đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
suy ra:\(\frac{ac}{bd}=\frac{bk.dk}{bd}=k.k=k^2\)
\(\frac{a^2+c^2}{b^2+d^2}=\frac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\frac{b^2k^2+d^2k^2}{b^2+d^2}=\frac{k^2.\left(b^2+d^2\right)}{b^2+d^2}=k^2\)
vậy \(\frac{ab}{bd}=\frac{a^2+c^2}{b^2+d^2}\)
Ta có:\(\frac{a}{b}=\frac{c}{d}=>\frac{a}{b}.\frac{c}{d}=\frac{c}{d}.\frac{c}{d}=>\frac{ac}{bd}=\frac{c^2}{d^2}\)
\(\frac{c}{d}=\frac{a}{b}=>\frac{a}{b}.\frac{c}{d}=\frac{a}{b}.\frac{a}{b}=>\frac{ac}{bd}=\frac{a^2}{b^2}\)
=>\(\frac{ac}{bd}=\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}\)
=>\(\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\)
\(\frac{a}{b}\).\(\frac{c}{d}\)=\(\frac{c}{d}.\frac{c}{d}\)là sao? và tương tự ở dưới???
Ta có:
\(\frac{a}{b}\)=\(\frac{c}{d}\)=> \(\frac{a^2}{b^2}\)= \(\frac{c^2}{d^2}\)= \(\frac{a^2+c^2}{b^2+d^2}\)(1)
\(\frac{a}{b}\)= \(\frac{c}{d}\)=>\(\frac{a}{b}\)* \(\frac{a}{b}\)= \(\frac{a}{b}\)* \(\frac{c}{d}\)=>\(\frac{a^2}{b^2}\)= \(\frac{a}{b}\)* \(\frac{c}{d}\)(2)
Từ (1) và (2) => \(\frac{ac}{bd}\)= \(\frac{a^2.c^2}{b^2.d^2}\)
ta có
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}.\frac{c}{d}=\frac{c}{d}.\frac{c}{d}\Rightarrow\frac{ac}{bd}=\frac{c^2}{d^2}\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}.\frac{c}{d}=\frac{a}{b}.\frac{a}{b}\Rightarrow\frac{ac}{bd}=\frac{a^2}{b^2}\)
\(\Rightarrow\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\)
Đặt a/b=c/d=k suy ra: a=bk,c=dk
Thay a=bk ,c=dk vào biểu thức ac/bd=a^2+c^2/b^2+d^2 ta có :
ac/bd=bk*dk/b*d=k*k=k^2 (1)
a^2+c^2/b^2+d^2=bk^2+dk^2/b^2+d^2=k^2*(b^2+d^2)/b^2+d^2=k^2 (2)
Từ (1) và (2) suy ra ac/bd=a^2+c^2/b^2+d^2 (=k^2)
Suy ra đpcm
Có \(\frac{a}{b}=\frac{b}{c}\Leftrightarrow\frac{a}{c}=\frac{b}{d}\)
Đặt \(\frac{a}{c}=\frac{b}{d}=k\Rightarrow a=c.k;b=d.k\)
\(\Rightarrow a^2=c^2.k^2;b^2=d^2.k^2\)
Khi đó \(\frac{a^2+c^2}{b^2+d^2}=\frac{c^2.k^2+c^2}{d^2.k^2+d^2}=\frac{c^2.\left(k^2+1\right)}{d^2.\left(k^2+1\right)}=\frac{c^2}{d^2}=\frac{a^2}{b^2}\)
I dont know ! I come from English but I live in VietNam
mik thấy bài này bình thường mà
