Issolvi (11x+3)^2-(6x+1)(6x-1)

$$( 11 x + 3 ) ^ { 2 } - ( 6 x + 1 ) ( 6 x - 1 )$$

Evalwa

$$85x^{2}+66x+10$$

Passi tas-Soluzzjonijiet

Uża teorema binomjali \left(a+b\right)^{2}=a^{2}+2ab+b^{2} biex tespandi \left(11x+3\right)^{2}.

$$121x^{2}+66x+9-\left(6x+1\right)\left(6x-1\right)$$

Ikkunsidra li \left(6x+1\right)\left(6x-1\right). Il-multiplikazzjoni tista' tiġi ttrasformata fid-differenza tal-kwadrati li jużaw ir-regola: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Ikkwadra 1.

$$121x^{2}+66x+9-\left(\left(6x\right)^{2}-1\right)$$

Espandi \left(6x\right)^{2}.

$$121x^{2}+66x+9-\left(6^{2}x^{2}-1\right)$$

Ikkalkula 6 bil-power ta' 2 u tikseb 36.

$$121x^{2}+66x+9-\left(36x^{2}-1\right)$$

Biex issib l-oppost ta' 36x^{2}-1, sib l-oppost ta' kull terminu.

$$121x^{2}+66x+9-36x^{2}+1$$

Ikkombina 121x^{2} u -36x^{2} biex tikseb 85x^{2}.

$$85x^{2}+66x+9+1$$

Żid 9 u 1 biex tikseb 10.

$$85x^{2}+66x+10$$

Espandi

$$85x^{2}+66x+10$$

Passi tas-Soluzzjonijiet

Uża teorema binomjali \left(a+b\right)^{2}=a^{2}+2ab+b^{2} biex tespandi \left(11x+3\right)^{2}.

$$121x^{2}+66x+9-\left(6x+1\right)\left(6x-1\right)$$

Ikkunsidra li \left(6x+1\right)\left(6x-1\right). Il-multiplikazzjoni tista' tiġi ttrasformata fid-differenza tal-kwadrati li jużaw ir-regola: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Ikkwadra 1.

$$121x^{2}+66x+9-\left(\left(6x\right)^{2}-1\right)$$

Espandi \left(6x\right)^{2}.

$$121x^{2}+66x+9-\left(6^{2}x^{2}-1\right)$$

Ikkalkula 6 bil-power ta' 2 u tikseb 36.

$$121x^{2}+66x+9-\left(36x^{2}-1\right)$$

Biex issib l-oppost ta' 36x^{2}-1, sib l-oppost ta' kull terminu.

$$121x^{2}+66x+9-36x^{2}+1$$

Ikkombina 121x^{2} u -36x^{2} biex tikseb 85x^{2}.

$$85x^{2}+66x+9+1$$

Żid 9 u 1 biex tikseb 10.

$$85x^{2}+66x+10$$