Issolvi 4(p+q)^2-(2p+q)^2

$$4 ( p + q ) ^ { 2 } - ( 2 p + q ) ^ { 2 }$$

Evalwa

$$q\left(4p+3q\right)$$

Passi tas-Soluzzjonijiet

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

$$4\left(p^{2}+2pq+q^{2}\right)-\left(2p+q\right)^{2}$$

Uża l-propjetà distributtiva biex timmultiplika 4 b'p^{2}+2pq+q^{2}.

$$4p^{2}+8pq+4q^{2}-\left(2p+q\right)^{2}$$

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

$$4p^{2}+8pq+4q^{2}-\left(4p^{2}+4pq+q^{2}\right)$$

Biex issib l-oppost ta' 4p^{2}+4pq+q^{2}, sib l-oppost ta' kull terminu.

$$4p^{2}+8pq+4q^{2}-4p^{2}-4pq-q^{2}$$

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

$$8pq+4q^{2}-4pq-q^{2}$$

Ikkombina 8pq u -4pq biex tikseb 4pq.

$$4pq+4q^{2}-q^{2}$$

Ikkombina 4q^{2} u -q^{2} biex tikseb 3q^{2}.

$$4pq+3q^{2}$$

Espandi

$$4pq+3q^{2}$$

Passi tas-Soluzzjonijiet

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

$$4\left(p^{2}+2pq+q^{2}\right)-\left(2p+q\right)^{2}$$

Uża l-propjetà distributtiva biex timmultiplika 4 b'p^{2}+2pq+q^{2}.

$$4p^{2}+8pq+4q^{2}-\left(2p+q\right)^{2}$$

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

$$4p^{2}+8pq+4q^{2}-\left(4p^{2}+4pq+q^{2}\right)$$

Biex issib l-oppost ta' 4p^{2}+4pq+q^{2}, sib l-oppost ta' kull terminu.

$$4p^{2}+8pq+4q^{2}-4p^{2}-4pq-q^{2}$$

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

$$8pq+4q^{2}-4pq-q^{2}$$

Ikkombina 8pq u -4pq biex tikseb 4pq.

$$4pq+4q^{2}-q^{2}$$

Ikkombina 4q^{2} u -q^{2} biex tikseb 3q^{2}.

$$4pq+3q^{2}$$