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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 \(\dfrac{a}{b}=\dfrac{a-c}{c-b}\)
\(\Leftrightarrow\dfrac{10a+b}{10b+c}=\dfrac{b}{c}\)
=>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
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\frac{1}{c}:\frac{1}{2}=\frac{1}{a}+\frac{1}{b}\)
\(\frac{2}{c}=\frac{a+b}{ab}\)
\(\Rightarrow2ab=ac+bc\)
\(\Rightarrow ac-ab=ab-bc\)
\(\Rightarrow a.\left(c-b\right)=b.\left(a-c\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\)( đpcm )
Câu 2:
Theo đề, ta có: \(\dfrac{10a+b}{a+b}=\dfrac{10b+c}{b+c}\)
=>10ab+10ac+b^2+bc=10ab+10b^2+ac+bc
=>9ac-9b^2=0
=>ac-b^2=0
=>ac=b^2
=>a/b=b/c
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
b+c+d/a=c+d+a/b=d+a+b/c=a+b+c/d=3(a+b+c+d)/a+b+c+d=3
suy ra k=3
taco:\(\dfrac{b+c+d}{a}=\dfrac{c+d+a}{b}+\dfrac{d+a+b}{c}=\dfrac{a+b+c}{d}=k\)=>\(\dfrac{b+c+d}{a}+1=\dfrac{c+d+a}{b}+1=\dfrac{a+b+d}{c}+1=\dfrac{a+b+c}{d}+1=k+1\)=>\(\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}=k+1=\dfrac{a+b+c+d+a+b+c+d+a+b+c+d}{a+b+c+d}=\dfrac{4.\left(a+b+c+d\right)}{a+b+c+d}=4\)
=>k+1=4
=>k=3
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 \(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\)
=>\(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\left(=k^2\right)\)
4/ \(\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{5}=\dfrac{z}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=\dfrac{y}{20}\\\dfrac{y}{20}=\dfrac{z}{24}\end{matrix}\right.\Leftrightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}=k\) (đặt k)
Suy ra \(x=15k;y=20k;z=24k\)
Thay vào,ta có:
\(M=\dfrac{2.15k+3.20k+4.24k}{3.15k+4.20k+5.24k}=\dfrac{186k}{245k}=\dfrac{186}{245}\)
Đặt :
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(VT=\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\left(1\right)\)
\(VP=\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
\(\dfrac{b+c}{bc}=\dfrac{2}{a}\) <=>\(ab+ac=2bc\)
<=>\(bc-ab=ac-bc\)
<=>\(b\left(c-a\right)=c\left(a-b\right)\)
<=>\(\dfrac{b}{c}=\dfrac{a-b}{c-a}\)(ĐPCM)
Chúc Bạn Học Tốt,đạt nhiều thành tích trong học tập :)
Từ \(\dfrac{b}{c}=\dfrac{a-b}{c-a}\)\(\Rightarrow b\left(c-a\right)=c\left(a-b\right)\)
\(\Rightarrow bc-ab=ac-bc\)
\(\Rightarrow2bc=ac+ab\)\(\Rightarrow2bc=a\left(b+c\right)\)
\(\Rightarrow\dfrac{b+c}{bc}=\dfrac{2}{a}\) (ĐPCM)