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\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)
\(\Rightarrow\frac{a+b-c}{c}+1=\frac{b+c-a}{a}+1=\frac{c+a-b}{b}+1\)
\(\Rightarrow\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}\)
+)Nếu a+b+c=0\(\Rightarrow a+b=-c;b+c=-a;c+a=-b\)
\(\Rightarrow B=\frac{a+b}{a}.\frac{c+a}{c}.\frac{b+c}{b}=\frac{-c}{a}.\frac{-b}{c}.\frac{-a}{b}=\frac{-\left(abc\right)}{abc}=-1\)
Nếu \(a+b+ c\ne0\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\Rightarrow a+b=2c\)
\(b+ c=2a\)
\(c+a=2b\)
\(\Rightarrow B=\frac{2c}{a}.\frac{2b}{c}.\frac{2a}{b}=2.2.2=8\)
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a) Ta có: \(A\left(x\right)=ax^2+bx+c\)
Thay \(A\left(-1\right)\) ta được:
\(A\left(-1\right)=a\left(-1\right)^2+b\left(-1\right)+c=a+c-b\)
\(=b-8-b=-8\)
b) \(\left\{{}\begin{matrix}A\left(0\right)=4\\A\left(1\right)=9\\A\left(2\right)=14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=4\\a+b+c=9\\4a+2b+c=14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=4\\a+b=5\\4a+2b=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=4\\a+b=5\\2a+b=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=4\\a=0\\b=5\end{matrix}\right.\)
c)
Ta có: \(\left\{{}\begin{matrix}A\left(2\right)=4a+2b+c\\A\left(-1\right)=a-b+c\end{matrix}\right.\)
\(\Leftrightarrow A\left(2\right)+A\left(-1\right)=5a+b+2c=0\)
\(\Leftrightarrow A\left(2\right)=-A\left(-1\right)\)
\(\Leftrightarrow A\left(2\right)\times A\left(-1\right)=-\left[A\left(2\right)\right]^2\le0\)
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\(\frac{1}{c}=\frac{1}{2}.\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\frac{1}{c}:\frac{1}{2}=\frac{1}{a}+\frac{1}{b}\)
\(\frac{2}{c}=\frac{a}{ab}+\frac{b}{ab}\)
\(\frac{2}{c}=\frac{a+b}{ab}\)
\(2ab=\left(a+b\right).c\)
\(ab+ab=ac+bc\)
\(ab-bc=ac-ab\)
\(b.\left(a-c\right)=a.\left(c-b\right)\)
\(\frac{a}{b}=\frac{a-c}{c-b}\)
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\(\frac{1}{c}=\frac{1}{2}.\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\frac{1}{c}:\frac{1}{2}=\frac{b}{ab}+\frac{a}{ab}\)
\(\frac{2}{c}=\frac{a+b}{ab}\)
\(\Rightarrow2ab=a.\left(b+c\right)\)
\(ab+ab=ac+cb\)
\(ab-cb=ac-ab\)
\(b.\left(a-c\right)=a.\left(c-b\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\)
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Câu hỏi của Nguyễn Thị Hồng Nhung - Toán lớp 7 - Học toán với OnlineMath
\(X=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}\cdot\frac{c+b}{c}\cdot\frac{c+a}{a}\)
Mà \(a+b+c=0\Rightarrow\hept{\begin{cases}a+b=-c\\a+c=-b\\c+b=-a\end{cases}}\)
\(\Rightarrow X=\frac{\left(-a\right)\cdot\left(-b\right)\cdot\left(-c\right)}{abc}=-1\)
nên ta đc X là 1 số nguyên