Toán Lớp 8: d) (x2 + x – 3).(x2 – x + 3)
e) (6×3 – 7×2 – x + 2) : (2x + 1) h) (27×3 +1) : (9×2 – 3x + 1) – (3x –19)
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Toán Lớp 8: d) (x2 + x – 3).(x2 – x + 3)
e) (6×3 – 7×2 – x + 2) : (2x + 1) h) (27×3 +1) : (9×2 – 3x + 1) – (3x –19)
giúp với ạ hứa vote
Comments ( 1 )
d)\left( {{x^2} + x – 3} \right).\left( {{x^2} – x + 3} \right)\\
= {\left( {{x^2}} \right)^2} – {\left( {x – 3} \right)^2}\\
= {x^4} – \left( {{x^2} – 6x + 9} \right)\\
= {x^4} – {x^2} + 6x – 9\\
e)\left( {6{x^3} – 7{x^2} – x + 2} \right):\left( {2x + 1} \right)\\
= \left( {6{x^3} + 3{x^2} – 10{x^2} – 5x + 4x + 2} \right):\left( {2x + 1} \right)\\
= \left( {2x + 1} \right)\left( {3{x^2} – 5x + 2} \right):\left( {2x + 1} \right)\\
= 3{x^2} – 5x + 2\\
h)\\
\left( {27{x^3} + 1} \right):\left( {9{x^2} – 3x + 1} \right) – \left( {3x – 19} \right)\\
= \left( {3x + 1} \right).\left( {9{x^2} – 3x + 1} \right):\left( {9{x^2} – 3x + 1} \right) – 3x + 19\\
= 3x + 1 – 3x + 19\\
= 20
\end{array}$