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\(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\Rightarrow\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có :
\(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}=\dfrac{a+b+a-b}{c+d+c-d}=\dfrac{2a}{2c}=\dfrac{a}{c}\) \(\left(1\right)\)
\(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}=\dfrac{a+b-a+b}{c+d-c+d}=\dfrac{2b}{2d}=\dfrac{b}{d}\) \(\left(2\right)\)
Từ \(\left(1\right),\left(2\right)\), ta có :
\(\dfrac{a}{c}=\dfrac{b}{d}\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}\)
Ta có: \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)
\(\rightarrow\left(a+b\right)\left(c-d\right)=\left(a-b\right)\left(c+d\right)\)
\(\rightarrow ac-ad+bc-bd=ac+ad-bc-bd\)
\(\rightarrow-ad+bc=ad-bc\)
\(\rightarrow bc+bc=ad+ad\)
\(\rightarrow2bc=2ad\)
\(\rightarrow bc=ad\)
\(\rightarrow\dfrac{a}{b}=\dfrac{c}{d}\left(đpcm\right)\)
Chúc bạn học tốt!
Ta có: \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\Leftrightarrow\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}.\)
Áp dụng tính chất dãy tỉ số bằng nhau có:
\(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}=\dfrac{a+b+a-b}{c+d+c-d}=\dfrac{a+a+b-b}{c+c+d-d}=\dfrac{2a}{2c}=\dfrac{a}{c}_{\left(1\right)}.\)
\(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}=\dfrac{a+b-a+b}{c+d-c+d}=\dfrac{a-a+b+b}{c-c+d+d}=\dfrac{2b}{2d}=\dfrac{b}{d}_{\left(2\right)}.\)
Từ \(_{\left(1\right)+\left(2\right)}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\Leftrightarrow\dfrac{a}{b}=\dfrac{c}{d}\) (t/c tỉ lệ thức).
\(\Rightarrowđpcm.\)
a=b*k
c=d*k
thì b*k+b/b*k-b=b*(k+1)/b*(k-1)=k+1/k-1
thì d*k+d/d*k-d=d*(k+1)/d*(k-1)=k+1/k-1
nen suy ra a+b/a-b=c+d/c-d
\(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\Leftrightarrow\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}=\dfrac{a+b+a-b}{c+d+c-d}=\dfrac{2a}{2c}=\dfrac{a}{c}\)(1)
\(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}=\dfrac{a+b-a+b}{c+d-c+d}=\dfrac{2b}{2d}=\dfrac{b}{d}\)(2)
Từ (1) và (2) ta có: \(\dfrac{a}{c}=\dfrac{b}{d}\Leftrightarrow\dfrac{a}{b}=\dfrac{c}{d}\left(đpcm\right)\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow a=bk,c=dk\)
Ta có: \(\dfrac{a+b}{a-b}=\dfrac{bk+b}{bk-b}=\dfrac{b\left(k+1\right)}{b\left(k-1\right)}=\dfrac{k+1}{k-1}\) (1)
\(\dfrac{c+d}{c-d}=\dfrac{dk+d}{dk-d}=\dfrac{d\left(k+1\right)}{d\left(k-1\right)}=\dfrac{k+1}{k-1}\) (2)
Từ (1) và (2) \(\Rightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)
Ta có: \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\).Theo tính chất của dãy tỉ số bằng nhau:
\(\Rightarrow\)\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=\(\dfrac{a+b}{c+d}\)=\(\dfrac{a-b}{c-d}\)
Vì \(\dfrac{a+b}{c+d}\)=\(\dfrac{a-b}{c-d}\)\(\Leftrightarrow\)\(\dfrac{a+b}{a-b}\)=\(\dfrac{c+d}{c-d}\)
Vậy \(\dfrac{a+b}{a-b}\)=\(\dfrac{c+d}{c-d}\)
Nếu bạn muốn làm cách cơ bản thì hãy làm theo mình.Còn nếu bạn học toán nâng cao thì làm theo cách bạn Linh hay hơn.Chúc bạn học tốt
C) đúng. Vì
\(\dfrac{a}{b}=\dfrac{c}{d}\)
=>\(\dfrac{a}{c}=\dfrac{b}{d}\)
=>\(\dfrac{c}{a}=\dfrac{d}{b}\)
b,
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{b}{d}=\dfrac{a}{c}=\dfrac{b+a}{d+c}\\ \Rightarrow\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
