Toán Lớp 8: Bài 1phân tích đa thức thành nhân tử
a) x² – y² – 9x +9
b) 8x – 8y -x² +2xy – y ²
c) x^4 + 81
Bài 2 tìm x
a) ( 2x + 5 )² – ( x + 2 )² = 0
b)( x + 1 )( 7x + 2 )- 2x² – 4x – 2 = 0
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: Bài 1phân tích đa thức thành nhân tử
a) x² – y² – 9x +9
b) 8x – 8y -x² +2xy – y ²
c) x^4 + 81
Bài 2 tìm x
a) ( 2x + 5 )² – ( x + 2 )² = 0
b)( x + 1 )( 7x + 2 )- 2x² – 4x – 2 = 0
Comments ( 1 )
1,\\
a,\\
\left( {x – y – 3} \right)\left( {x + y – 3} \right)\\
b,\\
\left( {x – y} \right).\left( {8 – x + y} \right)\\
c,\\
\left( {{x^2} – 3\sqrt 2 x + 9} \right)\left( {{x^2} + 3\sqrt 2 x + 9} \right)\\
2,\\
a,\\
\left[ \begin{array}{l}
x = – 3\\
x = – \dfrac{7}{3}
\end{array} \right.\\
b,\\
\left[ \begin{array}{l}
x = – 1\\
x = 0
\end{array} \right.
\end{array}\)
1,\\
a,\\
{x^2} – {y^2} – 6x + 9\\
= \left( {{x^2} – 6x + 9} \right) – {y^2}\\
= \left( {{x^2} – 2.x.3 + {3^2}} \right) – {y^2}\\
= {\left( {x – 3} \right)^2} – {y^3}\\
= \left[ {\left( {x – 3} \right) – y} \right].\left[ {\left( {x – 3} \right) + y} \right]\\
= \left( {x – y – 3} \right)\left( {x + y – 3} \right)\\
b,\\
8x – 8y – {x^2} + 2xy – {y^2}\\
= \left( {8x – 8y} \right) – \left( {{x^2} – 2xy + {y^2}} \right)\\
= 8\left( {x – y} \right) – {\left( {x – y} \right)^2}\\
= \left( {x – y} \right).\left[ {8 – \left( {x – y} \right)} \right]\\
= \left( {x – y} \right).\left( {8 – x + y} \right)\\
c,\\
{x^4} + 81\\
= \left( {{x^4} + 18{x^2} + 81} \right) – 18{x^2}\\
= \left[ {{{\left( {{x^2}} \right)}^2} + 2.{x^2}.9 + {9^2}} \right] – {\left( {3\sqrt 2 } \right)^2}{x^2}\\
= {\left( {{x^2} + 9} \right)^2} – {\left( {3\sqrt 2 x} \right)^2}\\
= \left[ {\left( {{x^2} + 9} \right) – 3\sqrt 2 x} \right].\left[ {\left( {{x^2} + 9} \right) + 3\sqrt 2 x} \right]\\
= \left( {{x^2} – 3\sqrt 2 x + 9} \right)\left( {{x^2} + 3\sqrt 2 x + 9} \right)\\
2,\\
a,\\
{\left( {2x + 5} \right)^2} – {\left( {x + 2} \right)^2} = 0\\
\Leftrightarrow \left[ {\left( {2x + 5} \right) – \left( {x + 2} \right)} \right].\left[ {\left( {2x + 5} \right) + \left( {x + 2} \right)} \right] = 0\\
\Leftrightarrow \left( {2x + 5 – x – 2} \right).\left( {2x + 5 + x + 2} \right) = 0\\
\Leftrightarrow \left( {x + 3} \right).\left( {3x + 7} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 3 = 0\\
3x + 7 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = – 3\\
x = – \dfrac{7}{3}
\end{array} \right.\\
b,\\
\left( {x + 1} \right)\left( {7x + 2} \right) – 2{x^2} – 4x – 2 = 0\\
\Leftrightarrow \left( {x + 1} \right)\left( {7x + 2} \right) – 2.\left( {{x^2} + 2x + 1} \right) = 0\\
\Leftrightarrow \left( {x + 1} \right)\left( {7x + 2} \right) – 2{\left( {x + 1} \right)^2} = 0\\
\Leftrightarrow \left( {x + 1} \right).\left[ {\left( {7x + 2} \right) – 2.\left( {x + 1} \right)} \right] = 0\\
\Leftrightarrow \left( {x + 1} \right).\left( {7x + 2 – 2x – 2} \right) = 0\\
\Leftrightarrow \left( {x + 1} \right).5x = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 1 = 0\\
x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = – 1\\
x = 0
\end{array} \right.
\end{array}\)