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\(A=\left(\frac{a+b}{b}\right).\left(\frac{b+c}{c}\right).\left(\frac{a+c}{a}\right)\)
Vì \(a+b+c=0\)
\(\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\a+c=-b\end{cases}}\)
\(\Rightarrow A=\frac{-c}{b}.\left(\frac{-a}{c}\right).\left(\frac{-b}{a}\right)\)
\(\Rightarrow A=-1\)
\(\frac{b+c-a}{a}+\frac{2a}{a}=\frac{a+c-b}{b}+\frac{2b}{b}=\frac{a+b-c}{c}+\frac{2c}{c}\)
\(\frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}\)
=> a=b=c
A=(1+1)(1+1)(1+1) = 2.2.2 =8
Ta có: a+b+c=0a+b+c=0
\Rightarrow b+a=-c⇒b+a=−c
\Rightarrow c+b=-a⇒c+b=−a
\Rightarrow a+c=-b⇒a+c=−b
Ta có: A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)A=(1+
b
a
)(1+
c
b
)(1+
a
c
)
\Rightarrow A=\left(\frac{b+a}{b}\right)\left(\frac{c+b}{c}\right)\left(\frac{a+c}{a}\right)⇒A=(
b
b+a
)(
c
c+b
)(
a
a+c
)
\Rightarrow A=\left(\frac{-c}{b}\right)\left(\frac{-a}{c}\right)\left(\frac{-b}{a}\right)⇒A=(
b
−c
)(
c
−a
)(
a
−b
)
\Rightarrow A=-1⇒A=−1
Đặt: \(\frac{a}{2013}=\frac{b}{2012}=\frac{c}{2011}=k\Rightarrow\hept{\begin{cases}a=2013k\\b=2012k\\c=2011k\end{cases}}\)
\(P=\frac{\left(a-c\right)^4}{\left(a-b\right)^2\left(b-c\right)^2}=\frac{\left(2013k-2011k\right)^4}{\left(2013k-2012k\right)^2\left(2012k-2011k\right)^2}=\frac{16k^4}{k^4}=16\)
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: \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(=\frac{1}{c}\times2=\frac{1}{a}+\frac{1}{b}\)
\(=\frac{2}{c}=\frac{1}{a}+\frac{1}{b}\)
\(=\frac{2}{c}=\frac{b+a}{ab}\)
= \(c\left(b+a\right)=ab\times2\)
= cb +ca = ab+ab
= ab - cb = ac-ab
\(=b\left(a-c\right)=a\left(c-b\right)\)
= \(\frac{a}{b}=\frac{a-c}{c-b}\)
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\frac{1}{c}=\frac{1}{2a}+\frac{1}{2b}\)
\(\frac{1}{c}=\frac{a+b}{2ab}\)
\(2ab=c\left(a+b\right)\)
\(ab+ab=ac+bc\)
\(ab-bc=ac-ab\)
\(b\left(a-c\right)=a\left(c-b\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\left(đpcm\right)\)
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{a+b+c}=1\Rightarrow a=b=c\)
\(\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)