Rozwiązanie poniżej ;-)
Wzór, który wykorzystam w zadaniu:
[tex](a-b)(a+b)=a^2-b^2[/tex]
a)
[tex](x-5)(x+5)=x^2-5^2=\boxed{x^2-25}[/tex]
b)
[tex](10+x)(10-x)=10^2-x^2=\boxed{100-x^2}[/tex]
c)
[tex]\left(a-\dfrac{1}{4}b\right)\left(a+\dfrac{1}{4}b\right)=a^2-\left(\dfrac{1}{4}b\right)^2=\boxed{a^2-\frac{1}{16}b^2}\\[/tex]
d)
[tex]\left(\dfrac{2}{3}a+b\right)\left(\dfrac{2}{3}a-b\right)=\left(\dfrac{2}{3}a\right)^2-b^2=\boxed{\frac{4}{9}a^2-b^2}\\[/tex]
e)
[tex](0,3s-2t)(2t+0,3s)=(0,3s)^2-(2t)^2=\boxed{0,09s^2-4t^2}\\[/tex]
f)
[tex](0,5s+0,6t)(0,6t-0,5s)=(0,6t)^2-(0,5s)^2=\boxed{0,36t^2-0,25s^2}\\[/tex]
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Rozwiązanie poniżej ;-)
Wzór, który wykorzystam w zadaniu:
[tex](a-b)(a+b)=a^2-b^2[/tex]
a)
[tex](x-5)(x+5)=x^2-5^2=\boxed{x^2-25}[/tex]
b)
[tex](10+x)(10-x)=10^2-x^2=\boxed{100-x^2}[/tex]
c)
[tex]\left(a-\dfrac{1}{4}b\right)\left(a+\dfrac{1}{4}b\right)=a^2-\left(\dfrac{1}{4}b\right)^2=\boxed{a^2-\frac{1}{16}b^2}\\[/tex]
d)
[tex]\left(\dfrac{2}{3}a+b\right)\left(\dfrac{2}{3}a-b\right)=\left(\dfrac{2}{3}a\right)^2-b^2=\boxed{\frac{4}{9}a^2-b^2}\\[/tex]
e)
[tex](0,3s-2t)(2t+0,3s)=(0,3s)^2-(2t)^2=\boxed{0,09s^2-4t^2}\\[/tex]
f)
[tex](0,5s+0,6t)(0,6t-0,5s)=(0,6t)^2-(0,5s)^2=\boxed{0,36t^2-0,25s^2}\\[/tex]