Toán Lớp 8: bai 1: rut gon cac bieu thuc sau:
a) (x+3) (x^2-3x+9)-(54+x^3)
b) (2x+y) (4x^2-2xy+y^2)-(2x-y) (4x^2+2xy+y^2)
bai2:chung minh rang:
a)a^3+b^3=(a+b)^3-3ab (a+b)
b)a^3-b^3=(a-b)^3+3ab (a-b)
Giup minh voi a =)))
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Toán Lớp 8: bai 1: rut gon cac bieu thuc sau:
a) (x+3) (x^2-3x+9)-(54+x^3)
b) (2x+y) (4x^2-2xy+y^2)-(2x-y) (4x^2+2xy+y^2)
bai2:chung minh rang:
a)a^3+b^3=(a+b)^3-3ab (a+b)
b)a^3-b^3=(a-b)^3+3ab (a-b)
Giup minh voi a =)))
Comments ( 1 )
1,\\
a,\,\,\, – 27\\
b,\,\,\,\,2{y^3}\\
2,\\
a,\\
{a^3} + {b^3} = {\left( {a + b} \right)^3} – 3ab\left( {a + b} \right)\\
b,\\
{a^3} – {b^3} = {\left( {a – b} \right)^3} + 3ab\left( {a – b} \right)
\end{array}\)
1,\\
a,\\
\left( {x + 3} \right)\left( {{x^2} – 3x + 9} \right) – \left( {54 + {x^3}} \right)\\
= \left( {x + 3} \right)\left( {{x^2} – x.3 + {3^2}} \right) – \left( {54 + {x^3}} \right)\\
= \left( {{x^3} + {3^3}} \right) – \left( {54 + {x^3}} \right)\\
= \left( {{x^3} + 27} \right) – \left( {54 + {x^3}} \right)\\
= {x^3} + 27 – 54 – {x^3}\\
= – 27\\
b,\\
\left( {2x + y} \right)\left( {4{x^2} – 2xy + {y^2}} \right) – \left( {2x – y} \right)\left( {4{x^2} + 2xy + {y^2}} \right)\\
= \left( {2x + y} \right).\left[ {{{\left( {2x} \right)}^2} – 2x.y + {y^2}} \right] – \left( {2x – y} \right).\left[ {{{\left( {2x} \right)}^2} + 2x.y + {y^2}} \right]\\
= \left[ {{{\left( {2x} \right)}^3} + {y^3}} \right] – \left[ {{{\left( {2x} \right)}^3} – {y^3}} \right]\\
= \left( {8{x^3} + {y^3}} \right) – \left( {8{x^3} – {y^3}} \right)\\
= 8{x^3} + {y^3} – 8{x^3} + {y^3}\\
= 2{y^3}\\
2,\\
a,\\
{a^3} + {b^3} = \left( {{a^3} + 3{a^2}b + 3a{b^2} + {b^3}} \right) – \left( {3{a^2}b + 3a{b^2}} \right)\\
= {\left( {a + b} \right)^3} – 3ab\left( {a + b} \right)\\
b,\\
{a^3} – {b^3} = \left( {{a^3} – 3{a^2}b + 3a{b^2} – {b^3}} \right) + \left( {3{a^2}b – 3a{b^2}} \right)\\
= {\left( {a – b} \right)^3} + 3ab\left( {a – b} \right)
\end{array}\)