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Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\left(1\right)\)
Thay (1) vào từng vế của đề bài:
\(VT=\dfrac{a^2+ac}{c^2-ac}=\dfrac{bk\left(bk+dk\right)}{dk\left(dk-bk\right)}=\dfrac{b\left(b+d\right)}{d\left(d-b\right)}\)
Vế phải đặt thừa số chung sẽ ra VT => đpcm.
Đặt :
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Leftrightarrow a=bk;c=dk\)
\(VT=\dfrac{ac}{bd}=\dfrac{bkdk}{bd}=\dfrac{bdk^2}{bd}=k^2\left(1\right)\)
\(VP=\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2.k^2+d^2.k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) =>\(a=bk,c=dk\)
=> \(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k.k=k^2\left(1\right)\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}\)
=\(\dfrac{k^2.\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(2\right)\)
Từ (1)và(2)=>\(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
Chúc Bạn Học Tốt
a)Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{a}{b}=\frac{c}{d}\) =>\(\frac{a}{c}=\frac{b}{d}\)
=>\(\frac{ac}{bd}=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\)
=>\(\frac{ac}{bd}=\frac{a^2+b^2}{c^2+d^2}\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}=\frac{a^2-c^2}{b^2-d^2}\)
\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}=\frac{a-c}{b-d}\Rightarrow\frac{\left(a+c\right)^2}{\left(b+d\right)^2}=\frac{\left(a-c\right)^2}{\left(b-d\right)^2}\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}=\frac{a^2-b^2}{c^2-d^2}\Rightarrow\frac{a^2+b^2}{a^2-b^2}=\frac{c^2+d^2}{c^2-d^2}\)
\(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{a^2+b^2}{c^2+d^2}\)
a) Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\dfrac{a}{b}=\dfrac{3a}{3b}\) ; \(\dfrac{c}{d}=\dfrac{2c}{2d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}=\dfrac{3a+2c}{3b+2d}\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{3a+2c}{3b+2d}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}\)=k (1)
=> a=bk ,c=dk
a.Có \(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=\dfrac{k\left(b+d\right)}{b+d}=k\left(2\right)\)
Từ (1) và (2)=>\(\dfrac{a+c}{b+d}=\dfrac{a}{b}\left(=k\right)\)
b. Có \(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\)
=>\(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\left(=k^2\right)\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Suy ra: \(VT=\dfrac{bk^2\left(b+d\right)}{dk^2\left(d-b\right)}=\dfrac{b\left(b+d\right)}{d\left(d-b\right)}\)
\(VP=\dfrac{b^2+bd}{d^2-bd}=\dfrac{b\left(b+d\right)}{d\left(d-b\right)}\)
\(\Rightarrow VT=VP\rightarrowđpcm.\)
a; Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2}{d^2}\)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2}{d^2}\)
Do đó: \(\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}\)
b: \(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=k^2\)
Do đó: \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
c: \(\dfrac{7a^2-3ab}{11a^2-8b^2}=\dfrac{7b^2k^2-3\cdot bk\cdot b}{11b^2k^2-8b^2}=\dfrac{b^2\left(7k^2-3k\right)}{b^2\left(11k^2-8\right)}=\dfrac{7k^2-3k}{11k^2-8}\)
\(\dfrac{7c^2-3cd}{11c^2-8d^2}=\dfrac{7d^2k^2-3kd^2}{11d^2k^2-8d^2}=\dfrac{7k^2-3k}{11k^2-8}\)
Do đó: \(\dfrac{7a^2-3ab}{11a^2-8b^2}=\dfrac{7c^2-3cd}{11c^2-8d^2}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
Ta có: \(\dfrac{a^2+ac}{c^2-ac}=\dfrac{b^2k^2+bk\cdot dk}{d^2k^2-bk\cdot dk}=\dfrac{bk^2\cdot\left(b+d\right)}{dk^2\cdot\left(d-b\right)}=\dfrac{b\left(b+d\right)}{d\left(d-b\right)}\left(1\right)\)
\(\dfrac{b^2+bd}{d^2-bd}=\dfrac{b\left(b+d\right)}{d\left(d-b\right)}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\dfrac{a^2+ac}{c^2-ac}=\dfrac{b^2+bd}{d^2-bd}\)