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\(\frac{x-a}{bc}+\frac{x-b}{ac}+\frac{x-c}{ab}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\frac{x-a}{bc}+\frac{x-b}{ac}+\frac{x-c}{ab}=\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\)
\(\frac{ax-a^2+bx-b^2+cx-c^2}{abc}=2\left(\frac{ab+bc+ac}{abc}\right)\)
\(ax-a^2+bx-b^2+cx-c^2=2\left(ab+bc+ac\right)\)
\(x\left(a+b+c\right)-\left(a^2+b^2+c^2\right)=2\left(ab+bc+ac\right)\)
\(x\left(a+b+c\right)=a^2+b^2+c^2+2ab+2bc+2ac\)
\(x=a+b+c\)
a) \(P=\frac{bc}{\left(a-b\right)\left(a-c\right)}+\frac{ac}{\left(b-c\right)\left(b-a\right)}+\frac{ab}{\left(c-a\right)\left(c-b\right)}\)
Đặt \(x=\frac{b}{c-a},y=\frac{c}{a-b},z=\frac{a}{b-c}\) , suy ra : \(P=-xy-yz-xz\)
Lại có : \(\left(x-1\right)\left(y-1\right)\left(z-1\right)=\left(x+1\right)\left(y+1\right)\left(z+1\right)\)
\(\Rightarrow xy+yz+xz=-1\Rightarrow P=1\)
\(Q=\frac{\left[\left(x+\frac{1}{x}\right)^2\right]^3-\left(x^3+\frac{1}{x^3}\right)^2}{\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)}=\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)\)
\(=3x+\frac{3}{x}=3\left(x+\frac{1}{x}\right)\)
a) \(\frac{a+b-x}{c}+\frac{b+c-x}{a}+\frac{c+a-x}{b}+\frac{4x}{a+b+c}=1\)
\(\Leftrightarrow\frac{a+b-x}{c}+1+\frac{b+c-x}{a}+1+\frac{c+a-x}{b}+1+\frac{4x}{a+b+c}-4=0\)
\(\Leftrightarrow\frac{a+b+c-x}{c}+\frac{a+b+c-x}{a}+\frac{a+b+c-x}{b}+\frac{4x-4\left(a+b+c\right)}{a+b+c}=0\)
\(\Leftrightarrow\left(x-a-b-x\right)\left(\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}\right)=0\)
b)đề bài như trên
\(\Leftrightarrow\left(\frac{x-a-b-c}{bc}\right)+\left(\frac{x-b}{ca}-\frac{1}{a}-\frac{1}{c}\right)+\left(\frac{x-c}{ab}-\frac{1}{a}-\frac{1}{b}\right)=0\)
\(\Leftrightarrow\left(x-a-b-c\right)\left(\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}\right)=0\)
pt \(\Rightarrow\frac{x}{a+b}+\frac{x}{a+c}+\frac{x}{b+c}=\left(\frac{ab}{a+b}+c\right)+\left(\frac{ac}{a+c}+b\right)+\left(\frac{bc}{b+c}+a\right)\)
\(\Rightarrow\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right).x=\frac{ab+ac+bc}{a+b}+\frac{ac+ab+bc}{a+c}+\frac{bc+ab+ac}{b+c}\)
\(\Rightarrow\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right).x=\left(ab+bc+ac\right)\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)\)
Nếu \(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\ne0\) => phương trình có 1 ngiệm x = ab + bc +ca
Nếu \(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}=0\) => phương trình có vô số nghiệm
Điều kiện a; b ; c khác 0
\(\Rightarrow x.\left(\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}\right)=2.\left(\frac{bc+ac+ab}{abc}\right)+\left(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\right)\)
\(\Rightarrow x.\left(\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}\right)=\frac{2bc+2ac+2ab}{abc}+\frac{a^2}{abc}+\frac{b^2}{abc}+\frac{c^2}{abc}\)
\(\Rightarrow x.\left(\frac{a+b+c}{abc}\right)=\frac{\left(a+b+c\right)^2}{abc}\)
\(\Rightarrow x.\left(a+b+c\right)=\left(a+b+c\right)^2\)
Nếu a+ b+ c khác 0 => phương trình có nghiệm duy nhất là \(\Rightarrow x=a+b+c\)
Nếu a+ b + c = 0 => x. 0 = 0 => pt có vô số nghiêm
x=a+b+c