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Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) (a ≠ c; b ≠ d)
CMR: \(\dfrac{a+c}{a-c}=\dfrac{b+d}{b-d}\)
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Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=> \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có:
\(\dfrac{a+c}{a-c}=\dfrac{bk+dk}{bk=dk}=\dfrac{\left(b+d\right)k}{\left(b-d\right)k}=\dfrac{b+d}{b-d}\)
Mà \(\dfrac{b+d}{b-d}=\dfrac{b+d}{b-d}\)
=> \(\dfrac{a+c}{a-c}=\dfrac{b+d}{b-d}\) (đpcm)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\) (*)
a) Từ (*) ta có:
\(\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\) (1)
\(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\) (2)
Từ (1) và (2) suy ra \(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)
b) Từ (*) ta có:
\(\dfrac{a}{b}=\dfrac{bk}{b}=k\) (3)
\(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=\dfrac{k\left(b+d\right)}{b+d}=k\) (4)
Từ (3) và (4) suy ra \(\dfrac{a}{b}=\dfrac{a+c}{b+d}\)
c) Từ (*) ta có:
\(\dfrac{a}{3a+b}=\dfrac{bk}{3bk+b}=\dfrac{bk}{b\left(3k+1\right)}=\dfrac{k}{3k+1}\) (5)
\(\dfrac{c}{3c+d}=\dfrac{dk}{3dk+d}=\dfrac{dk}{d\left(3k+1\right)}=\dfrac{k}{3k+1}\) (6)
Từ (5) và (6) suy ra \(\dfrac{a}{3a+b}=\dfrac{c}{3c+d}\)
d) Từ (*) ta có:
\(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\) (7)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2.k^2+d^2.k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\) (8)
Từ (7) và (8) suy ra \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
e) Từ (*) ta có:
\(\dfrac{ab}{cd}=\dfrac{bk.b}{dk.d}=\dfrac{b^2}{d^2}=\dfrac{b}{d}\) (9)
\(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{b^2.k^2-b^2}{d^2.k^2-d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\dfrac{b}{d}\) (10)
Từ (9) và (10) suy ra \(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)
f) Từ (*) ta có:
\(\dfrac{ab}{cd}=\dfrac{bk.b}{dk.d}=\dfrac{b^2}{d^2}=\dfrac{b}{d}\) (11)
\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\dfrac{\left[b\left(k-1\right)\right]^2}{\left[d\left(k-1\right)\right]^2}=\dfrac{b}{d}\) (12)
Từ (11) và (12) suy ra \(\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
\(a.\)\(\frac{a}{b}=\frac{c}{d}\)
\(\Rightarrow\)\(\frac{a}{b}+1=\frac{c}{d}+1\)
\(\Rightarrow\)\(\frac{a+b}{b}=\frac{c+d}{d}\left(đpcm\right)\)
\(b.\)\(\frac{a}{b}=\frac{c}{d}\)
\(\Rightarrow\)\(\frac{a}{b}-1=\frac{c}{d}-1_{ }\)
\(\Rightarrow\)\(\frac{a-b}{b}=\frac{c-d}{d}\)\(\left(đpcm\right)\)
\(c.\)\(\frac{a}{b}=\frac{c}{d}\)
\(\Rightarrow\)\(\frac{b}{a}=\frac{d}{c}\)
\(\Rightarrow\)\(\frac{b}{a}+1=\frac{d}{c}+1\)
\(\Rightarrow\)\(\frac{b+a}{a}=\frac{d+c}{c}\)hay \(\frac{a+b}{a}=\frac{c+d}{d}\left(đpcm\right)\)
\(d.\)Tương tự \(c\) nhé bn. Chúc bn học tốt!
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
=> \(\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
a, Ta có:\(\frac{a-b}{a+b}=\frac{bk-b}{bk+b}=\frac{b.\left(k-1\right)}{b.\left(k+1\right)}=\frac{k-1}{k+1}\left(1\right)\)
Lại có \(\frac{c-d}{c+d}=\frac{dk-d}{dk+d}=\frac{d.\left(k-1\right)}{d.\left(k+1\right)}=\frac{k-1}{k+1}\left(2\right)\)
Từ (1) và (2) => ĐPCM
b, Ta có \(\frac{a.b}{c.d}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\left(1\right)\)
Lại có \(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\frac{b^2.\left(k+1\right)^2}{d^2.\left(k+1\right)^2}=\frac{b^2}{d^2}\left(2\right)\)
Từ (1) và (2) => ĐPCM
a) (a-b)+(c-d)=(a+c)-(b+d)
=>a-b+c-d=a+c-b+d
=>a+c-b+d=a+c-b+d
b) (a-b)-(c-d)=(a+d)-(b+c)
=>a-b-c+d=a+d-b-c
=>a+d-b-b-c=a+d-b-c
a,(a-b) + (c - d)= (a+c) - (b+d)
=>a - b + c - d= a+c - b - d (phá ngoặc)
=> - a + c-b- d=a+c - b - d
=>(a-b) + (c - d)= (a+c) - (b+d) (đpcm)