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\(b^2=ac\Rightarrow\frac{b}{c}=\frac{a}{b}=\frac{2010a}{2010b}=\frac{2011b}{2011c}=\frac{2010a+2011b}{2010b+2011c}\)
\(\Rightarrow\frac{b}{c}.\frac{a}{b}=\left(\frac{2010a+2011b}{2010b+2011c}\right).\left(\frac{2010a+2011b}{2010b+2011c}\right)\)
\(\Rightarrow\frac{a}{c}=\frac{\left(2010a+2011b\right)^2}{\left(2010b+2011c\right)^2}\)
7)
\(\left(0,36\right)^8=\left(0,6^2\right)^8=0,6^{16}.\)
8)
a) \(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\)
\(\Rightarrow\left(\frac{3}{5}\right)^n=\left[\left(\frac{3}{5}\right)^2\right]^5\)
\(\Rightarrow\left(\frac{3}{5}\right)^n=\left(\frac{3}{5}\right)^{10}\)
\(\Rightarrow n=10\)
Vậy \(n=10.\)
b) \(\left(-0,25\right)^p=\frac{1}{256}\)
\(\Rightarrow\left(-0,25\right)^p=\left(\frac{1}{4}\right)^4\)
\(\Rightarrow\left(-0,25\right)^p=\left(0,25\right)^4\)
\(\Rightarrow p=4\)
Vậy \(p=4.\)
Chúc bạn học tốt!
a,b,c tỉ lệ với m, m+n, m+2n => \(\frac{a}{m}=\frac{b}{m+n}=\frac{c}{m+2n}=k\)
=> \(a=mk;\)\(b=\left(m+n\right)k=mk+nk\); \(c=\left(m+2n\right)k=mk+2nk\)
Ta có: \(VT=4\left(a-b\right)\left(b-c\right)=4\left(mk-mk-nk\right)\left(mk+nk-mk-2nk\right)\)
\(=4\left(-nk\right)\left(-nk\right)=4n^2k^2\)
\(VP=\left(c-a\right)^2=\left(mk+2nk-mk\right)^2=\left(2nk\right)^2=4n^2k^2\)
suy ra: đpcm
\(\frac{3a+b+c}{a}=\frac{a+3b+c}{b}=\frac{a+b+3c}{c}=\frac{\left(3a+b+c\right)+\left(a+3b+c\right)+\left(a+b+3c\right)}{a+b+c}\)
\(=\frac{5\left(a+b+c\right)}{a+b+c}=5\)
\(\Rightarrow\frac{3a+b+c}{a}=5\Rightarrow3a+b+c=5a\Rightarrow b+c=2a\)
Tương tự ta có : \(a+c=2b;a+b=2c\)
\(\Rightarrow B=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)
\(=\frac{a+b}{b}.\frac{b+c}{c}.\frac{a+c}{a}=\frac{2c}{b}.\frac{2a}{c}.\frac{2b}{a}\)
\(=\frac{8abc}{abc}=8\)
Ta có:
\(\frac{3a+b+c}{a}=\frac{a+3b+c}{b}=\frac{3c+b+a}{c}\)
\(\Rightarrow3+\frac{b+c}{a}=3+\frac{a+c}{b}=3+\frac{b+a}{c}\)
\(\Rightarrow\frac{b+c}{a}=\frac{a+c}{b}=\frac{b+a}{c}=\frac{2\left(a+b+c\right)}{a+b+c}\)
TH1:\(a+b+c=0\)\(\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\a+c=-b\end{matrix}\right.\)
Thay vào B, ta có:
\(B=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=-1\)
TH2:\(a+b+c\ne0\)
\(\Rightarrow\frac{b+c}{a}=\frac{a+c}{b}=\frac{b+a}{c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\Rightarrow\left\{{}\begin{matrix}b+c=2a\\a+c=2b\\b+a=2c\end{matrix}\right.\)
Thay vào B, ta có:
\(B=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=8\)
Vậy \(\left[{}\begin{matrix}B=-1\\B=8\end{matrix}\right.\)
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\(\frac{2a+a+b+c}{a}=\frac{2b+a+b+c}{b}=\frac{2c+a+b+c}{c}\)
\(\Rightarrow2+\frac{a+b+c}{a}=2+\frac{a+b+c}{b}=2+\frac{a+b+c}{c}\)
\(\Rightarrow\frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}\Rightarrow a=b=c\)
\(\Rightarrow B=\left(1+1\right)\left(1+1\right)\left(1+1\right)\)
Phân số nghịch đảo của phân số \(\frac{m}{n}\) là: \(\frac{n}{m}\)