Odpowiedź :
1)
[tex](4x)^2=4^2\cdot x^2=16\cdot x^2=16x^2\\[/tex]
2)
[tex](\frac{1}{3}x^2y)^2=(\frac{1}{3})^2\cdot (x^2)^2\cdot y^2=\frac{1}{9}\cdot x^4\cdot y^2=\frac{1}{9}x^4y^2\\[/tex]
3)
[tex](-\frac{2xy}{3})^4=\frac{(2xy)^4}{3^4}=\frac{2^4x^4y^4}{81}=\frac{16x^4y^4}{81}\\[/tex]
Cześć!
Szczegółowe wyjaśnienie:
a) (4x)² = 4x² · x² = 16 · x² = 16x²
b)
[tex]( \frac{1}{3} x {}^{2} y) {}^{2} \\ ( \frac{1}{3}) {}^{2} \cdot \: (x {}^{2} ) {}^{2} \cdot \: y {}^{2} = \frac{1}{9} x {}^{4} y {}^{2} [/tex]
c)
[tex]( - \frac{2xy}{3} ) {}^{4} = ( \frac{2xy}{3} ){}^{4} = \frac{(2xy {}^{} ) {}^{4} }{3 {}^{4} } = \frac{16x {}^{4}y {}^{4} }{81} [/tex]