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\(\left(\dfrac{2x+2y-z}{3}\right)^2+\left(\dfrac{2y+2z-x}{3}\right)^2+\left(\dfrac{2z+2x-y}{3}\right)^2\)
\(=\dfrac{4y^2+4x^2+z^2+8xy-4xz-4yz+4y^2+4z^2+x^2+8yz-4xy-4xz}{9}+\dfrac{\left(2z+2x-y\right)^2}{9}\)
\(=\dfrac{8y^2+5x^2+5z^2+4xy-8xz+4yz+4z^2+4x^2+y^2+8xz-4yz-4xy}{9}\)
\(=\dfrac{9y^2+9z^2+9x^2}{9}=x^2+y^2+z^2\)
a, Xét tử thức \(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)
\(=x^2\left(y-z\right)-y^2\left(x-z\right)+z^2\left[\left(x-z\right)-\left(y-z\right)\right]\)
\(=x^2\left(y-z\right)-y^2\left(x-z\right)+z^2\left(x-z\right)-z^2\left(y-z\right)\)
\(=\left(x^2-z^2\right)\left(y-z\right)-\left(y^2-z^2\right)\left(x-z\right)\)
\(=\left(x-z\right)\left(x+z\right)\left(y-z\right)-\left(y-z\right)\left(y+z\right)\left(x-z\right)\)
\(=\left(x-z\right)\left(xy-xz+yz-z^2-y^2-yz+yz+z^2\right)\)
\(=\left(x-z\right)\left(xy-xz+yz-y^2\right)=\left(x-z\right)\left[x\left(y-z\right)-y\left(y-z\right)\right]\)
\(=\left(x-z\right)\left(x-y\right)\left(y-z\right)\)
Mẫu thức \(x^2y-x^2z+y^2z-y^3=x^2\left(y-z\right)-y^2\left(y-z\right)=\left(x-y\right)\left(x+y\right)\left(y-z\right)\)
Vậy \(\frac{x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)}{x^2y-x^2z+y^2z-y^3}=\frac{\left(x-y\right)\left(y-z\right)\left(x-z\right)}{\left(x-y\right)\left(x+y\right)\left(y-z\right)}=\frac{x-z}{x+y}\)
b, \(\frac{x^5+x+1}{x^3+x^2+x}=\frac{x^5-x^2+x^2+x+1}{x\left(x^2+x+1\right)}=\frac{x^2\left(x-1\right)\left(x^2+x+1\right)+x^2+x+1}{x\left(x^2+x+1\right)}=\frac{\left(x^2+x+1\right)\left(x^3-x^2+1\right)}{x\left(x^2+x+1\right)}=\frac{x^3-x^2+1}{x}\)
Câu a:
Xét tử số:
\(x^2(y-z)+y^2(z-x)+z^2(x-y)\)
\(=x^2(y-z)-y^2[(y-z)+(x-y)]+z^2(x-y)\)
\(=x^2(y-z)-y^2(y-z)-y^2(x-y)+z^2(x-y)\)
\(=(x^2-y^2)(y-z)-(y^2-z^2)(x-y)\)
\(=(x-y)(y-z)[(x+y)-(y+z)]=(x-y)(y-z)(x-z)\)
Xét mẫu số:
\(x^2y-x^2z+y^2z-y^3=x^2(y-z)-y^2(y-z)=(x^2-y^2)(y-z)\)
\(=(x-y)(x+y)(y-z)\)
Do đó:
\(\frac{x^2(y-z)+y^2(z-x)+z^2(x-y)}{x^2y-x^2z+y^2z-y^3}=\frac{(x-y)(y-z)(x-z)}{(x-y)(x+y)(y-z)}=\frac{x-z}{x+y}\)
Câu b:
Xét tử số:
\(x^5+x+1=x^5-x^2+x^2+x+1=x^2(x^3-1)+x^2+x+1\)
\(=x^2(x-1)(x^2+x+1)+(x^2+x+1)\)
\(=(x^2+x+1)(x^3-x^2+1)\)
Xét mẫu số:
\(x^3+x^2+x=x(x^2+x+1)\)
Do đó: \(\frac{x^5+x+1}{x^3+x^2+1}=\frac{(x^2+x+1)(x^3-x^2+1)}{x(x^2+x+1)}=\frac{x^3-x^2+1}{x}\)
d)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+.....+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+.....-\dfrac{1}{x+99}+\dfrac{1}{x+100}\)=\(\dfrac{1}{x}-\dfrac{1}{x+100}\)
=\(\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)
=\(\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{100}{x\left(x+100\right)}\)
1: \(=\dfrac{\left(x^2+2xy+y^2\right)-1}{\left(x^2+2x+1\right)-y^2}\)
\(=\dfrac{\left(x+y+1\right)\left(x+y-1\right)}{\left(x+1-y\right)\left(x+1+y\right)}=\dfrac{x+y-1}{x-y+1}\)
2: \(=\dfrac{\left(x^2-y^2\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}\)
\(=\dfrac{\left(x-y\right)\left(x^2+y^2\right)}{x^2-xy+y^2}\)
3: \(=\dfrac{\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz}{2x^2+2y^2+2z^2-2xy-2yz-2xz}\)
\(=\dfrac{\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)}{2\left(x^2+y^2+z^2-xy-yz-xz\right)}\)
\(=\dfrac{x+y+z}{2}\)
ĐKXĐ: \(z\ne2\)
\(\left(\dfrac{z^2+2z+4}{z-2}\right)^2+7+\dfrac{\left(z-2\right)\left(z^2+2x+4\right)}{\left(z-2\right)^2}=0\)
\(\Leftrightarrow\left(\dfrac{z^2+2z+4}{z-2}\right)^2+\dfrac{z^2-2z+4}{z-2}+7=0\)
Đặt \(\dfrac{z^2+2z+4}{z-2}=x\)
\(\Rightarrow x^2+x+7=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{27}{4}=0\)
Pt đã cho vô nghiệm
Em cảm ơn ạ.