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Ta có :\(\frac{a}{b}=\frac{b}{c}\)
=> \(\frac{a^2}{b^2}=\frac{b^2}{c^2}=\frac{a^2+b^2}{b^2+c^2}\)
=> \(\frac{a^2}{b^2}=\frac{a^2+b^2}{b^2+c^2}\)
=> \(\frac{a}{b}.\frac{a}{b}=\frac{a^2+b^2}{b^2+c^2}\)
=> \(\frac{a}{b}.\frac{b}{c}=\frac{a^2+b^2}{b^2+c^2}\)
=> \(\frac{a}{c}=\frac{a^2+b^2}{b^2+c^2}\left(\text{đpcm}\right)\)
ta có: \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(=\frac{1}{c}\times2=\frac{1}{a}+\frac{1}{b}\)
\(=\frac{2}{c}=\frac{1}{a}+\frac{1}{b}\)
\(=\frac{2}{c}=\frac{b+a}{ab}\)
= \(c\left(b+a\right)=ab\times2\)
= cb +ca = ab+ab
= ab - cb = ac-ab
\(=b\left(a-c\right)=a\left(c-b\right)\)
= \(\frac{a}{b}=\frac{a-c}{c-b}\)
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\frac{1}{c}=\frac{1}{2a}+\frac{1}{2b}\)
\(\frac{1}{c}=\frac{a+b}{2ab}\)
\(2ab=c\left(a+b\right)\)
\(ab+ab=ac+bc\)
\(ab-bc=ac-ab\)
\(b\left(a-c\right)=a\left(c-b\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\left(đpcm\right)\)
Có \(a^2+ab+\frac{b^2}{3}=c^2+\frac{b^2}{3}+a^2+ac+c^2\left(=25\right)\)
\(\Rightarrow a^2+ab+\frac{b^2}{3}=2c^2+\frac{b^2}{3}+a^2+ac\\ \Rightarrow ab=2c^2+ac\\ \Rightarrow ab+ac=2c^2+2ac\\ \Rightarrow a\left(b+c\right)=2c\left(a+c\right)\\ \Rightarrow\frac{2c}{a}=\frac{b+c}{a+c}\)
Từ \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\Rightarrow\frac{1}{c}=\frac{1}{2}\left(\frac{a+b}{ab}\right)\)
\(\Rightarrow\frac{1}{c}=\frac{a+b}{2ab}\)
\(\Rightarrow2ab=c.\left(a+b\right)\)
\(\Rightarrow ab+ab=ac+bc\)
\(\Rightarrow ab-bc=ac-ab\)
\(\Rightarrow b.\left(a-c\right)=a.\left(c-b\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\)
Ta có\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
<=> cd(a2 + b2) = ab(c2 + d2)
<=> a2cd + b2cd - abc2 - abd2 = 0
<=> (a2cd - abc2) + (b2cd - abd2) = 0
<=> ac(ad - bc) + bd(bc - ad) = 0
<=> ac(ad - bc) - bd(ad - bc) = 0
<=> (ac - bd)(ad - bc) = 0
<=> \(\orbr{\begin{cases}ac-bd=0\\ad-bc=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}ac=bd\\ad=bc\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{a}{d}=\frac{b}{c}\\\frac{a}{b}=\frac{c}{d}\end{cases}}\left(\text{đpcm}\right)\)
\(\frac{a}{b}=\frac{b}{c}\)\(\Rightarrow\left(\frac{a}{b}\right)^2=\left(\frac{b}{c}\right)^2=\frac{a^2}{b^2}=\frac{b^2}{c^2}=\frac{a^2+b^2}{b^2+c^2}\)
mà \(\left(\frac{a}{b}\right)^2=\frac{a}{b}.\frac{a}{b}=\frac{a}{b}.\frac{b}{c}=\frac{a}{c}\)
\(\Rightarrow\frac{a^2+b^2}{b^2+c^2}=\frac{a}{c}\)
Từ a/b = b/c
Suy ra : bb = ac
b2 = ac
vậy : a2 + b2 / b2+ c2 = a2 + ac / ac + c2 = a(a+c) / c(a+c) = a/c
Vậy : Ta có được cái cần chứng minh :))
Lớp mình vừa kiểm tra 15' bài này xong .
cảm ơn bạn