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\(\Leftrightarrow\dfrac{10a+b}{10b+c}=\dfrac{b}{c}\)
=>10ac+bc=10b^2+cb
=>10ac=10b^2
=>ac=b^2
=>a/b=b/c=k
=>a=bk; b=ck
=>a=ck*k=k^2*c
\(\dfrac{a}{c}=\dfrac{k^2c}{c}=k^2\)
\(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{b^2k^2+b^2}{c^2k^2+c^2}=\dfrac{b^2}{c^2}=\dfrac{c^2k^2}{c^2}=k^2\)
=>ĐPCM
Ta có : \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=> a = b.k ; c = d.k
Ta lại có : \(\dfrac{a-b}{a+b}=\dfrac{b.k-b}{b.k+b}=\dfrac{b.\left(k-1\right)}{b.\left(k+1\right)}=\dfrac{k-1}{k+1}\)
\(\dfrac{c-d}{c+d}=\dfrac{d.k-d}{d.k+d}=\dfrac{d.\left(k-1\right)}{d.\left(k+1\right)}=\dfrac{k-1}{k+1}\)
Vì \(\dfrac{a-b}{a+b}=\dfrac{k-1}{k+1}\) ; \(\dfrac{c-d}{c+d}=\dfrac{k-1}{k+1}\) nên \(\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)
Vậy \(\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)
\(\Leftrightarrow\dfrac{10a+b}{a+b}=\dfrac{10b+c}{b+c}\)
=>10ac+bc=10b^2+cb
=>10ac=10b^2
=>ac=b^2
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 \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\left(đpcm\right)\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}\)=k (1)
=> a=bk ,c=dk
a.Có \(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=\dfrac{k\left(b+d\right)}{b+d}=k\left(2\right)\)
Từ (1) và (2)=>\(\dfrac{a+c}{b+d}=\dfrac{a}{b}\left(=k\right)\)
b. Có \(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\)
=>\(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\left(=k^2\right)\)
Giải:
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)
\(\Rightarrow\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\Rightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\left(đpcm\right)\)
Vậy...
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(\left\{{}\begin{matrix}a=b.k\\c=d.k\end{matrix}\right.\) (1)
Thay (1) vào:
\(\dfrac{a+b}{a-b}=\dfrac{b.k+b}{b.k-b}=\dfrac{b.\left(k+1\right)}{b.\left(k-1\right)}=\dfrac{k+1}{k-1}\) (2)
\(\dfrac{c+d}{c-d}=\dfrac{d.k+d}{d.k-d}=\dfrac{d.\left(k+1\right)}{d.\left(k-1\right)}=\dfrac{k+1}{k-1}\) (3)
Từ (2) và (3) =>\(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}=\dfrac{k+1}{k-1}\)
Ta có :
\(\dfrac{1}{c}=\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{c}:\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{c}\cdot\dfrac{2}{1}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{2}{c}\)
\(\Rightarrow\dfrac{b}{ab}+\dfrac{a}{ab}=\dfrac{2}{c}\)
\(\Rightarrow\dfrac{a+b}{ab}=\dfrac{2}{c}\)
\(\Rightarrow2ab=\left(a+b\right)c\)
\(\Rightarrow ab+ab=ac+bc\)
\(\Rightarrow ac-ab=ab-bc\)
\(\Rightarrow a\left(c-b\right)=b\left(a-c\right)\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{a-c}{c-b}\)
Vậy \(\dfrac{a}{b}=\dfrac{a-c}{c-b}\)
Áp dụng tính chất dãy tỉ số bằng nhau ; ta có :
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\frac{1}{c}:\frac{1}{2}=\frac{1}{a}+\frac{1}{b}\)
\(\frac{2}{c}=\frac{a+b}{ab}\)
\(\Rightarrow2ab=ac+bc\)
\(\Rightarrow ac-ab=ab-bc\)
\(\Rightarrow a.\left(c-b\right)=b.\left(a-c\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\)( đpcm )
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a+b+b}{a+b-c}=\dfrac{a-b+c}{a-b-c}=\dfrac{a+b+c-\left(a-b+c\right)}{a+b-c-\left(a-b-c\right)}=\dfrac{a+b+c-a+b-c}{a+b-c-a+b+c}=\dfrac{2b}{2b}=1\)
\(\Rightarrow a+b+c=a+b-c\)
\(\Rightarrow c=-c\)
\(\Rightarrow c-\left(-c\right)=0\)
\(\Rightarrow c+c=0\)
\(\Rightarrow2c=0\)
\(\Rightarrow c=0\)
\(\Rightarrow\) Đpcm.