Cho x+y+z=a
x^2+y^2+z^2=b^2
;1/x=1/y+1/z=1/c
tính x^3+y^3+z^3 theo a,b,c
Cho x, y, z thỏa: x+y+z=a ; x^2+y^2+z^2=b ; 1/x+1/y+1/z=1/c Tính xy + yz +xz và x^3+y^3+z^3 theo a,b,c
ta có: \(x+y+z=a\Rightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)=a^2\)
\(\Rightarrow b+2\left(xy+yz+xz\right)=a^2\Rightarrow xy+yz+xz=\frac{a^2-b}{2}\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{c}\Rightarrow\frac{xy+yz+xz}{xyz}=\frac{1}{c}\Rightarrow c\left(xy+yz+xz\right)=xyz\)
Ta có:\(x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)+3xyz\)
\(=a\left(b-\frac{a^2-b}{2}\right)+\frac{3c\left(a^2-b\right)}{2}\)
Cho các số x, y, z thoả mãn: \(\left\{{}\begin{matrix}x+y+z=a\\\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{c}\\x^2+y^2+z^2=b^2\end{matrix}\right.\)
Tính \(P=x^3+y^3+z^3\) theo a, b, c.
Lời giải:
$xy+yz+xz=\frac{1}{2}[(x+y+z)^2-(x^2+y^2+z^2)]=\frac{1}{2}(a^2-b^2)$
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{c}$
$\Rightarrow xyz=c(xy+yz+xz)=\frac{1}{2}c(a^2-b^2)$
Khi đó:
$P=(x+y+z)^3-3(x+y)(y+z)(x+z)$
$=(x+y+z)^3-3[(x+y+z)(xy+yz+xz)-xyz]=(x+y+z)^3-3(xy+yz+xz)(x+y+z)+3xyz$
$=a^3-\frac{3}{2}a(a^2-b^2)+\frac{3}{2}c(a^2-b^2)$
cho x+y+z=a ;x^2+ y^2 + z^2=b^2 ; 1/x+1/y+1/z=1/c . Tinh x^3+y^3+z^3 theo a,b,c
1)Phân tích thành nhân tử:
a. (((x^2)+(y^2))^2)((y^2)-(x^2))+(((y^2)+(z^2))^2)((z^2)-(y^2))+(((z^2)+(x^2))^2)((x^2)-(z^2))
b. ((x-a)^4)+4a^4
c. (x^4)-(8x^2)+4
d. (x^8)+(x^4)+1
e. x((y^2)-(z^2))+y((z^2)-(x^2))+z((x^2)-(y^2))
f. (8x^3)(y+z)-(y^3)(z+2x)-(z^3)(2x-y)
g. (12x-1)(6x-1)(4x-1)(3x-1)-5
2) Cho (a^3)+(b^3)+(c^3)=3abc và abc khác 0. Tính A=(1+a/b)(1+b/c)(1+c/a).
3) Rút gọn phân thức:
((x^3)+(y^3)+(z^3)-3xyz)/(((x-y)^2)+((y-z)^2)+((z-x)^2))
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
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:
\(x^3+y^3+z^3\)
\(=\left(x+y+z\right)^3-3\left(x^2z+xyz+xz^2+x^2y+xyz+xy^2+y^2z+xyz+yz^2\right)+3xyz\)
\(=\left(x+y+z\right)^3-3\left[xz\left(x+y+z\right)+xy\left(x+y+z\right)+yz\left(x+y+z\right)\right]+3xyz\)
\(=\left(x+y+z\right)^3-3\left[\left(xy+yz+zx\right)\left(x+y+z\right)\right]+3xyz\)
\(=a^3-3\left[\dfrac{\left(a^2-b^2\right)}{c}.a\right]+3\left(\dfrac{a^2-b^2}{2c}\right)\)
\(=a^3-\dfrac{3a\left(a^2-b^2\right)}{c}+\dfrac{3\left(a^2-b^2\right)}{2c}\)
\(=a^3-\dfrac{6a\left(a^2-b^2\right)}{2c}+\dfrac{3\left(a^2-b^2\right)}{2c}\)
\(=a^3-\dfrac{6a\left(a^2-b^2\right)+3\left(a^2-b^2\right)}{2c}\)
\(=a^3-\dfrac{3\left(a^2-b^2\right)\left(2a+1\right)}{2c}\)
Cho x+y+z=a;x^2+y^2+z^2=b^2 ;1/x+1/y+1/z=1/c tính x^3+y^3+z^3?
Cho các số thực x ; y ; z thỏa mãn x^2-y=a ; y^2-z=b ; z^2-x=c .Tính giá trị biểu thức sau theo a; b; c.
P=x^3 (z-y^2) + y^3 (x-z^2) + z^3 (y-x^2) + xyz (xyz-1)
cho \(\hept{\begin{cases}x+y+z=a\\x^2+y^2+z^2=b\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{c}\end{cases}}\) . Tính \(x^3+y^3+z^3\) theo a, b, c
\(x+y+z=a\Rightarrow\left(x+y+z\right)^2=a^2\Rightarrow xy+yz+zx=\frac{a^2-b}{2}\\ \)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{c}\Rightarrow\frac{xy+yz+xz}{xyz}=\frac{1}{c}\Rightarrow xyz=\frac{\left(a^2-b\right)c}{2}\)
Ta có
\(x^3+y^3+z^3-3xyz=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz\)
\(=\left(x+y+z\right)\left(\left(x+y\right)^2-\left(x+y\right)z+z^2\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-\left(xy+yz+zx\right)\right)\)
\(\Rightarrow x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-\left(xy+yz+xz\right)\right)+3xyz\)
\(=a\left(b-\frac{a^2-b}{2}\right)+3\frac{\left(a^2-b\right)c}{2}\)
cho x+y+z=a ;x2+ y2 + z2=b2 ; 1/x+1/y+1/z=1/c . Tinh x3+y3+z3 theo a,b,c