tính
a, (2/3x^2 + 1/3)^3
b, (4x^2 – 5 + 3y) (3y – 4x^2 – 5)
c, (x-2y) (x^2 +2xy +4y^2)
d, 63x^3 + 27y^6
so sánh
a, A= 2021.2025 và B=2023^2
b, B= 2^16 và B= 3( 2^2 +1) (2^4 +1) (2^8 +1)
tính
a, (2/3x^2 + 1/3)^3
b, (4x^2 – 5 + 3y) (3y – 4x^2 – 5)
c, (x-2y) (x^2 +2xy +4y^2)
d, 63x^3 + 27y^6
so sánh
a, A= 2021.2025 và B=2023^2
b, B= 2^16 và B= 3( 2^2 +1) (2^4 +1) (2^8 +1)
@Queen
$\begin{array}{l}
a){\left( {\dfrac{2}{3}{x^2} + \dfrac{1}{3}} \right)^3}\\
= \dfrac{{{{\left( {2{x^2} + 1} \right)}^3}}}{{{3^3}}}\\
= \dfrac{{{{\left( {2{x^2}} \right)}^3} + 3.{{\left( {2{x^2}} \right)}^2}.1 + 3.2{x^2} + 1}}{{27}}\\
= \dfrac{{8{x^6} + 12{x^4} + 6{x^2} + 1}}{{27}}\\
b)\left( {4{x^2} – 5 + 3y} \right)\left( {3y – 4{x^2} – 5} \right)\\
= \left( {3y – 5 + 4{x^2}} \right)\left( {3y – 5 – 4{x^2}} \right)\\
= {\left( {3y – 5} \right)^2} – {\left( {4{x^2}} \right)^2}\\
= 9{y^2} – 2.3y.5 + {5^2} – 16{x^4}\\
= 9{y^2} – 30y + 25 – 16{x^4}\\
c)\left( {x – 2y} \right)\left( {{x^2} + 2xy + 4{y^2}} \right)\\
= {x^3} – {\left( {2y} \right)^3}\\
= {x^3} – 8{y^3}\\
d)63{x^3} + 27{y^6}\\
= 9\left( {7{x^3} + 3{y^6}} \right)\\
a)A = 2021.2025\\
= \left( {2023 – 2} \right)\left( {2023 + 2} \right)\\
= {2023^2} – {2^2}\\
= {2023^2} – 4 < {2023^2}\\
\Rightarrow A < B\\
b)B = 3.\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\\
= \left( {{2^2} – 1} \right)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\\
= \left( {{2^4} – 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\\
= \left( {{2^8} – 1} \right).\left( {{2^8} + 1} \right)\\
= {2^{16}} – 1 < {2^{16}}\\
\Rightarrow A > B
\end{array}$
Đáp án:
$\begin{array}{l}
a){\left( {\dfrac{2}{3}{x^2} + \dfrac{1}{3}} \right)^3}\\
= \dfrac{{{{\left( {2{x^2} + 1} \right)}^3}}}{{{3^3}}}\\
= \dfrac{{{{\left( {2{x^2}} \right)}^3} + 3.{{\left( {2{x^2}} \right)}^2}.1 + 3.2{x^2} + 1}}{{27}}\\
= \dfrac{{8{x^6} + 12{x^4} + 6{x^2} + 1}}{{27}}\\
b)\left( {4{x^2} – 5 + 3y} \right)\left( {3y – 4{x^2} – 5} \right)\\
= \left( {3y – 5 + 4{x^2}} \right)\left( {3y – 5 – 4{x^2}} \right)\\
= {\left( {3y – 5} \right)^2} – {\left( {4{x^2}} \right)^2}\\
= 9{y^2} – 2.3y.5 + {5^2} – 16{x^4}\\
= 9{y^2} – 30y + 25 – 16{x^4}\\
c)\left( {x – 2y} \right)\left( {{x^2} + 2xy + 4{y^2}} \right)\\
= {x^3} – {\left( {2y} \right)^3}\\
= {x^3} – 8{y^3}\\
d)63{x^3} + 27{y^6}\\
= 9\left( {7{x^3} + 3{y^6}} \right)\\
a)A = 2021.2025\\
= \left( {2023 – 2} \right)\left( {2023 + 2} \right)\\
= {2023^2} – {2^2}\\
= {2023^2} – 4 < {2023^2}\\
\Rightarrow A < B\\
b)B = 3.\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\\
= \left( {{2^2} – 1} \right)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\\
= \left( {{2^4} – 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\\
= \left( {{2^8} – 1} \right).\left( {{2^8} + 1} \right)\\
= {2^{16}} – 1 < {2^{16}}\\
\Rightarrow A > B
\end{array}$