Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
7,\\
a,\\
x\left( {y + z - yz} \right) - y\left( {z + x - zx} \right) + z\left( {y - x} \right)\\
 = \left( {xy + xz - xyz} \right) - \left( {yz + yx - zxy} \right) + \left( {zy - zx} \right)\\
 = xy + xz - xyz - yz - xy + xyz + yz - xz\\
 = \left( {xy - xy} \right) + \left( {xz - xz} \right) + \left( { - xyz + xyz} \right) + \left( { - yz + yz} \right)\\
 = 0\\
b,\\
\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right) - x\left( {x - 4} \right)\left( {x - 5} \right) - 3.\left( {5{x^2} - 3x + 2} \right)\\
 = \left( {{x^2} + 2x + x + 2} \right)\left( {x + 3} \right) - \left( {{x^2} - 4x} \right)\left( {x - 5} \right) - \left( {15{x^2} - 9x + 6} \right)\\
 = \left( {{x^2} + 3x + 2} \right)\left( {x + 3} \right) - \left( {{x^3} - 5{x^2} - 4{x^2} + 20x} \right) - \left( {15{x^2} - 9x + 6} \right)\\
 = \left( {{x^3} + 3{x^2} + 3{x^2} + 9x + 2x + 6} \right) - \left( {{x^3} - 9{x^2} + 20x} \right) - 15{x^2} + 9x - 6\\
 = \left( {{x^3} + 6{x^2} + 11x + 6} \right) - {x^3} + 9{x^2} - 20x - 15{x^2} + 9x - 6\\
 = \left( {{x^3} - {x^2}} \right) + \left( {6{x^2} + 9{x^2} - 15{x^2}} \right) + \left( {11x - 20x + 9x} \right) + \left( {6 - 6} \right)\\
 = 0\\
8,\\
a,\\
a\left( {b - c} \right) - b\left( {a + c} \right) + c\left( {a - b} \right)\\
 = \left( {ab - ac} \right) - \left( {ba + bc} \right) + \left( {ca - cb} \right)\\
 = ab - ac - ab - bc + ca - cb\\
 = \left( {ab - ab} \right) + \left( { - ac + ca} \right) + \left( { - bc - cb} \right)\\
 =  - 2bc\\
b,\\
\left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)\\
 = a.\left( {{a^2} - ab + {b^2}} \right) + b\left( {{a^2} - ab + {b^2}} \right)\\
 = {a^3} - {a^2}b + a{b^2} + {a^2}b - a{b^2} + {b^3}\\
 = {a^3} + \left( { - {a^2}b + {a^2}b} \right) + \left( {a{b^2} - a{b^2}} \right) + {b^3}\\
 = {a^3} + {b^3}\\
c,\\
\left( {{x^3} + {x^2}y + x{y^2} + {y^3}} \right)\left( {x - y} \right)\\
 = x.\left( {{x^3} + {x^2}y + x{y^2} + {y^3}} \right) - y.\left( {{x^3} + {x^2}y + x{y^2} + {y^3}} \right)\\
 = {x^4} + {x^3}y + {x^2}{y^2} + x{y^3} - {x^3}y - {x^2}{y^2} - x{y^3} - {y^4}\\
 = {x^4} + \left( {{x^3}y - {x^3}y} \right) + \left( {{x^2}{y^2} - {x^2}{y^2}} \right) + \left( {x{y^3} - x{y^3}} \right) - {y^4}\\
 = {x^4} - {y^4}
\end{array}\)

Câu 7:

a)x(y+z-yz)-y(z+xz)+z(y-x)

=xy+xz-xyz-yz-xy+xyz+zy-zx

=(xy-xy)+(xz-zx)-(xyz-xyz)-(yz-zy)

=0

b)(x+1)(x+2)(x+3)-x(x-4)(x-5)-3(5x^2-3x+2)

=(x^2+2x+x+2)(x+3)-(x^2-4x)(x-5)-(15x^2-9x+6)

=(x^2+3x+2)(x+3)-(x^3-5x^2-4x^2+20x)-(15x^2-9x+6)

Câu 8:

a)(b-c)-b(a+c)+c(a-b)

=ab-ac-ab-bc+ca-cb

=-2bc

b)(a+b)(a^2-ab+b^2)

=a^3-a^2b+ab^2+a^2b-ab^2+b^3

=a^3+b^3

c)(x^3+x^2y+xy^2+y^3)(x-y)

=x^4+x^3y+x^2y^2+xy^3-x^3y-x^2y^2-xy^3-y^4

=x^4-y^4