Toán Lớp 8: 1) phân tích đa thức thành nhân tử heeeeeeeeeeeeepppppppppppppppppppppp
a) 9x+9xy
b)6,4x^2-16x+16
c)x^3+27
d)7x^2-14x+7-7y^2
e)x^2-y^2-y-1/4
g)x^2+12x-13
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Toán Lớp 8: 1) phân tích đa thức thành nhân tử heeeeeeeeeeeeepppppppppppppppppppppp
a) 9x+9xy
b)6,4x^2-16x+16
c)x^3+27
d)7x^2-14x+7-7y^2
e)x^2-y^2-y-1/4
g)x^2+12x-13
Comments ( 1 )
a)9x + 9xy = 9x\left( {1 + y} \right)\\
b)4{x^2} – 16x + 16\\
= 4\left( {{x^2} – 4x + 4} \right)\\
= 4.{\left( {x – 2} \right)^2}\\
c){x^3} + 27\\
= {x^3} + {3^3}\\
= \left( {x + 3} \right)\left( {{x^2} – 3x + 9} \right)\\
d)7{x^2} – 14x + 7 – 7{y^2}\\
= 7\left( {{x^2} – 2x + 1 – {y^2}} \right)\\
= 7.\left( {{{\left( {x – 1} \right)}^2} – {y^2}} \right)\\
= 7.\left( {x – 1 – y} \right)\left( {x – 1 + y} \right)\\
e){x^2} – {y^2} – y – \dfrac{1}{4}\\
= {x^2} – \left( {{y^2} + y + \dfrac{1}{4}} \right)\\
= {x^2} – {\left( {y + \dfrac{1}{2}} \right)^2}\\
= \left( {x – y – \dfrac{1}{2}} \right)\left( {x + y + \dfrac{1}{2}} \right)\\
g){x^2} + 12x – 13\\
= {x^2} – x + 13x – 13\\
= \left( {x – 1} \right)\left( {x + 13} \right)
\end{array}$