Toán Lớp 8: giúp với ạ
phân tích đa thức sau thành nhân tử
a)(xy+1)^2-(x+y)^2
b)(x+y)^3-(x-y)^3
c)3x^4y^2+3x^3y^2+3xy^2+3y^2
d)4(x^2-y^2)-8(x-ay)-4(a^2-1)
e)(x+y)^3-1-3xy(x+y-1)
Leave a reply
Register Now
Login
Lost Password
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Toán Lớp 8: giúp với ạ
phân tích đa thức sau thành nhân tử
a)(xy+1)^2-(x+y)^2
b)(x+y)^3-(x-y)^3
c)3x^4y^2+3x^3y^2+3xy^2+3y^2
d)4(x^2-y^2)-8(x-ay)-4(a^2-1)
e)(x+y)^3-1-3xy(x+y-1)
Comments ( 1 )
a){\left( {xy + 1} \right)^2} – {\left( {x + y} \right)^2}\\
= \left( {xy + 1 + x + y} \right)\left( {xy + 1 – x – y} \right)\\
= \left( {y + 1} \right)\left( {x + 1} \right).\left( {x – 1} \right)\left( {y – 1} \right)\\
b){\left( {x + y} \right)^3} – {\left( {x – y} \right)^3}\\
= {x^3} + 3{x^2}y + 3x{y^2} + {y^3}\\
– {x^3} + 3{x^2}y – 3x{y^2} + {y^3}\\
= 6{x^2}y + 2{y^3}\\
= 2y\left( {3{x^2} + {y^2}} \right)\\
c)3{x^4}{y^2} + 3{x^3}{y^2} + 3x{y^2} + 3{y^2}\\
= 3{y^2}\left( {{x^4} + {x^3} + x + 1} \right)\\
= 3{y^2}\left( {x + 1} \right)\left( {{x^3} + 1} \right)\\
= 3{y^2}\left( {x + 1} \right)\left( {x + 1} \right)\left( {{x^2} – x + 1} \right)\\
= 3{y^2}{\left( {x + 1} \right)^2}\left( {{x^2} – x + 1} \right)\\
d)4\left( {{x^2} – {y^2}} \right) – 8\left( {x – ay} \right) – 4\left( {{a^2} – 1} \right)\\
= 4.\left( {{x^2} – {y^2} – 2x + 2ay – {a^2} + 1} \right)\\
= 4.\left( {{x^2} – 2x + 1 – \left( {{y^2} – 2ay + {a^2}} \right)} \right)\\
= 4.\left[ {{{\left( {x – 1} \right)}^2} – {{\left( {y – a} \right)}^2}} \right]\\
= 4.\left( {x – 1 – y + a} \right)\left( {x – 1 + y – a} \right)\\
e){\left( {x + y} \right)^3} – 1 – 3xy\left( {x + y – 1} \right)\\
= \left( {x + y – 1} \right)\left[ {{{\left( {x + y} \right)}^2} + \left( {x + y} \right) + 1} \right] – 3xy\left( {x + y – 1} \right)\\
= \left( {x + y – 1} \right)\left( {{x^2} + 2xy + {y^2} + x + y + 1 – 3xy} \right)\\
= \left( {x + y – 1} \right)\left( {{x^2} – xy + {y^2} + x + y + 1} \right)
\end{array}$