Câu hỏi của Chu Hoàng THủy Tiên - Toán lớp 7 - Học toán với OnlineMath

TH1 : a + b + c ≠ 0

Áp dụng t/c dãy tỉ số bằng nhau ta có

\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{a+b+b+c+a+c}{a+b+c}=2\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\a+c=2b\end{matrix}\right.\)

Khi đó \(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)

\(=\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{a+c}{a}=\dfrac{2c}{b}.\dfrac{2a}{c}.\dfrac{2b}{a}=8\)

TH2 : a + b + c = 0

\(\Rightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)

Khi đó \(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)

\(=\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{a+c}{a}=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}=-1\)

Xét 2 TH sau:

TH1: a+b+c=0

Khi đó:

\(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\\ =\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{c+a}{a}\\ =\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}\\ =-1\)

TH2: a+b+c khác 0

Ta có:

\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)

Suy ra: a+b=2c; b+c=2a; c+a=2b

Do đó:

\(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\\ =\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{c+a}{a}\\ =\dfrac{2c}{b}.\dfrac{2a}{c}.\dfrac{2b}{a}\\ =8\)

Xét 2 TH sau:

TH1: a+b+c=0

Khi đó:

\(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\\ =\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{c+a}{a}\\ =\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}\\ =-1\)

TH2: a+b+c khác 0

Ta có:

\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)

Suy ra: a+b=2c; b+c=2a; c+a=2b

Do đó:

\(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\\ =\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{c+a}{a}\\ =\dfrac{2c}{b}.\dfrac{2a}{c}.\dfrac{2b}{a}\\ =8\)

Đặt x=a-b; y=b-c; z=c-a

=>x+y+z=0

a+b+c=0

=>a+b=-c; b+c=-a; a+c=-b

Đặt \(X = \frac{a-b}{c} + \frac{b-c}{a} + \frac{c-a}{b}\)

\(=\frac{ab(a-b) + bc(b-c) + ca(c-a)}{abc}=\frac{-\left(a-b\right)\left(b-c\right)\left(c-a\right)}{abc}\)

Đặt \(Y = \frac{c}{a-b} + \frac{a}{b-c} + \frac{b}{c-a}\)

\(=\frac{c}{x}+\frac{a}{y}+\frac{b}{z}=\frac{cyz+axz+bxy}{xyz}\)

Đặt T=cyz+axz+bxy

\(=cyz+(-b-c)xz+bxy=cyz-bxz-cxz+bxy\)

\(=cz(y-x)+bx(y-z)=-3abc\)

Do đó: \(Y = \frac{-3abc}{(a-b)(b-c)(c-a)}\)

A=X*Y

\(=\frac{-\left(a-b\right)\cdot\left(b-c\right)\left(c-a\right)}{abc}\cdot\frac{-3abc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

=3

a) Ta có : \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}\Leftrightarrow\frac{a+b}{c}+1=\frac{b+c}{a}+1=\frac{c+a}{b}+1\)

\(\Rightarrow\frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}\)

• TH1: Nếu a + b + c = 0 \(\Rightarrow P=\frac{a+b}{b}.\frac{b+c}{c}.\frac{a+c}{a}=\frac{-c}{b}.\frac{-a}{c}.\frac{-b}{a}=\frac{-\left(abc\right)}{abc}=-1\)
• TH2 : Nếu \(a+b+c\ne0\) \(\Rightarrow a=b=c\)

\(\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

b) Đề bài sai ^^

Dựa vào tính chất dãy tỉ số bằng nhau:

  \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{\left(a+b\right)+\left(b+c\right)+\left(c+a\right)}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)

Suy ra:

   \(a+b=2c;b+c=2a;c+a=2b\)

Từ đẳng thức đầu a + b = 2 c  => a = 2c - b thay vào 2 đẳng thức cuối ta có:

   \(b+c=2\left(2c-b\right)\)  và \(c+\left(2c-b\right)=2b\)

=> b = c => a = c

Vậy a = b = c

Khi đó:

  \(P=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

Ta có:

\(\left\{{}\begin{matrix}a^2+b=b^2+c\\b^2+c=c^2+a\\a^2+b=c^2+a\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a^2-b^2=c-b\\b^2-c^2=a-c\\a^2-c^2=a-b\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)\left(a+b\right)=c-b\\\left(b-c\right)\left(b+c\right)=a-c\\\left(a-c\right)\left(a+c\right)=a-b\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a+b=\dfrac{c-b}{a-b}\\b+c=\dfrac{a-c}{b-c}\\a+c=\dfrac{a-b}{a-c}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b-1=\dfrac{c-a}{a-b}\\b+c-1=\dfrac{a-b}{b-c}\\a+c-1=\dfrac{c-b}{a-c}\end{matrix}\right.\)

\(\Rightarrow T=\left(a+b-1\right)\left(b+c-1\right)\left(a+c-1\right)\)

\(=\dfrac{\left(c-a\right)\left(a-b\right)\left(c-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\dfrac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=1\)

Tham khảo:

https://olm.vn/hoi-dap/detail/264403587120.html