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Ta có:
\(\frac{x}{\left(a-b\right)\left(a-c\right)}+\frac{x}{\left(b-a\right)\left(b-c\right)}+\frac{x}{\left(c-a\right)\left(c-b\right)}=2\)
\(\Leftrightarrow x\left(\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\right)=2\)
\(\Leftrightarrow0x=2\)
Vậy PT vô nghiệm
\(\frac{b+c+d}{\left(b-a\right)\left(c-a\right)\left(d-a\right)\left(x-a\right)}=\frac{\left(a+b+c+d-x\right)+\left(x-a\right)}{\left(b-a\right)\left(c-a\right)\left(d-a\right)\left(x-a\right)}\)\(=\frac{\left(a+b+c+d-x\right)}{\left(b-a\right)\left(c-a\right)\left(d-a\right)\left(x-a\right)}+\frac{1}{\left(b-a\right)\left(c-a\right)\left(d-a\right)}\)
Áp dụng hoán vị vòng \(b\rightarrow c\rightarrow d\rightarrow a\rightarrow b\) vào VT , ta được :
\(\left(a+b+c+d-x\right)\)[\(\frac{1}{\left(a-b\right)\left(a-c\right)\left(a-d\right)\left(a-x\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)\left(b-d\right)\left(b-x\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)\left(c-d\right)\left(c-x\right)}\)\(+\frac{1}{\left(d-a\right)\left(d-b\right)\left(d-c\right)\left(d-x\right)}\).
Quy đồng mẫu thức và tính toán biểu thức trong [ ] ta được :
\(\frac{-1}{\left(x-a\right)\left(x-b\right)\left(x-c\right)\left(x-d\right)}\)
Vậy ...............
Lời giải:
\(\text{VT}=\frac{b-c}{b+c}+\frac{c-a}{c+a}+\frac{a-b}{a+b}=\left(\frac{b}{b+c}-\frac{b}{a+b}\right)+\left(\frac{c}{c+a}-\frac{c}{c+b}\right)+\left(\frac{a}{a+b}-\frac{a}{a+c}\right)\)
\(=\frac{b(a-c)}{(b+c)(a+b)}+\frac{c(b-a)}{(c+a)(c+b)}+\frac{a(c-b)}{(a+b)(a+c)}\)
\(=\frac{b(a-c)(a+c)+c(b-a)(b+a)+a(c-b)(c+b)}{(a+b)(b+c)(c+a)}=\frac{b(a^2-c^2)+c(b^2-a^2)+a(c^2-b^2)}{(a+b)(b+c)(c+a)}\)
\(=\frac{(a^2b+b^2c+c^2a)-(ab^2+bc^2+ca^2)}{(a+b)(b+c)(c+a)}(*)\)
Và:
\(\text{VP}=\frac{(b^2-c^2)(b+c)+(c^2-a^2)(c+a)+(a^2-b^2)(a+b)}{(a+b)(b+c)(c+a)}\)
\(=\frac{(a^2b+b^2c+c^2a)-(ab^2+bc^2+ca^2)}{(a+b)(b+c)(c+a)}(**)\)
Từ $(*); (**)\Rightarrow $ đpcm
Ap dụng hằng đẳng thức.
\(A=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-c\right)\left(b-a\right)}+\frac{b^2}{\left(a-c\right)\left(b-a\right)}+\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)
\(=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-c\right)\left(c-a\right)}+\frac{c^2}{\left(c-a\right)\left(b-c\right)}\)
\(=\frac{\left(a+b\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(b+c\right)\left(b-c\right)}{\left(b-c\right)\left(c-a\right)}\)
\(=\frac{a+b}{a-c}+\frac{b+c}{c-a}=\frac{a+b}{a-c}-\frac{b+c}{a-c}=1\left(đpcm\right)\)
Ta có
\(B=\frac{\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(x-c\right)\left(x-a\right)}{\left(b-c\right)\left(b-a\right)}+\frac{\left(x-a\right)\left(x-b\right)}{\left(c-a\right)\left(c-b\right)}\)
\(=\frac{\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(x-c\right)\left(x-a\right)}{\left(b-c\right)\left(c-a\right)}+\frac{\left(x-c\right)\left(x-a\right)}{\left(a-c\right)\left(b-a\right)}+\frac{\left(x-a\right)\left(x-b\right)}{\left(c-a\right)\left(c-b\right)}\)
\(=\frac{\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(x-c\right)\left(x-a\right)}{\left(b-c\right)\left(c-a\right)}-\frac{\left(x-c\right)\left(x-a\right)}{\left(a-c\right)\left(a-b\right)}+\frac{\left(x-a\right)\left(x-b\right)}{\left(a-c\right)\left(c-b\right)}\)
\(=\frac{\left(x-b\right)\left(x-c\right)-\left(x-c\right)\left(x-a\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(x-c\right)\left(x-a\right)-\left(x-b\right)\left(x-a\right)}{\left(b-c\right)\left(c-a\right)}\)
\(=\frac{\left(x-c\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(x-a\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)}\).
\(=\frac{x-c}{a-c}-\frac{x-a}{a-c}=\frac{x-c-x+a}{a-c}\)
\(=1\)