Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

\(\dfrac{2a+3b}{2a-3b}=\dfrac{2bk+3b}{2bk-3b}=\dfrac{b\left(2k+3\right)}{b\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\left(1\right)\)

\(\dfrac{2c+3d}{2c-3d}=\dfrac{2dk+3d}{2dk-3d}=\dfrac{d\left(2k+3\right)}{d\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\left(2\right)\)

Từ (1) và (2) suy ra \(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3b}\left(=\dfrac{2k+3}{2k-3}\right)\)

Áp dụng tính chất dãy tỉ số băng nhau,ta có:

\(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{c}=\dfrac{b}{d}=>\dfrac{2a}{2c}=\dfrac{3b}{3d}=>\dfrac{2a+3b}{2c+3d}=\dfrac{2a-3d}{2c-3d}=>\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\left(đpcm\right)\)

a) Có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{4a}{3b}=\frac{4c}{3d}\)

Áp dụng dãy tỉ số bằng nhau ta có :

\(\frac{4a}{3b}=\frac{4c}{3d}\Rightarrow\frac{4a-3b}{4a+3b}=\frac{4c-3d}{4c+3d}\Rightarrow\frac{4a-3d}{4c-3d}=\frac{4a+3b}{4c+3d}\)

b) Có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{2a}{3b}=\frac{2c}{3d}\)

Áp dụng dãy tỉ số bằng nhau ta có :

\(\frac{2a}{3b}=\frac{2c}{2d}\Rightarrow\frac{2a-3b}{2a+3b}=\frac{2c-3d}{2c+3d}\)

a) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có:

\(\dfrac{2a+3b}{2a-3b}=\dfrac{2bk+3b}{2bk-3b}=\dfrac{b\left(2k+3\right)}{b\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\) (1)

\(\dfrac{2c+3d}{2c-3d}=\dfrac{2dk+3d}{2dk-3d}=\dfrac{d\left(2k+3\right)}{d\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\) (2)

Từ (1) và (2) suy ra \(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\)

b) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=q\Rightarrow\left\{{}\begin{matrix}a=bq\\c=dq\end{matrix}\right.\)

Ta có:

\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bq+b}{dq+d}\right)^2=\left[\dfrac{b\left(q+1\right)}{d\left(q+1\right)}\right]^2=\dfrac{b}{d}\) (1)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bq\right)^2+b^2}{\left(dq\right)^2+d^2}=\dfrac{b^2.q^2+b^2}{d^2.q^2+d^2}=\dfrac{b^2\left(q^2+1\right)}{d^2\left(q^2+1\right)}=\dfrac{b}{d}\) (2)

Từ (1) và (2) suy ra \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)

\(\dfrac{a}{b}=\dfrac{c}{d}\) => \(\dfrac{a}{c}=\dfrac{b}{d}\)

áp dụng tính chất dãy tỉ số = nhau ta có

\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a+3b}{2c+3d}=\dfrac{2a-3b}{2c-3d}\)

= \(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\) (đpcm)

Bạn vào câu hỏi tương tự nhé  Dương Diệu Linh

a, Vì \(\dfrac{a}{c}=\dfrac{c}{b}\Rightarrow ab=c^2\)

Ta có :

\(\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{\left(b+a\right)\left(b-a\right)}{a^2+ab}=\dfrac{\left(b+a\right)\left(b-a\right)}{a\left(a+b\right)}=\dfrac{b-a}{a}\)

Vậy \(\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{b-a}{a}\)

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