Cho a, b, c thỏa mãn a+b+c=0 và \(abc\ne0\) . Chứng minh rằng: \(\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)=9\)
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23 tháng 11 2020
\(x^3+3x^2+\frac{2x}{x^4}+x\)
\(=x\left(x^2+3x+\frac{2}{x^4}+1\right)\)
TT
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23 tháng 11 2020
a, \(\left(x-1\right)^2+x\left(5-x\right)=8\Leftrightarrow x^2-2x+1+5x-x^2=8\)
\(\Leftrightarrow3x+1=8\Leftrightarrow x=\frac{7}{3}\)
b, \(\left(12x^4-6x\right):6x+2x\left(2+x\right)\left(2-x\right)=7\)
Ta có : \(\left(12x^4-6x\right):6x=2x^3-1\)
\(\Leftrightarrow2x^3-1+2x\left(4-x^2\right)=7\)
\(\Leftrightarrow2x^3-1+8x-2x^3=7\Leftrightarrow x=1\)
Đặt: \(A=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\)
\(\Leftrightarrow A=\frac{ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)}{abc}\)
\(\Leftrightarrow abc\cdot A=ab\left(a-b\right)+bc\left[\left(b-a\right)+\left(a-c\right)\right]+ca\left(c-a\right)\)
\(\Leftrightarrow abc\cdot A=ab\left(a-b\right)-bc\left(a-b\right)-bc\left(c-a\right)+ca\left(c-a\right)\)
\(\Leftrightarrow abc\cdot A=b\left(a-b\right)\left(a-c\right)+c\left(c-a\right)\left(a-b\right)\)
\(\Leftrightarrow abc\cdot A=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)
\(\Rightarrow A=-\frac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{abc}\) (1)
Đặt: \(B=\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\)
\(\Leftrightarrow B=\frac{a\left(a-b\right)\left(c-a\right)+b\left(a-b\right)\left(b-c\right)+c\left(b-c\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(\Leftrightarrow\left(a-b\right)\left(b-c\right)\left(c-a\right)\cdot B=a\left(a-b\right)\left(c-a\right)-\left(c+a\right)\left(a-b\right)\left(b-c\right)+c\left(b-c\right)\left(c-a\right)\)
\(=a\left(a-b\right)\left(c-a\right)-c\left(a-b\right)\left(b-c\right)-a\left(a-b\right)\left(b-c\right)+c\left(b-c\right)\left(c-a\right)\)
\(=a\left(a-b\right)\left(2c-a-b\right)+c\left(b-c\right)\left(b+c-2a\right)\)
\(=3ac\left(a-b\right)-3ac\left(b-c\right)\)
\(=3ac\left(a+c-2b\right)=-9abc\)
\(\Rightarrow B=-\frac{9abc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(\Rightarrow A\cdot B=9\)