B8:
A) (X + 1 )( 1+ x – x mũ 2 + x mũ 3 – x mũ 4 ) + ( x- 1 )( 1+ x + x + x mũ 2 + x mũ 3 + x mũ 4 )
B) ( 2b mũ 2 – 2 – 5b + 6b mũ 3 )( 3+ 3b mũ 2 – b )
B9 :
A) ( y -5 )( y + 8 ) – ( y+4 )( y- 1)
B) y mũ 4 – ( y mũ 2 – 1 )( y mũ 2 + 1 )
Nhanh nhất đc 5 sao với tlhn ạ 😉
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
8,\\
a,\\
\left( {x + 1} \right)\left( {1 + x – {x^2} + {x^3} – {x^4}} \right) + \left( {x – 1} \right)\left( {1 + x + {x^2} + {x^3} + {x^4}} \right)\\
= \left( {x + {x^2} – {x^3} + {x^4} – {x^5} + 1 + x – {x^2} + {x^3} – {x^4}} \right) + \left( {x + {x^2} + {x^3} + {x^4} + {x^5} – 1 – x – {x^2} – {x^3} – {x^4}} \right)\\
= \left( {2x + 1 – {x^5}} \right) + \left( {{x^5} – 1} \right)\\
= 2x\\
b,\\
\left( {2{b^2} – 2 – 5b + 6{b^3}} \right)\left( {3 + 3{b^2} – b} \right)\\
= \left( {6{b^2} + 6{b^4} – 2{b^3}} \right) + \left( { – 6 – 6{b^2} + 2b} \right) + \left( { – 15b – 15{b^3} + 5{b^2}} \right) + \left( {18{b^3} + 18{b^5} – 6{b^4}} \right)\\
= 18{b^5} + {b^3} + 5{b^2} – 13b – 6\\
9,\\
a,\\
\left( {y – 5} \right)\left( {y + 8} \right) – \left( {y + 4} \right)\left( {y – 1} \right)\\
= \left( {{y^2} + 3y – 40} \right) – \left( {{y^2} + 3y – 4} \right)\\
= – 36\\
b,\\
{y^4} – \left( {{y^2} – 1} \right)\left( {{y^2} + 1} \right)\\
= {y^4} – \left( {{y^4} – 1} \right)\\
= 1
\end{array}\)