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Lời giải:
Bạn tham khảo cách làm tương tự tại đây:
https://hoc24.vn/cau-hoi/cho-dfracab-2017ccdfracbc-2017aadfracca-2017bbvoi-a-b-c-ne0-tinhp-left1dfracabrightleft1dfracb.161494910584
Kết quả $P=8$ hoặc $P=-1$
E xin lỗi, e ko nhận câu trả lời này vì có chứa link tới các web khác
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\(\dfrac{b+c-5}{a}=\dfrac{a+c+2}{b}=\dfrac{a+b+3}{c}=\dfrac{2a+2b+2c}{a+b+c}=2\\ \Rightarrow\left\{{}\begin{matrix}b+c-5=2a\\a+c+2=2b\\a+b+3=2c\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b+c=a+5\\a+b+c=b-2\\a+b+c=c-3\end{matrix}\right.\)
Lại có \(\dfrac{1}{a+b+c}=2\Rightarrow a+b+c=\dfrac{1}{2}\Rightarrow\left\{{}\begin{matrix}a+5=\dfrac{1}{2}\\b-2=\dfrac{1}{2}\\c-3=\dfrac{1}{2}\end{matrix}\right.\)
Từ đó tự giải ra
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1: Nhân chéo
Bài 2:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
\(\Rightarrow\left(\dfrac{a}{b}\right)^3=\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}\)
\(\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)
\(\Rightarrowđpcm\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a+b+c}{a+b-c}=\dfrac{a-b+c}{a-b-c}\)
\(=\dfrac{a+b+c-a+b-c}{a+b-c-a+b+c}\)
\(=\dfrac{\left(a-a\right)+\left(b+b\right)+\left(c-c\right)}{\left(a-a\right)+\left(b+b\right)+\left(c-c\right)}\)
\(=\dfrac{2b}{2b}=1\)
\(\Rightarrow a+b+c=a+b-c\)
\(\Rightarrow c=-c\)
\(\Rightarrow c+c=0\)
\(\Rightarrow2c=0\Rightarrow c=0\)
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a.b.c}{b.c.d}=\dfrac{a}{d}\left(1\right)\)
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\Rightarrow\left(\dfrac{a}{b}\right)^3=\left(\dfrac{b}{c}\right)^3=\left(\dfrac{c}{d}\right)^3\)
\(=\left(\dfrac{a+b+c}{b+c+d}\right)^3\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) ta có:
\(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`VT = (b-c)/((a-b)(a-c)) + (c-a)/((b-c)(b-a)) +(a-b)/((c-a)(c-b)) = 2/(a-b) + 2/(b-c) + 2/(c-a)`
`=-((a-b-a+c)/((a-b)(a-c))+(b-c-b+a)/((b-c)(b-a))+(c-a-c+b)/((c-a)(c-b)))`
`=-((a-b)/((a-b)(a-c))-(a-c)/((a-b)(a-c))+(b-c)/((b-c)(b-a))-(b-a)/((b-c)(b-a))+(c-a)/((c-a)(c-b))-(c-b)/((c-a)(c-b)))`
`= 1/(c-a)+1/(a-b)+1/(a-b)+1/(b-c)+1/(b-c)+1/(c-a)`
`=2/(a-b)+2/(b-c)+2/(c-a)=VP(đpcm)`
![](https://rs.olm.vn/images/avt/0.png?1311)
Theo đề bài thì:
\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\)
\(=\dfrac{a+b-c+b+c-a+c+a-b}{c+a+b}\)
\(=\dfrac{\left(a+b+b+c+c+a\right)-a-b-c}{c+a+b}\)
\(=\dfrac{a+b+c}{c+a+b}=1\)
Nên: \(\left\{{}\begin{matrix}a+b-c=c\\b+c-a=a\\c+a-b=b\end{matrix}\right.\)
Mà
\(P=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{c}{b}\right)\left(1+\dfrac{a}{c}\right)\)
\(P=\left(\dfrac{a}{a}+\dfrac{b}{a}\right)\left(\dfrac{b}{b}+\dfrac{c}{b}\right)\left(\dfrac{c}{c}+\dfrac{a}{c}\right)\)
\(P=\left(\dfrac{a+b}{a}\right)\left(\dfrac{b+c}{b}\right)\left(\dfrac{c+a}{c}\right)\)
\(P=\left(\dfrac{b+c-a+c+a-b}{a}\right)\left(\dfrac{c+a-b+a+b-c}{b}\right)\left(\dfrac{a+b-c+b+c-a}{c}\right)\)
\(P=\dfrac{2c}{a}.\dfrac{2a}{b}.\dfrac{2b}{c}=\dfrac{8ab}{abc}=8\)
Vậy \(P=8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b)\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}\)
Ta có:
\(\dfrac{a+b}{c}=\dfrac{b+c}{a}\) và \(\dfrac{b+c}{a}=\dfrac{c+a}{b}\)
\(\Rightarrow1+\dfrac{a+b}{c}=1+\dfrac{b+c}{a}\)và \(1+\dfrac{b+c}{a}=1 +\dfrac{c+a}{b}\)
\(\Rightarrow\dfrac{c}{c}+\dfrac{a+b}{c}=\dfrac{a}{a}+\dfrac{b+c}{a}\)và \(\dfrac{a}{a}+\dfrac{b+c}{a}=\dfrac{b}{b}+\dfrac{c+a}{b}\)
\(\Rightarrow\dfrac{a+b+c}{c}=\dfrac{a+b+c}{a}\)và \(\dfrac{a+b+c}{a}=\dfrac{a+b+c}{b}\)
\(\Rightarrow\dfrac{a+b+c}{c}-\dfrac{a+b+c}{a}=0\) \(\Rightarrow\left(a+b+c\right)\cdot\left(\dfrac{1}{c}-\dfrac{1}{a}\right)=0\)
và \(\dfrac{a+b+c}{a}-\dfrac{a+b+c}{b}=0\)
\(\Rightarrow\left(a+b+c\right)\cdot\left(\dfrac{1}{a}-\dfrac{1}{b}\right)=0\)
+) Vì a,b,c đôi một khác 0
\(\Rightarrow a+b+c=0\)
\(\rightarrow a+b=\left(-c\right)\)
\(\rightarrow a+c=\left(-b\right)\)
\(\rightarrow b+c=\left(-a\right)\)
+) Ta có:
\(M=\left(1+\dfrac{a}{b}\right)\cdot\left(1+\dfrac{b}{c}\right)\cdot\left(1+\dfrac{c}{a}\right)\)
\(=\left(\dfrac{a+b}{b}\right)\cdot\left(\dfrac{b+c}{a}\right)\cdot\left(\dfrac{c+a}{c}\right)\)
\(=\dfrac{-c}{b}\cdot\dfrac{-a}{c}\cdot\dfrac{-b}{a}\)
\(=\left(-1\right)\)
Ta có:
\(\dfrac{1}{c}=\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\Leftrightarrow\dfrac{1}{c}.2=\dfrac{1}{a}+\dfrac{1}{b}\)
\(\Leftrightarrow\dfrac{2}{c}=\dfrac{a+b}{ab}\Leftrightarrow2ab=\left(a+b\right)c\)
\(\Leftrightarrow ab+ab=ac+bc\Leftrightarrow ab-bc=ac-ab\)
\(\Leftrightarrow b\left(a-c\right)=a\left(c-b\right)\Leftrightarrow\dfrac{a}{b}=\dfrac{a-c}{c-b}\)
thak