Cho xyz=1 Tính 5/x+xy+1 + 5/y+yz+1 + 5/z+zx+1
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(x+y+z)²=x²+y²+z²+2(xy+yz+zx)
→ x²+y²+z²=(1/2)²-2.(-2)=17/4
(x+y+z)³=x³+y³+z³+3(x+y)(y+z)(z+x)
=x³+y³+z³+3(x+y+z)(xy+yz+zx)-3xyz
→ x³+y³+z³=(1/2)³+3.(-1/2)-3.1/2.(-2)=13/8
(xy+yz+zx)²=x²y²+y²z²+z²x²+2xyz(x+y+z)
→ x²y²+y²z²+z²x²=(-2)²-2.1/2.(-1/2)=9/2
(x²+y²+z²)(x³+y³+z³)=x^5+y^5+z^5+(x²y²+y²z²+z²x²)(x+y+z)-xyz(xy+yz+zx)
→ x^5+y^5+z^5=17/4.13/8+(-2).(-1/2)-9/2.1/2=181/32
\(=\dfrac{xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(z-1\right)}{xy\left(z+1\right)+y\left(z+1\right)-x\left(z+1\right)-\left(z+1\right)}\\ =\dfrac{\left(z-1\right)\left(xy-y-x+1\right)}{\left(z+1\right)\left(xy+y-x-1\right)}=\dfrac{\left(z-1\right)\left(x-1\right)\left(y-1\right)}{\left(z+1\right)\left(x+1\right)\left(y-1\right)}=\dfrac{\left(z-1\right)\left(x-1\right)}{\left(z+1\right)\left(x+1\right)}\\ =\dfrac{\left(5003-1\right)\left(5001-1\right)}{\left(5003+1\right)\left(5001+1\right)}=\dfrac{5002\cdot5000}{5004\cdot5002}=\dfrac{5000}{5004}=\dfrac{1250}{1251}\)
Ta có:
\(xy+yz+zx=-5;xz=-5\)
\(\Rightarrow xy+yz=0\)
\(\Rightarrow y\left(x+z\right)=0\)
\(\Rightarrow\orbr{\begin{cases}y=0\\x+z=0\end{cases}}\)
Nếu \(y=0\) ta có:
\(x+0+z=2\Rightarrow x+z=2\)
\(A=x^3+y^3+z^3=\left(x+z\right)\left[\left(x+z\right)^2-3xz\right]+y^3=2\cdot\left(2^2+3\cdot5\right)+0=38\)
Nếu \(x+z=0\Rightarrow y=2\),ta có:
\(A=x^3+y^3+z^3=\left(x+z\right)\left[\left(x+z\right)^2-3xz\right]+y^3=8\)
Vậy \(A=8\left(h\right)A=38\)
a: A=y(x-4)-5(x-4)
=(x-4)(y-5)
Khi x=14 và y=5,5 thì A=(14-4)(5,5-5)=0,5*10=5
b: \(B=x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\)
Khi x=5,2 và y=4,8 thì B=(5,2+4,8)(5,2-5)
=0,2*10=2
d: Khi x=5,75 và y=4,25 thì
D=5,75^3-5,75^2*4,25+4,25^3
=8087/64
Ta có : \(\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+zx}=\frac{z}{z+xz+xyz}+\frac{xz}{xz+xyz+xyz^2}+\frac{1}{1+z+xz}\)
\(=\frac{z}{z+xz+1}+\frac{xz}{xz+1+z}+\frac{1}{1+z+xz}=\frac{1+xz+z}{1+xz+z}=1\)
a: A=yx-4y-5x+20
=y(x-4)-5(x-4)
=(x-4)(y-5)
Khi x=14 và y=5,5 thì A=(14-4)(5,5-5)=0,5*10=5
b: \(B=x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\)
Khi x=5,2 và y=4,8 thì B=(5,2+4,8)(5,2-5)
=0,2*10=2
d: Khi x=5,75 và y=4,25 thì
D=5,75^3-5,75^2*4,25+4,25^3
=8087/64
c: \(D=xyz-xy-yz-xz+x+y+z-1\)
=xy(z-1)-yz+y-xz+z+x-1
=xy(z-1)-y(z-1)-z(x-1)+(x-1)
=(z-1)(xy-y)-(x-1)(z-1)
=(z-1)(xy-y-1)
=(11-1)(9*10-10-1)
=10*79=790
\(\dfrac{1}{1+x+xy}+\dfrac{1}{1+y+yz}+\dfrac{1}{1+z+zx}\)
\(=\dfrac{1}{1+x+xy}+\dfrac{x}{x+xy+xyz}+\dfrac{xy}{xy+xyz+xyzx}\)
\(=\dfrac{1}{1+x+xy}+\dfrac{x}{x+xy+1}+\dfrac{xy}{xy+1+x}\) (Do xyz = 1)
\(=1\).