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Ta có
a3 + b3 + c3 - 3abc = 0
<=> (a + b)3 + c3 - 3ab(a + b) - 3abc = 0
<=> (a + b + c)(a2 + b2 + c2 + 2ab - ac - bc) - 3ab(a + b + c) = 0
<=> (a + b + c)(a2 + b2 + c2 - ab - ac - bc) = 0
<=> (a2 + b2 + c2 - ab - ac - bc) = 0
<=> (a2 - 2ab + b2) + (a2 - 2ac - c2) + (b2 - 2bc + c2) = 0
<=> (a - b)2 + (a - c)2 + (b - c)2 = 0
<=> a = b = c
=> P = (1 + 1)(1 + 1)(1 +1) = 8
1)
\(x+2+\frac{3}{x-2}\)
\(=\frac{\left(x+2\right)\left(x-2\right)}{x-2}+\frac{3}{x-2}\)
\(=\frac{x^2-4}{x-2}+\frac{3}{x-2}\)
\(=\frac{x^2-4+3}{x-2}\)
\(=\frac{x^2-1}{x-2}\)
2)
\(\frac{x^2}{\left(x-y\right)\left(x-z\right)}+\frac{y^2}{\left(y-x\right)\left(y-z\right)}+\frac{z^2}{\left(z-x\right)\left(z-y\right)}\)
\(=\frac{x^2}{\left(x-y\right)\left(x-z\right)}-\frac{y^2}{\left(x-y\right)\left(y-z\right)}+\frac{z^2}{\left(x-z\right)\left(y-z\right)}\)
\(=\frac{x^2\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}-\frac{y^2\left(x-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}+\frac{z^2\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(=\frac{x^2\left(y-z\right)-y^2\left(x-z\right)+z^2\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(=\frac{x^2y-x^2z-xy^2+y^2z+xz^2-yz^2}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(=\frac{x^2y-x^2z-xy^2+y^2z+xz^2-yz^2}{\left(x^2-xy-xz+yz\right)\left(y-z\right)}\)
\(=\frac{x^2y-x^2z-xy^2+y^2z+xz^2-yz^2}{x^2y-xy^2-xyz+y^2z-x^2z+xyz+xz^2-yz^2}\)
\(=\frac{x^2y-x^2z-xy^2+y^2z+xz^2-yz^2}{x^2y-x^2z-xy^2+y^2z+xz^2-yz^2}\)
\(=1\)
Lời giải:
Từ điều kiện $xyz=1$ ta có:
\(x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
\(\Leftrightarrow x+y+z=xy+yz+xz\)
\(\Leftrightarrow x+y+z-xy-yz-xz+xyz-1=0\)
\(\Leftrightarrow x(1-y)+(y+z-yz-1)+(xyz-xz)=0\)
\(\Leftrightarrow x(1-y)+(1-y)(z-1)-xz(1-y)=0\)
\(\Leftrightarrow (1-y)(x+z-1-xz)=0\)
\(\Leftrightarrow (1-y)(1-x)(z-1)=0\)
\(\Leftrightarrow (x-1)(y-1)(z-1)=0\)
Khi đó:
\(P=(x^{19}-1)(y^5-1)(z^{1890}-1)=(x-1)(x^{18}+x^{17}+...+1)(y-1)(y^4+...+1)(z-1)(z^{1889}+...+1)\)
\(=(x-1)(y-1)(z-1).A=0\)