Ta có

\(B=\frac{\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(x-c\right)\left(x-a\right)}{\left(b-c\right)\left(b-a\right)}+\frac{\left(x-a\right)\left(x-b\right)}{\left(c-a\right)\left(c-b\right)}\)

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

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

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

\(=\frac{\left(x-c\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(x-a\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)}\).

\(=\frac{x-c}{a-c}-\frac{x-a}{a-c}=\frac{x-c-x+a}{a-c}\)

\(=1\)

mk không hiểu

đề đúng mà bn

đề đúng thì giải giùm ik bạn ơi

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

\(\frac{(b-c)(1+a)^2}{x+a^2}+\frac{(c-a)(1+b)^2}{x+b^2}+\frac{(a-b) (1+c)^2}{x+c^2}=0\)

\(\Leftrightarrow \sum (b-c)(1+a)^2(x+b^2)(x+c^2)=0\)

\(\Leftrightarrow (a-b)(b-c)(c-a)(x^2+(-2a-ca-ba-cb-2c-2b-1)x+ba+2acb+cb+ca)=0\)

\(\Leftrightarrow x^2+(-2a-ca-ba-cb-2c-2b-1)x+ba+2acb+cb+ca=0\)

Xét phương trình  \(x^2+(-2a-ca-ba-cb-2c-2b-1)x+ba+2acb+cb+ca=0\)

Ta thấy \(\Delta=(2a+2b+2c+ab+bc+ca-1)^2+8(a+b+c-abc)\)

Nếu \(\Delta <0\) thì phương trình vô nghiệm

Nếu \(\Delta =0\) thì phương trình có nghiệm kép

Nếu \(\Delta >0\) thì phương trình có hai nghiệm 

Quy đồng lên, lấy MTC là (a-b)(b-c)(a-c)

\(a\left(y+z\right)=b\left(z+x\right)=c\left(x+y\right)\Leftrightarrow\frac{y+z}{\frac{1}{a}}=\frac{z+x}{\frac{1}{b}}=\frac{x+y}{\frac{1}{c}}=\)

\(=\frac{y+z-\left(z+x\right)}{\frac{1}{a}-\frac{1}{b}}=\frac{z+x-\left(x+y\right)}{\frac{1}{b}-\frac{1}{c}}=\frac{x+y-\left(y+z\right)}{\frac{1}{c}-\frac{1}{a}}=\frac{y-x}{\frac{b-a}{ab}}=\frac{z-y}{\frac{c-b}{bc}}=\frac{x-z}{\frac{a-c}{ac}}\)

Chia các vế của 3 tỷ lệ thức cuối cho abc ta có:

\(\frac{y-x}{\frac{b-a}{ab}\cdot abc}=\frac{z-y}{\frac{c-b}{bc}\cdot abc}=\frac{x-z}{\frac{a-c}{ac}\cdot abc}=\frac{y-x}{c\left(b-a\right)}=\frac{z-y}{a\left(c-b\right)}=\frac{x-z}{b\left(a-c\right)}\)

Hay: \(\frac{x-y}{c\left(a-b\right)}=\frac{y-z}{a\left(b-c\right)}=\frac{z-x}{b\left(c-a\right)}\)đpcm