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\(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)
\(=\frac{c-b}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{a-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{b-a}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{c-b+b-a+a-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
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 : \(\left(x-1\right)\left(y-1\right)\left(z-1\right)=\left(x+1\right)\left(y+1\right)\left(z+1\right)\)
\(\Rightarrow xy+yz+xz=-1\Rightarrow P=1\)
\(Q=\frac{\left[\left(x+\frac{1}{x}\right)^2\right]^3-\left(x^3+\frac{1}{x^3}\right)^2}{\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)}=\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)\)
\(=3x+\frac{3}{x}=3\left(x+\frac{1}{x}\right)\)
1) \(M=a^2b^2c^2\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)
Em chú ý bài toán sau nhé: Nếu a+b+c=0 <=> \(a^3+b^3+c^3=3abc\)
CM: có:a+b=-c <=> \(\left(a+b\right)^3=-c^3\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)\)
Chú ý: a+b=-c nên \(a^3+b^3+c^3=3abc\)
Do \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
Thay vào biểu thwusc M ta được M=3abc (ĐPCM)
2, em có thể tham khảo trong sách Nâng cao phát triển toán 8 nhé, anh nhớ không nhầm thì bài này trong đó
Nếu không thấy thì em có thể quy đồng lên mà rút gọn
Ta có
\(B=\frac{\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(x-c\right)\left(x-a\right)}{\left(b-c\right)\left(b-a\right)}+\frac{\left(x-a\right)\left(x-b\right)}{\left(c-a\right)\left(c-b\right)}\)
\(=\frac{\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(x-c\right)\left(x-a\right)}{\left(b-c\right)\left(c-a\right)}+\frac{\left(x-c\right)\left(x-a\right)}{\left(a-c\right)\left(b-a\right)}+\frac{\left(x-a\right)\left(x-b\right)}{\left(c-a\right)\left(c-b\right)}\)
\(=\frac{\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(x-c\right)\left(x-a\right)}{\left(b-c\right)\left(c-a\right)}-\frac{\left(x-c\right)\left(x-a\right)}{\left(a-c\right)\left(a-b\right)}+\frac{\left(x-a\right)\left(x-b\right)}{\left(a-c\right)\left(c-b\right)}\)
\(=\frac{\left(x-b\right)\left(x-c\right)-\left(x-c\right)\left(x-a\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(x-c\right)\left(x-a\right)-\left(x-b\right)\left(x-a\right)}{\left(b-c\right)\left(c-a\right)}\)
\(=\frac{\left(x-c\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(x-a\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)}\).
\(=\frac{x-c}{a-c}-\frac{x-a}{a-c}=\frac{x-c-x+a}{a-c}\)
\(=1\)