Câu hỏi của Nguyễn Thanh Liêm

Question

Cho x + y + z = a ; x^2 + y^2 + z^2 = b^2 và 1/x+1/y+1/z= c. Tính giá trị của biểu thức x^3 + y^3 + z^3 theo a, b, c

Yukru · Accepted Answer

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:

$x^3+y^3+z^3$

$=\left(x+y+z\right)^3-3\left(x^2z+xyz+xz^2+x^2y+xyz+xy^2+y^2z+xyz+yz^2\right)+3xyz$

$=\left(x+y+z\right)^3-3\left[xz\left(x+y+z\right)+xy\left(x+y+z\right)+yz\left(x+y+z\right)\right]+3xyz$

$=\left(x+y+z\right)^3-3\left[\left(xy+yz+zx\right)\left(x+y+z\right)\right]+3xyz$

$=a^3-3\left[\dfrac{\left(a^2-b^2\right)}{c}.a\right]+3\left(\dfrac{a^2-b^2}{2c}\right)$

$=a^3-\dfrac{3a\left(a^2-b^2\right)}{c}+\dfrac{3\left(a^2-b^2\right)}{2c}$

$=a^3-\dfrac{6a\left(a^2-b^2\right)}{2c}+\dfrac{3\left(a^2-b^2\right)}{2c}$

$=a^3-\dfrac{6a\left(a^2-b^2\right)+3\left(a^2-b^2\right)}{2c}$

$=a^3-\dfrac{3\left(a^2-b^2\right)\left(2a+1\right)}{2c}$

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