Issolvi 1(3a+2b)^2-(3a-2b)^2

$$1 ( 3 a + 2 b ) ^ { 2 } - ( 3 a - 2 b ) ^ { 2 }$$

Evalwa

$$24ab$$

Passi tas-Soluzzjonijiet

Uża teorema binomjali \left(p+q\right)^{2}=p^{2}+2pq+q^{2} biex tespandi \left(3a+2b\right)^{2}.

$$1\left(9a^{2}+12ab+4b^{2}\right)-\left(3a-2b\right)^{2}$$

Uża l-propjetà distributtiva biex timmultiplika 1 b'9a^{2}+12ab+4b^{2}.

$$9a^{2}+12ab+4b^{2}-\left(3a-2b\right)^{2}$$

Uża teorema binomjali \left(p-q\right)^{2}=p^{2}-2pq+q^{2} biex tespandi \left(3a-2b\right)^{2}.

$$9a^{2}+12ab+4b^{2}-\left(9a^{2}-12ab+4b^{2}\right)$$

Biex issib l-oppost ta' 9a^{2}-12ab+4b^{2}, sib l-oppost ta' kull terminu.

$$9a^{2}+12ab+4b^{2}-9a^{2}+12ab-4b^{2}$$

Ikkombina 9a^{2} u -9a^{2} biex tikseb 0.

$$12ab+4b^{2}+12ab-4b^{2}$$

Ikkombina 12ab u 12ab biex tikseb 24ab.

$$24ab+4b^{2}-4b^{2}$$

Ikkombina 4b^{2} u -4b^{2} biex tikseb 0.

$$24ab$$

Espandi

$$24ab$$

Passi tas-Soluzzjonijiet

Uża teorema binomjali \left(p+q\right)^{2}=p^{2}+2pq+q^{2} biex tespandi \left(3a+2b\right)^{2}.

$$1\left(9a^{2}+12ab+4b^{2}\right)-\left(3a-2b\right)^{2}$$

Uża l-propjetà distributtiva biex timmultiplika 1 b'9a^{2}+12ab+4b^{2}.

$$9a^{2}+12ab+4b^{2}-\left(3a-2b\right)^{2}$$

Uża teorema binomjali \left(p-q\right)^{2}=p^{2}-2pq+q^{2} biex tespandi \left(3a-2b\right)^{2}.

$$9a^{2}+12ab+4b^{2}-\left(9a^{2}-12ab+4b^{2}\right)$$

Biex issib l-oppost ta' 9a^{2}-12ab+4b^{2}, sib l-oppost ta' kull terminu.

$$9a^{2}+12ab+4b^{2}-9a^{2}+12ab-4b^{2}$$

Ikkombina 9a^{2} u -9a^{2} biex tikseb 0.

$$12ab+4b^{2}+12ab-4b^{2}$$

Ikkombina 12ab u 12ab biex tikseb 24ab.

$$24ab+4b^{2}-4b^{2}$$

Ikkombina 4b^{2} u -4b^{2} biex tikseb 0.

$$24ab$$