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Ta có :
\(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7b^2+5bd}{7b^2-5bd}\Leftrightarrow\dfrac{7a^2+5ac}{7b^2+5bd}=\dfrac{7a^2-5ac}{7b^2-5bd}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\\ Thaya=bk;c=dk,tacó:\)
\(\dfrac{7a^2+5ac}{7b^2+5bd}=\dfrac{7\cdot b^2\cdot k^2+5\cdot bk\cdot dk}{7b^2+5bd}=\dfrac{k^2\cdot\left(7b^2+5ac\right)}{7b^2+5ac}=k^2\left(1\right)\)
\(\dfrac{7a^2-5ac}{7b^2-5bd}=\dfrac{7\cdot b^2\cdot k^2-5\cdot bk\cdot dk}{7b^2-5bd}=\dfrac{k^2\cdot\left(7b^2-5ac\right)}{7b^2-5ac}=k^2\left(2\right)\)
từ (1) và (2) \(\RightarrowĐpcm\)
ta có \(\frac{7a^2+5ac}{7a^2-5ac}=\frac{\frac{7a^2+5ac}{a^2}}{\frac{7a^2-5ac}{a^2}}=\frac{7+5\frac{c}{a}}{7-5\frac{c}{a}}\)
tương tự ta có \(\frac{7b^2+5bd}{7b^2-5bd}=\frac{\frac{7b^2+5bd}{b^2}}{\frac{7b^2-5bd}{b^2}}=\frac{7+5\frac{d}{b}}{7-5\frac{d}{b}}\)
Mà \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{c}{a}=\frac{d}{b}\Rightarrow\frac{7+5\frac{c}{a}}{7-5\frac{c}{a}}=\frac{7+5\frac{d}{b}}{7-5\frac{d}{b}}\) Nên \(\frac{7a^2+5ac}{7a^2-5ac}=\frac{7b^2+5bd}{7b^2-5bd}\)
hok trường chuyên mak dell bt bài ni ak:))
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk;c=dk\)
Thay vào ta được:\(\frac{7a^2+5ac}{7a^2-5ac}=\frac{7b^2k^2+5bk\cdot dk}{7b^2k^2-5bk\cdot dk}=\frac{bk^2\left(7b+5d\right)}{bk^2\left(7b-5d\right)}=\frac{7b+5d}{7b-5d}\left(1\right)\)
\(\frac{7b^2+5bd}{7b^2-5bd}=\frac{b\left(7b+5d\right)}{b\left(7b-5d\right)}=\frac{7b+5d}{7b-5d}\left(2\right)\)
Từ (1) và (2) \(\Rightarrowđpcm\)
Ta có : a/b = c/d => a/c = b/d
Đặt \(\frac{a}{c}=\frac{b}{d}=k\) => \(\hept{\begin{cases}a=ck\\b=dk\end{cases}}\)
Khi đó, ta có: \(\frac{7.\left(ck\right)^2+5c^2k}{7\left(ck\right)^2-5c^2k}=\frac{7.c^2.k^2+5.c^2.k}{7.c^2.k^2-5.c^2.k}=\frac{\left(7k+5\right).c^2.k}{\left(7k-5\right).c^2.k}=\frac{7k+5}{7k-5}\)(1)
\(\frac{7.\left(dk\right)^2+5.d^2.k}{7\left(dk\right)^2-5.d^2.k}=\frac{7.d^2.k^2+5.d^2.k}{7.d^2.k^2-5.d^2.k}=\frac{\left(7k+5\right).d^2.k}{\left(7k-5\right).d^2.k}=\frac{7k+5}{7k-5}\) (2)
Từ (1) và (2) suy ra (Đpcm)
Với \(\dfrac{a}{b}=\dfrac{c}{d}\)
=> \(\dfrac{a}{b}.\)\(\dfrac{c}{d}=\dfrac{ac}{bd}=\dfrac{aa}{bb}=\dfrac{a^2}{b^2}\)
Ta có : \(\dfrac{a^2}{b^2}=\dfrac{ac}{bd}\)
=> \(\dfrac{7a^2}{7b^2}=\dfrac{5ac}{5bd}\)
Áp dụng t/c dãy tỉ số bằng nhau:
\(\dfrac{7a^2}{7b^2}=\dfrac{5ac}{5bd}=\dfrac{7a^2+5ac}{7b^2+5bd}=\dfrac{7a^2-5ac}{7b^2-5bd}\) (1)
Từ (1) => \(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7b^2-5bd}{7b^2-5bd}\) (ĐPCM)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk,c=dk\)
a, Ta có: \(\left(\frac{a+b}{c+d}\right)^2=\left(\frac{bk+b}{dk+d}\right)^2=\left[\frac{b\left(k+1\right)}{d\left(k+1\right)}\right]^2=\frac{b^2}{d^2}\left(1\right)\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\frac{b^2k^2+b^2}{d^2k^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\left(2\right)\)
Từ (1) và (2) => \(\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)
b, thay vào giống a là đc