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Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=kb\\c=kd\end{matrix}\right.\) . Sau đó thế vào biểu thức tính rồi suy ra đpcm
Ví dụ bài đầu tiên: Thế a = kb; c=kd vào biểu thức,ta có:
\(\dfrac{a}{a+b}=\dfrac{kb}{kb+b}=\dfrac{kb}{b\left(k+1\right)}=\dfrac{k}{k+1}\) (1)
\(\dfrac{c}{c+d}=\dfrac{kd}{kd+d}=\dfrac{kd}{d\left(k+1\right)}=\dfrac{k}{k+1}\) (2)
Từ (1) và (2) ,ta có đpcm: \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
Các bài sau làm tương tự:Thế a=kb ; c=kd vào biểu thức rồi tính từng vế . Sau đó so sánh hai vế. Thấy hai vế = nhau => đpcm
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) ⇒ a = b.k ; c = d.k
\(\dfrac{3a-2c}{5a+4c}=\dfrac{3.b.k-2.d.k}{5.b.k+4.d.k}=\dfrac{k\left(3.b-2.d\right)}{k\left(5b+4d\right)}=\dfrac{3b-2d}{5b+4d}\)
\(\dfrac{3b-2d}{5b+4d}=\dfrac{3b-2d}{5b+4d}\Rightarrow\dfrac{3a-2c}{5a+4c}=\dfrac{3b-2d}{5b+4d}\)
Đặt:
\(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\dfrac{a+2b}{b}=\dfrac{bk+2b}{b}=\dfrac{b\left(k+2\right)}{b}=k+2\)
\(\dfrac{c+2d}{d}=\dfrac{dk+2d}{d}=\dfrac{d\left(k+2\right)}{d}=k+2\)
Vậy \(\dfrac{a+2b}{b}=\dfrac{c+2d}{d}\Rightarrowđpcm\)
\(\dfrac{a+b}{a-b}=\dfrac{bk+b}{bk-b}=\dfrac{b\left(k+1\right)}{b\left(k-1\right)}=\dfrac{k+1}{k-1}\)
\(\dfrac{c+d}{c-d}=\dfrac{dk+d}{dk-d}=\dfrac{d\left(k+1\right)}{d\left(k-1\right)}=\dfrac{k+1}{k-1}\)
Vậy \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\Rightarrowđpcm\)
a) ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ad=bc\Leftrightarrow ad+2bd=bc+2bd\)
\(\Leftrightarrow d\left(a+2b\right)=b\left(c+2d\right)\Leftrightarrow\dfrac{a+2b}{b}=\dfrac{c+2d}{d}\left(đpcm\right)\)
b) ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ad=bc\Leftrightarrow2ad=2bc\Leftrightarrow ad+ad=bc+bc\)
\(\Leftrightarrow ad-bc=bc-ad\Leftrightarrow ac+ad-bc-bd=ac+bc-ad-bd\)
\(\Leftrightarrow a\left(c+d\right)-b\left(c+d\right)=c\left(a+b\right)-d\left(a+b\right)\)
\(\Leftrightarrow\left(a-b\right)\left(c+d\right)=\left(c-d\right)\left(a+b\right)\Leftrightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\left(đpcm\right)\)
a, ta có :
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{2b}{2d}\)
áp dụng tính chất dă y tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{2b}{2d}=\dfrac{a+2b}{c+2d}=\dfrac{2a-b}{2c-d}\)
\(\Rightarrow\dfrac{a+2b}{c+2d}=\dfrac{2a-b}{2c-d}\Rightarrow\dfrac{a+2b}{2a-b}=\dfrac{c+2d}{2c-d}\) (ĐPCM)
b, ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{3c}{3d}\)
áp dụng tính chất dă tỉ số bằng nhau ta có :
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{3c}{3d}=\dfrac{a+3c}{b+3d}=\dfrac{a-c}{b-d}\)
\(\Rightarrow\dfrac{a+3c}{b+3d}=\dfrac{a-c}{b-d}\)
\(\Rightarrow\left(a+3c\right)\left(b-d\right)=\left(b+3d\right)\left(a-c\right)\) (ĐPCM)
\(\dfrac{a+b}{c+d}=\dfrac{a-2b}{c-2d}\Rightarrow\left(a+b\right)\left(c-2d\right)=\left(c+d\right)\left(a-2b\right)\\ ac+bc-2ad-2bd=ac+ad-2bc-2bd\\ bc-2ad=ad-2bc\\ 3bc=3ad\\ bc=ad\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}\left(đpcm\right)\)
\(\dfrac{a+b}{c+d}=\dfrac{a-2b}{c-2d}\)
\(\Leftrightarrow\left(a+b\right)\left(c-2d\right)=\left(c+d\right)\left(a-2b\right)\)
\(\Leftrightarrow ac-2ad+bc-2bd=ac-2bc+ad-2bd\)
\(\Leftrightarrow2ad+ad=2bc+bc\)
\(\Leftrightarrow3ad=3bc\)
\(\Leftrightarrow ad=bc\rightarrowđpcm\)
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:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}=\dfrac{3a+2c}{3b+2d}\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{3a+2c}{3b+2d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}=\frac{2a+b+c+d-a-2b-c-d}{a-b}=1\)
\(\Rightarrow\left\{\begin{matrix}-a=b+c+d\\-b=a+c+d\\-c=b+c+d\\-d=a+b+c\end{matrix}\right.\)\(\Rightarrow a=b=c=d\)
\(M=\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{a+d}{b+c}\)
\(\Rightarrow M=\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}\)
\(\Rightarrow M=1+1+1+1\)
\(\Rightarrow M=4\)
Vậy \(M=4\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=b.k;b=d.k\)
Thay :
(1) : \(\dfrac{3a+2b}{3a-2b}=\dfrac{3bk+2b}{3bk-2b}=\dfrac{b.\left(3.k+2\right)}{b.\left(3.k-2\right)}=\dfrac{3.k+2}{3.k-2}\)
(2) : \(\dfrac{3c+2d}{3c-2d}=\dfrac{3dk+2d}{3dk-2d}=\dfrac{d.\left(3.k+2\right)}{d.\left(3.k-2\right)}=\dfrac{3.k+2}{3.k-2}\)
Do đó : \(\dfrac{3a+2b}{3a-2b}=\dfrac{3c+2d}{3c-2d}\)
a) ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{b}=\dfrac{4c}{4d}=\dfrac{a+4c}{b+4d}\left(đpcm\right)\)
b;c;d tương tự hết
b: a/b=c/d
nên 3a/3b=2c/2d
=>a/b=c/d=(3a+2c)/(3b+2d)
c: a/c=b/d nên a/c=2b/2d=(a-2b)/(c-2d)
d: a/c=b/d
nên 5a/5c=2b/2d
=>a/c=b/d=(5a-2b)/(5c-2d)