c,
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
ta có: \(a=bk;c=dk\)
\(\Rightarrow\dfrac{2a+3c}{2b+3d}=\dfrac{2bk+3dk}{2b+3d}=\dfrac{k^2.\left(2b+3d\right)}{2b+3d}=k^2\\ \Rightarrow\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k^2.\left(2b-3d\right)}{2b-3d}=k^2\\ \Rightarrow\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
d,
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
ta có:\(a=bk;c=dk\)
\(\Rightarrow\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\\ \Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{k^2.\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\\ \Rightarrow\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
e,
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
Ta có:\(a=bk;c=dk\)
\(\Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{k^2.\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\\ \Rightarrow\dfrac{a^2-c^2}{b^2-d^2}=\dfrac{k^2.\left(b-d\right)^2}{\left(b-d\right)^2}=k^2\\ \Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{a^2-c^2}{b^2-d^2}\)
f,
(để hôm sau lm nha, mỏi tay quá)
a, \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=> \(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{a+b}{c+d}\)=\(\dfrac{a-b}{c-d}\)(1)
\(\dfrac{a+b}{c+d}\)=\(\dfrac{a-b}{c-d}\)=> \(\dfrac{a+b}{a-b}\)=\(\dfrac{c+d}{c-d}\)
Còn các phần còn lại làm giống thế
Ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\) suy ra \(\dfrac{a}{c}=\dfrac{b}{d}\)
Theo tính chất dãy tỉ số bằng nhau ta có
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)
Suy ra: \(\dfrac{a+b}{a-c}=\dfrac{c+d}{c-d}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow a=bk\) và \(c=dk\)
Nên \(\dfrac{a+b}{c-d}=\dfrac{bk+b}{dk-d}=\dfrac{b\left(k+1\right)}{d\left(k-1\right)}=\dfrac{k+1}{k-1}\)
\(\dfrac{c+d}{c-d}=\dfrac{dk+d}{dk-d}=\dfrac{d\left(k+1\right)}{d\left(k-1\right)}=\dfrac{k+1}{k-1}\)
\(\Rightarrow\dfrac{a+b}{c-d}=\dfrac{c+d}{c-d}\) (với \(a-b\ne0,c-d\ne0\))
Vậy \(\dfrac{a}{b}=\dfrac{c}{d}thì\)\(\dfrac{a+b}{c-d}=\dfrac{c+d}{c-d}\) ( \(a-b\ne0,c-d\ne0\))
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Suy ra: \(\dfrac{a}{b}=\dfrac{bk}{b}=k\left(1\right)\)
\(Và:\) \(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=\dfrac{k\left(b+d\right)}{b+d}=k\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra \(\dfrac{a}{b}=\dfrac{a+c}{b+d}\)
Vậy \(\dfrac{a}{b}=\dfrac{a+c}{b+d}\) \(\left(ĐPCM\right)\)
Ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\)
Áp dụng t/c' dãy tỉ số bằng nhau , ta có :
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\)
Vậy \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\left(đpcm\right)\)
Lần sau khi hỏi nhớ tìm xem có câu nào tương tự không nhé.
Ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)
\(\Rightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)
Vậy nếu \(\dfrac{a}{b}=\dfrac{c}{d}\) ( a khác b, c khác d ) thì \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)
Vì : \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\) ⇒ \(\dfrac{a}{c}\) = \(\dfrac{b}{d}\)
Hay: a+b/c+d
Và: a-b/c-d
cùng = a/c=b/d
vậy : \(\dfrac{a+b}{c+d}\) = \(\dfrac{a-b}{c-d}\) (đpcm)