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Đặt \(\frac{a+b}{c+d}=\frac{a-2b}{c-2d}=k\)
=> \(\begin{cases}a+b=k.\left(c+d\right)=k.c+k.d\\a-2b-k.\left(c-2d\right)=k.c-k.2d\end{cases}\)
=> (a + b) - (a - 2b) = (k.c + k.d) - (k.c - k.2d)
=> a + b - a + 2b - k.c + k.d - k.c + k.2d
=> 3b = 3kd
=> b = kd
Mà a + b = k.c + k.d
=> a = k.c
=> \(\frac{a}{b}=\frac{k.c}{k.d}=\frac{c}{d}\left(đpcm\right)\)
Cách 2:
Ta có: \(\frac{a+b}{c+d}=\frac{a-2b}{c-2d}\)
=> (a + b).(c - 2d) = (c + d).(a - 2b)
=> (a + b).c - (a + b).2d = (c + d).a - (c + d).2b
=> ac + bc - 2ad - 2bd = ac + ad - 2bc - 2bd
=> ac + bc - 2ad - 2bd - ac - ad + 2bc + 2bd = 0
=> 3bc - 3ad = 0
=> 3bc = 3ad
=> bc = ad
=> \(\frac{a}{b}=\frac{c}{d}\left(đpcm\right)\)
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Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{b+c+d+a}=1\)
=> a = b = c = d
=> \(D=\frac{2a-a}{2a-a}+\frac{2a-a}{2a-a}+\frac{2a-a}{2a-a}+\frac{2a-a}{2a-a}\)
D = 1 + 1 + 1 + 1 = 4
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\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{4a}{4c}=\frac{2b}{2d}=\frac{4a-2b}{4c-2d}=\frac{5a}{5c}=\frac{2b}{2d}=\frac{5a+2b}{5c+2d}\)
Suy ra \(\frac{4a-2b}{4c-2d}=\frac{5a+2b}{5c+2d}\)Suy ra điều phải chứng minh: \(\frac{4a-2b}{5a+2b}=\frac{4c-2d}{5c+2d}\)
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Ta có : 2a + b + c+ d / a - 1 = a + 2b + c + d / b - 1 = a + b + 2c + d / c - 1 = a + b + c +2d / d - 1
=> a + b + c + d / a = a + b + c + d / b = a + b + c + d / c = a + b + c + d / d
Xét 2 trường hợp :
TH1: a + b + c + d = 0
=> a + b = - ( c + d ) ; b + c = - ( a + d ) ; c + d = - ( a + b )
Khi đó M = ( -1 ) . 4 = -4
TH2 : a + b + c + d khác 0
=> a = b = c = d
Khi đó M = 1 . 4 = 4
Vậy M = 4 hoặc M = - 4
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a) \(\hept{\begin{cases}\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{5a}{5c}=\frac{3b}{3d}=\frac{5a-3b}{5c-3d}\\\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{3a}{3c}=\frac{2b}{2d}=\frac{3a+2b}{3c+2d}\end{cases}}\)
\(\Rightarrow\frac{5a-3b}{5c-3d}=\frac{3a+2b}{3c+2d}\)
\(\Rightarrow\frac{5a-3b}{3a+2b}=\frac{5c-3d}{3c+2d}\)
b) Chứng minh tương tự
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\(\frac{a+b}{c+d}=\frac{a-2b}{c-2d}\\ \Leftrightarrow\left(a+b\right)\left(c-2d\right)=\left(c+d\right)\left(a-2b\right)\)
\(\Leftrightarrow ac-2ad+bc-2bd=ac+ad-2bc-2bd\)
\(\Leftrightarrow3bc=3ad\)
\(\Leftrightarrow bc=ad\)
\(\Leftrightarrow\frac{a}{b}=\frac{c}{d}\left(đpcm\right)\)
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a,Cách 1: \(\frac{a+b}{b}=\frac{c+d}{d}\)
=> (a+b)d = b(c+d)
=> ad + bd = bc + bd
=> ad = bc
=> \(\frac{a}{b}=\frac{c}{d}\)
Cách 2:
\(\frac{a+b}{b}=\frac{c+d}{d}\Rightarrow\frac{a}{b}+1=\frac{c}{d}+1\Rightarrow\frac{a}{b}=\frac{c}{d}\)
b,\(\frac{a}{a-2b}=\frac{c}{c-2d}\Rightarrow a\left(c-2d\right)=c\left(a-2b\right)\Rightarrow ac-2ad=ac-2bc\Rightarrow-2ad=-2bc\Rightarrow ad=bc\Rightarrow\frac{a}{b}=\frac{c}{d}\)
(a+b).(c-2d)=(a-2b).(c+d)
ac-2ad+bc-2bd=ac+ad-2bc-2bd
ac-ac+bc+2bc=2ad+ad+2bd-2bd
3bc=3ad
bc=ad
=> a/b=c/d