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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 \(\dfrac{a}{b}=\dfrac{c}{d}thì\)\(\dfrac{a+b}{c-d}=\dfrac{c+d}{c-d}\) ( \(a-b\ne0,c-d\ne0\))
a/ Đặt :
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có :
\(VT=\dfrac{a-b}{a+b}=\dfrac{bk-b}{bk+b}=\dfrac{b\left(k-1\right)}{b\left(k+1\right)}=\dfrac{k-1}{k+1}\)\(\left(1\right)\)
\(VP=\dfrac{c-d}{c+d}=\dfrac{dk-d}{dk+d}=\dfrac{d\left(k-1\right)}{d\left(k+1\right)}=\dfrac{k-1}{k+1}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
b/ Đặt :
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(VT=\dfrac{2a+5b}{3a-4b}=\dfrac{2bk+5b}{3bk-4b}=\dfrac{b\left(2k+5\right)}{b\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\left(1\right)\)
\(VP=\dfrac{2c+5d}{3c-4d}=\dfrac{2dk+5d}{3dk-4d}=\dfrac{d\left(2k+5\right)}{d\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
a) Từ \(\dfrac{a}{b}=\dfrac{c}{d}\) \(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)
Từ \(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\) \(\Rightarrow\dfrac{c-d}{c+d}=\dfrac{a-b}{a+b}\)
b) Từ \(\dfrac{a}{b}=\dfrac{c}{d}\) \(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{3a}{3c}=\dfrac{4b}{4d}=\dfrac{5b}{5d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{2a}{2c}=\dfrac{3a}{3c}=\dfrac{4b}{4d}=\dfrac{5b}{5d}=\dfrac{2a+5b}{2c+5d}=\dfrac{3a-4b}{3c-4d}\)
Từ \(\dfrac{2a+5b}{2c+5d}=\dfrac{3a-4b}{3c-4d}\) \(\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
a, Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) ( k # 0 )
\(\Rightarrow\) \(a=b.k\)
\(c=d.k\)
Ta có: \(\dfrac{a+b}{b}=\dfrac{b.k+b}{b}=\dfrac{b.\left(k+1\right)}{b}=k+1\) (1)
\(\dfrac{c+d}{d}=\dfrac{d.k+d}{d}=\dfrac{d.\left(k+1\right)}{d}=k+1\) (2)
Từ (1) và (2) \(\Rightarrow\) \(\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
b,
, Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) ( k # 0 )
\(\Rightarrow\) \(a=b.k\)
\(c=d.k\)
Ta có: \(\dfrac{a}{a+b}=\dfrac{b.k}{b.k+b}=\dfrac{b.k}{b.\left(k+1\right)}=\dfrac{k}{k+1}\) (1)
\(\dfrac{c}{c+d}=\dfrac{d.k}{d.k+d}=\dfrac{d.k}{d.\left(k+1\right)}=\dfrac{k}{k+1}\) (2)
Từ (1) và (2) \(\Rightarrow\) \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
Theo đề bài, ta có:
\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{a+b}{c+d}\)=\(\left(\dfrac{a+b}{c+d}\right)^2\)(*)
=> \(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{a^2}{c^2}\)=\(\dfrac{b^2}{d^2}\)=\(\dfrac{a^2+b^2}{c^2+d^2}\)(**)
Từ (*) và (**) suy ra:
\(\left(\dfrac{a+b}{c+d}\right)^2\)=\(\dfrac{a^2+b^2}{c^2+d^2}\)(đpcm)
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\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\Rightarrow\dfrac{a.b}{c.d}=\dfrac{a+b}{c+d}.\dfrac{a+b}{c+b}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
Ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\Rightarrow\dfrac{a.b}{c.d}=\dfrac{a+b}{c+d}.\dfrac{a+b}{c+d}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
\(\left(a-2c\right)\left(b+2d\right)=\left(b-2d\right)\left(a+2c\right)\)
\(\Leftrightarrow ab+2ad-2bc-4cd=ab+2bc-2ad-4cd\)
\(\Leftrightarrow2ad+2ad=2bc+2bc\Leftrightarrow4ab=4bc\)
\(\Leftrightarrow ad=bc\Rightarrow\dfrac{a}{b}=\dfrac{c}{d},\left(a,b,c,d\ne0\right)\)
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{a^2-b^2}{ab}=\dfrac{b^2k^2-b^2}{bk\cdot b}=\dfrac{b^2\left(k^2-1\right)}{b^2k}=\dfrac{k^2-1}{k}\)
\(\dfrac{c^2-d^2}{cd}=\dfrac{d^2k^2-d^2}{dk\cdot d}=\dfrac{d^2\left(k^2-1\right)}{d^2\cdot k}=\dfrac{k^2-1}{k}\)
Do đó: \(\dfrac{a^2-b^2}{ab}=\dfrac{c^2-d^2}{cd}\)
b: \(\dfrac{\left(a+b\right)^2}{a^2+b^2}=\dfrac{\left(bk+b\right)^2}{b^2k^2+b^2}=\dfrac{b^2\cdot\left(k+1\right)^2}{b^2\left(k^2+1\right)}=\dfrac{\left(k+1\right)^2}{k^2+1}\)
\(\dfrac{\left(c+d\right)^2}{c^2+d^2}=\dfrac{\left(dk+d\right)^2}{d^2k^2+d^2}=\dfrac{\left(k+1\right)^2}{k^2+1}\)
Do đó: \(\dfrac{\left(a+b\right)^2}{a^2+b^2}=\dfrac{\left(c+d\right)^2}{c^2+d^2}\)
Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)
\(\Rightarrow\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)
\(\Rightarrow\dfrac{a+c}{a-c}=\dfrac{b+d}{b-d}\left(đpcm\right)\)
Vậy...
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)
Từ đó suy ra : \(\dfrac{a+c}{a-c}=\dfrac{b+d}{b-d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+b}{c+d}\\ \Leftrightarrow\left(\dfrac{a}{b}\right)^{2011}=\left(\dfrac{c}{d}\right)^{2011}=\left(\dfrac{a+b}{c+d}\right)^{2011}\\ \Leftrightarrow\dfrac{a^{2011}}{b^{2011}}=\dfrac{c^{2011}}{d^{2011}}=\dfrac{\left(a+b\right)^{2011}}{\left(c+d\right)^{2011}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a^{2011}}{b^{2011}}=\dfrac{c^{2011}}{d^{2011}}=\dfrac{\left(a+b\right)^{2011}}{\left(c+d\right)^{2011}}=\dfrac{a^{2011}+c^{2011}}{b^{2011}+d^{2011}}\)
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\\ \Rightarrow\dfrac{a^{2011}}{b^{2011}}=\dfrac{c^{2011}}{d^{2011}}=\left(\dfrac{a+c}{b+d}\right)^{2011}\\ \dfrac{a^{2011}}{b^{2011}}=\dfrac{c^{2011}}{d^{2011}}=\dfrac{a^{2011}+c^{2011}}{b^{2011}+d^{2011}}\\ \Rightarrow\dfrac{a^{2011}+c^{2011}}{b^{2011}+d^{2011}}=\left(\dfrac{a+c}{b+d}\right)^{2011}\)
Nhanh nha
a: Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{k}{k-1}\)
\(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{k}{k-1}\)
Do đó: \(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)