Toán Lớp 8: B1: PT đa thức thành phân tử
a, 9 ( x + 1 ) ^ 2 – ( 3x -2 )
b, 9 ( x + 1 ) ^2 – ( 3x – 2 ) ^2
c, 49 – a^ 2 + 2ab – b^2
d, 4x^3 + 24x^2 – 12xy^2
e, 15a^m+2 b – 45 a ^m b
Mong các thánh giỏi toán giúp e
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Toán Lớp 8: B1: PT đa thức thành phân tử
a, 9 ( x + 1 ) ^ 2 – ( 3x -2 )
b, 9 ( x + 1 ) ^2 – ( 3x – 2 ) ^2
c, 49 – a^ 2 + 2ab – b^2
d, 4x^3 + 24x^2 – 12xy^2
e, 15a^m+2 b – 45 a ^m b
Mong các thánh giỏi toán giúp e
Comments ( 1 )
a)9{\left( {x + 1} \right)^2} – \left( {3x – 2} \right)\\
= 9{x^2} + 18x + 9 – 3x + 2\\
= 9{x^2} + 15x + 11??\\
b)9{\left( {x + 1} \right)^2} – {\left( {3x – 2} \right)^2}\\
= {\left( {3x + 3} \right)^2} – {\left( {3x – 2} \right)^2}\\
= \left( {3x + 3 – 3x + 2} \right)\left( {3x + 3 + 3x – 2} \right)\\
= 5.\left( {6x + 1} \right)\\
c)49 – {a^2} + 2ab – {b^2}\\
= 49 – \left( {{a^2} – 2ab + {b^2}} \right)\\
= {7^2} – {\left( {a – b} \right)^2}\\
= \left( {7 – a + b} \right)\left( {7 + a – b} \right)\\
d)4{x^3} + 24{x^2} – 12x{y^2}\\
= 4x\left( {{x^2} + 6x – 3{y^2}} \right)\\
e)15{a^m} + 2b – 45{a^m}b\\
??
\end{array}$