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sai đề rồi nhé , đề phải là :
\(\frac{x^3-y^3+z^3+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\frac{\left(x-y\right)^3+3xy.\left(x-y\right)+z^3+3xyz}{x^2+2xy+y^2+y^2+2yz+z^2+z^2-2xz+x^2}\)
\(=\frac{\left(x-y+z\right).\left[\left(x-y\right)^2-\left(x-y\right).z+z^2\right]+3xy.\left(x-y+z\right)}{2x^2+2y^2+2z^2+2xy+2yz-2xz}\)
\(=\frac{\left(x-y+z\right).\left(x^2-2xy+y^2-xz+yz+z^2+3xy\right)}{2.\left(x^2+y^2+z^2+xy+yz-xz\right)}\)
\(=\frac{\left(x-y+z\right).\left(x^2+y^2+z^2+xy+yz-xz\right)}{2.\left(x^2+y^2+z^2+xy+yz-xz\right)}\)
\(=\frac{x-y+z}{2}\)
\(\dfrac{5xy^3\left(x-y\right)}{10xy\left(x-y\right)^3}=\dfrac{y^2}{2\left(x-y\right)^2}\)
\(\dfrac{5xy^3\left(x-y\right)}{10xy\left(x-y\right)^3}=\dfrac{y^2}{2\left(x-y\right)^2}\)
\(a,\dfrac{21x^2y^3}{24x^3y^2}=\dfrac{7y}{8x}\)
\(b,\dfrac{15xy^3\left(x^2-y^2\right)}{20x^2y\left(x+y\right)^2}=\dfrac{15xy^3\left(x-y\right)\left(x+y\right)}{20x^2y\left(x+y\right)^2}=\dfrac{3y^2\left(x-y\right)}{4x\left(x+y\right)}=\dfrac{3xy^2-3y^3}{4x^2+4xy}\)
a) Ta có: \(\dfrac{21x^2y^3}{24x^3y^2}\)
\(=\dfrac{21x^2y^3:3x^2y^2}{24x^3y^2:3x^2y^2}\)
\(=\dfrac{7y}{8x}\)
\(b,=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}\left(x\ne-y\right)\\ c,=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{2\left(x-2\right)}{x+2}\left(x\ne-1;x\ne\pm2;x\ne0\right)\)
b: \(\dfrac{x^3-x^2y+xy^2}{x^3+y^3}=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}\)
c: \(\dfrac{\left(2x^2+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{2\left(x-2\right)}{x+2}\)
a) \(\left(3-xy^2\right)^2-\left(2+xy^2\right)^2\)
\(=\left(3-xy^2+2+xy^2\right)\left(3-xy^2-2-xy^2\right)\)
\(=5.\left(-2xy^2\right)\)
\(=-10xy^2\)
b) \(\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=x^3-y^3\)
c) \(\left(x-3\right)^3+\left(2-x\right)^3\)
\(=x^3-3x^2.3+3x.3^2-3^3+2^3-3.2^2.x+3.2.x^2-x^3\)
\(=x^3-9x^2+27x-27+8-12x+6x^2-x^3\)
\(=\left(x^3-x^3\right)+\left(-9x^2+6x^2\right)+\left(27x-12x\right)+\left(-27+8\right)\)
\(=-3x^2+15x-19\)
\(\frac{x^3+y^3+z^3-3xyz}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
\(=\frac{x^3+y^3+z^3-3xyz}{x^2-2xy+y^2+y^2-2yz+z^2+z^2-2xz+x^2}=\frac{\left(x+y+z\right).\left(x^2+y^2+z^2-xy-yz-zx\right)}{2.\left(x^2+y^2+z^2-xy-yz-zx\right)}=\frac{x+y+z}{2}\)
p/s: áp dụng 7 hàng đẳng thức là làm đc =)
