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![](https://rs.olm.vn/images/avt/0.png?1311)
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, \(\frac{x^5+x+1}{x^3+x^2+x}=\frac{x^5-x^2+x^2+x+1}{x\left(x^2+x+1\right)}=\frac{x^2\left(x-1\right)\left(x^2+x+1\right)+x^2+x+1}{x\left(x^2+x+1\right)}=\frac{\left(x^2+x+1\right)\left(x^3-x^2+1\right)}{x\left(x^2+x+1\right)}=\frac{x^3-x^2+1}{x}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)
b) \(=\dfrac{2y}{3\left(x+y\right)^2}=\dfrac{2y}{3x^2+6xy+3y^2}\)
c) \(=\dfrac{2x\left(x+1\right)}{x+1}=2x\)
d) \(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)
e) \(=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}=-9\left(x-2\right)^2=-9x^2+36x-36\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Thay $x=\sqrt{\frac{1}{2,5}}; y=z=\sqrt{\frac{1}{0,25}}$ ta thấy đề sai bạn nhé!
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(\frac{x^5+x+1}{x^3+x^2+1}=\frac{(x^2+x+1)(x^3-x^2+1)}{x(x^2+x+1)}=\frac{x^3-x^2+1}{x}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu a:
Xét tử số:
\(x^3-y^3+z^3+3xyz=(x-y)^3+3xy(x-y)+z^3+3xyz\)
\(=(x-y)^3+z^3+3xy(x-y+z)\)
\(=(x-y+z)[(x-y)^2-z(x-y)+z^2]+3xy(x-y+z)\)
\(=(x-y+z)(x^2+y^2+z^2-2xy-xz+yz)+3xy(x-y+z)\)
\(=(x-y+z)(x^2+y^2+z^2+xy+yz-xz)\)
Xét mẫu số:
\((x+y)^2+(y+z)^2+(z-x)^2\)
\(x^2+2xy+y^2+y^2+2yz+z^2+z^2-2zx+x^2\)
\(2(x^2+y^2+z^2+xy+yz-xz)\)
Do đó: \(\frac{x^3-y^3+z^3+3xyz}{(x+y)^2+(y+z)^2+(z-x)^2}=\frac{x-y+z}{2}\)
Câu b:
Xét tử số:
\((x^2-y)(y+1)+x^2y^2-1\)
\(=x^2y+x^2-y^2-y+x^2y^2-1\)
\(=(x^2y-y)+(x^2-1)+(x^2y^2-y^2)\)
\(=y(x^2-1)+(x^2-1)+y^2(x^2-1)=(x^2-1)(y^2+y+1)\)
Xét mẫu số:
\((x^2+y)(y+1)+x^2y^2+1\)
\(=x^2y+x^2+y^2+y+x^2y^2+1\)
\(=(x^2y+y)+(x^2+1)+(x^2y^2+y^2)\)
\(=y(x^2+1)+(x^2+1)+y^2(x^2+1)\)
\(=(x^2+1)(y+1+y^2)\)
Do đó:
\(\frac{(x^2-y)(y+1)+x^2y^2-1}{(x^2+y)(y+1)+x^2y^2+1}=\frac{(x^2-1)(y^2+y+1)}{(x^2+1)(y^2+y+1)}=\frac{x^2-1}{x^2+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(=\dfrac{\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz}{2x^2+2y^2+2z^2-2xy-2yz-2xz}\)
\(=\dfrac{\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)}{2\left(x^2+y^2+z^2-xy-yz-xz\right)}\)
\(=\dfrac{x+y+z}{2}\)
Bn ko hiểu chỗ nào... Để mk giải thik cho...