Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
4.a
\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\\ \Leftrightarrow\left(3x-y\right).4=3\left(x+y\right)\\ \Rightarrow12x-4y=3x+3y\\ \Rightarrow12x-3x=4y+3y\\ \Rightarrow9x=7y\\ \Rightarrow\dfrac{x}{y}=\dfrac{7}{9}\)
Với: \(a+b+c=0\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\a+c=-b\end{matrix}\right.\)
Khi đó: \(P=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{-abc}{abc}=-1\)
Với \(a+b+c\ne0\) áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a+b-c}{c}=\dfrac{a-b+c}{b}=\dfrac{-a+b+c}{a}=\dfrac{a+b-c+a-b+c-a+b+c}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\a+c=2b\\b+c=2a\end{matrix}\right.\)
Khi đó: \(P=\dfrac{8abc}{abc}=8\)
TH1: Với \(a+b+c\ne0\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a+b-c}{c}=\dfrac{a-b+c}{b}=\dfrac{-a+b+c}{a}=\dfrac{a+b-c+a-b+c-a+b+c}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)=> a + b = 2c ; a + c = 2b và b + c = 2a
\(\Rightarrow P=\dfrac{2c.2a.2b}{abc}=\dfrac{8abc}{abc}=8\)
TH2: Với a + b + c = 0
=> a + b = -c ; a + c = -b và c + b = -a
\(\Rightarrow P=\dfrac{\left(-c\right)\left(-a\right)\left(-b\right)}{abc}=-1\)
Vậy P = 8 hoặc P = -1
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 \(P=8\)
\(\dfrac{a+b-c}{c}=\dfrac{a+c-b}{b}=\dfrac{b+c-a}{a}\)
\(\Rightarrow\dfrac{a+b-c}{c}+2=\dfrac{a+c-b}{b}+2=\dfrac{b+c-a}{a}+2\)
\(\Rightarrow\dfrac{a+b-c}{c}+\dfrac{2c}{2}=\dfrac{a+c-b}{b}+\dfrac{2b}{b}=\dfrac{b+c-a}{a}+\dfrac{2a}{a}\)
\(\Rightarrow\dfrac{a+b+c}{c}=\dfrac{a+c+b}{b}=\dfrac{b+c+a}{a}\)
\(\Rightarrow a=b=c\) Thay vào A ta được :
\(A=\dfrac{\left(a+a\right)\left(a+a\right)\left(a+a\right)}{a.a.a}=\dfrac{2a.2a.2a}{a^3}=\dfrac{8.a^3}{a^3}=8\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có :
\(\dfrac{a+b-c}{c}=\dfrac{a+c-b}{b}=\dfrac{b+c-a}{a}=\dfrac{a+b-c+a+c-b+b+c-a}{c+b+a}=\dfrac{2a+2b+2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{a+b-c}{c}=2\\\dfrac{a+c-b}{b}=2\\\dfrac{b+c-a}{a}=2\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}a+b-c=2c\\a+c-b=2b\\b+c-a=2a\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}a+b=2c+c\\a+c=2b+b\\b+c=2a+a\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}a+b=3c\\a+c=3b\\b+c=3a\end{matrix}\right.\)
Ta có :
\(A=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\\ \Rightarrow A=\dfrac{3c\cdot3a\cdot3b}{abc}=\dfrac{27abc}{abc}=27\)
Ta có:\(\dfrac{b+c-3a}{a}=\dfrac{a+c-3b}{b}=\dfrac{a+c-3c}{c}\)
\(\Rightarrow\dfrac{b+c}{a}=\dfrac{a+c}{b}=\dfrac{a+b}{c}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{b+c}{a}=\dfrac{a+c}{b}=\dfrac{a+b}{c}=\dfrac{2\left(a+b+c\right)}{a+b+c}\)
Nếu a+b+c=0
\(\Rightarrow\)\(\left\{{}\begin{matrix}b+c=a\\a+c=b\\a+b=c\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{matrix}\right.\)
\(\Rightarrow M=-1\)
Nếu a+b+c\(\ne\)0
\(\Rightarrow\dfrac{b+c}{a}=\dfrac{a+c}{b}=\dfrac{a+b}{c}=2\)
\(\Rightarrow\left\{{}\begin{matrix}b+c=2a\\a+c=2b\\a+b=2c\end{matrix}\right.\)
\(\Rightarrow M=8\)
\(P=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{c}{b}\right)=\dfrac{\left(a+b\right)\left(b+c\right)\left(a+c\right)}{abc}\)
Với \(a+b+c=0\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\a+c=-b\end{matrix}\right.\)
Khi đó \(P=\dfrac{-abc}{abc}=-1\)
Với \(a+b+c\ne0\) ,áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=\dfrac{a+b-c+b+c-a+c+a-b}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\a+c=2b\end{matrix}\right.\)
Khi đó \(P=\dfrac{8abc}{abc}=8\)
Có: \(\dfrac{a+b-c}{c}=\dfrac{a+c-b}{b}=\dfrac{b+c-a}{a}\)
Áp dụng tính chất của dãy tỉ số bằng nhau , ta được:
\(\dfrac{a+b-c}{c}=\dfrac{a+c-b}{b}=\dfrac{b+c-a}{a}=\dfrac{a+b-c+a+c-b+b+c-a}{c+b+a}\)
\(=\dfrac{a+b+c}{a+b+c}\)
Xét: a + b + c = 0 \(\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)(1)
Thay (1) vào A, ta có:
\(A=\dfrac{-c.\left(-a\right).\left(-b\right)}{abc}=-1\)
Xét a + b + c ≠ 0:
\(\Rightarrow\dfrac{a+b-c}{c}=\dfrac{a+c-b}{b}=\dfrac{b+c-a}{a}=1\)
\(\Rightarrow\dfrac{a+b}{c}-1=\dfrac{a+c}{b}-1=\dfrac{b+c}{a}-1=1\)
\(\Rightarrow\dfrac{a+b}{c}=\dfrac{a+c}{b}=\dfrac{b+c}{a}=2\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\a+c=2b\\b+c=2a\end{matrix}\right.\)(2)
Thay (2) vào A, ta có:
\(A=\dfrac{2c.2a.2b}{abc}=8\)
Vậy...