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Ta có:
\(x+y+z=a\)
\(\Rightarrow\left(x+y+z\right)^2=a^2\)
Ta lại có:
\(x^2+y^2+z^2=b^2\)
\(\Rightarrow\left(x+y+z\right)^2-\left(x^2+y^2+z^2\right)=a^2-b^2\)
\(\Rightarrow x^2+y^2+z^2+2\left(xy+xz+yz\right)-x^2-y^2-z^2=a^2-b^2\)
\(\Rightarrow2\left(xy+xz+yz\right)=a^2-b^2\)
\(\Rightarrow xy+xz+yz=\dfrac{a^2-b^2}{2}\left(1\right)\)
Lại có:
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=c\)
\(\Rightarrow\dfrac{yz}{xyz}+\dfrac{xz}{xyz}+\dfrac{xy}{xyz}=c\)
\(\Rightarrow\dfrac{yz+xz+xy}{xyz}=c\)
\(\Rightarrow yz+xz+xy=c.xyz\left(2\right)\)
Từ (1) và (2) suy ra:
\(\dfrac{a^2-b^2}{2}=c.xyz\)
\(\Rightarrow\dfrac{a^2-b^2}{2c}=xyz\)
Như vậy ta có:
\(\left\{{}\begin{matrix}x+y+z=a\\xy+yz+zx=\dfrac{a^2-b^2}{2}\\xyz=\dfrac{a^2-b^2}{2c}\end{matrix}\right.\)
Ta có:
\(x^3+y^3+z^3\)
\(=\left(x+y+z\right)^3-3\left(x^2z+xyz+xz^2+x^2y+xyz+xy^2+y^2z+xyz+yz^2\right)+3xyz\)
\(=\left(x+y+z\right)^3-3\left[xz\left(x+y+z\right)+xy\left(x+y+z\right)+yz\left(x+y+z\right)\right]+3xyz\)
\(=\left(x+y+z\right)^3-3\left[\left(xy+yz+zx\right)\left(x+y+z\right)\right]+3xyz\)
\(=a^3-3\left[\dfrac{\left(a^2-b^2\right)}{c}.a\right]+3\left(\dfrac{a^2-b^2}{2c}\right)\)
\(=a^3-\dfrac{3a\left(a^2-b^2\right)}{c}+\dfrac{3\left(a^2-b^2\right)}{2c}\)
\(=a^3-\dfrac{6a\left(a^2-b^2\right)}{2c}+\dfrac{3\left(a^2-b^2\right)}{2c}\)
\(=a^3-\dfrac{6a\left(a^2-b^2\right)+3\left(a^2-b^2\right)}{2c}\)
\(=a^3-\dfrac{3\left(a^2-b^2\right)\left(2a+1\right)}{2c}\)
\(\left\{\begin{matrix}x+y+z=a\left(1\right)\\x^2+y^2+z^2=b^2\left(2\right)\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{c}\left(3\right)\end{matrix}\right.\)
HĐT ta có\(x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-\left(xy+xz+yz\right)\right)-3xyz\)
từ (3)=> (xy+xz+yz)/(xyz)=1/c(*)
(1) bình phường=>2(xy+xz+yz)=(a^2-b^2 )
(*)=> xyz=(a^2-b^2).c/2
Thay hết vào biểu thức trên => đáp số