}b^2=ac\\c^2=bd\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{b}{c}\\\dfrac{b}{c}=\dfrac{c}{d}\end{matrix}\right.\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
Đặt:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=k\) \(\Rightarrow\left\{{}\begin{matrix}a=bk\\b=ck\\c=dk\end{matrix}\right.\)
Thay vào r tính
Ta có: \(b^2=ac;c^2=bd\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{b}{c}\\\dfrac{b}{c}=\dfrac{c}{d}\end{matrix}\right.\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b-c}{b+c-d}\)
Đặt: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b-c}{b+c-d}=l\) ta có:
\(\left\{{}\begin{matrix}\left(\dfrac{a+b-c}{b+c-d}\right)^3=l^3\\\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3-c^3}{b^3+c^3-d^3}=l^3\end{matrix}\right.\Rightarrowđpcm\)
a) \(\dfrac{2a+3c}{2b+3d}\) = \(\dfrac{2a-3c}{2b-3d}\)
Từ \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\) = k ( k \(\in\) Q, k \(\ne\) 0 )
=> \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
VP = \(\dfrac{2a+3c}{2b+3d}\) = \(\dfrac{2.b.k+3.d.k}{2b+3d}\) = \(\dfrac{k.\left(2b+3d\right)}{2b+3d}\) = k (1)
VT = \(\dfrac{2a-3c}{2b-3d}\) = \(\dfrac{2.b.k-3.d.k}{2b-3d}\) = \(\dfrac{k.\left(2b-3d\right)}{2b-3d}\) = k (2)
Từ (1) và (2) ta có: \(\dfrac{2a+3c}{2b+3d}\) = \(\dfrac{2a-3c}{2b-3d}\)
hay: (2a+3c).(3b-3d) = (2a-3c).(2b+3d)
Ta có: \(b^2=a\cdot c\)
=>\(\frac{a}{b}=\frac{b}{c}\) (1)
Ta có: \(c^2=bd\)
=>\(\frac{b}{c}=\frac{c}{d}\) (2)
Từ (1),(2) suy ra \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\)
=>\(\begin{cases}c=dk\\ b=ck=dk\cdot k=dk^2\\ a=bk=dk^2\cdot k=dk^3\end{cases}\)
a: \(\frac{a^3+b^3-c^3}{b^3+c^3-d^3}=\frac{\left(dk^3\right)^3+\left(dk^2\right)^3-\left(dk\right)^3}{\left(dk^2\right)^3+\left(dk\right)^3-d^3}=\frac{d^3k^3\left(k^6+k^3-1\right)}{d^3\left(k^6+k^3-1\right)}=k^3\)
\(\left(\frac{a+b-c}{b+c-d}\right)^3=\left(\frac{dk^3+dk^2-dk}{dk^2+dk-d}\right)^3\)
\(=\left\lbrack\frac{dk\left(k^2+k-1\right)}{d\left(k^2+k-1\right)}\right\rbrack^3=k^3\)
Do đó: \(\frac{a^3+b^3-c^3}{b^3+c^3-d^3}=\left(\frac{a+b-c}{b+c-d}\right)^3\)
b: \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)
\(=\frac{\left(dk^3\right)^3+\left(dk^2\right)^3+\left(dk\right)^3}{\left(dk^2\right)^3+\left(dk\right)^3+d^3}=\frac{d^3k^3\left(k^6+k^3+1\right)}{d^3\left(k^6+k^3+1\right)}=k^3\)
\(\frac{a}{d}=\frac{dk^3}{d}=k^3\)
Do đó: \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}\)