Issolvi (sqrt{2}+1)^2+(sqrt{2}-1)^2

$$( \sqrt { 2 } + 1 ) ^ { 2 } + ( \sqrt { 2 } - 1 ) ^ { 2 }$$

Evalwa

$$6$$

Passi tas-Soluzzjonijiet

Uża teorema binomjali \left(a+b\right)^{2}=a^{2}+2ab+b^{2} biex tespandi \left(\sqrt{2}+1\right)^{2}.

$$\left(\sqrt{2}\right)^{2}+2\sqrt{2}+1+\left(\sqrt{2}-1\right)^{2}$$

Il-kwadrat ta' \sqrt{2} huwa 2.

$$2+2\sqrt{2}+1+\left(\sqrt{2}-1\right)^{2}$$

Żid 2 u 1 biex tikseb 3.

$$3+2\sqrt{2}+\left(\sqrt{2}-1\right)^{2}$$

Uża teorema binomjali \left(a-b\right)^{2}=a^{2}-2ab+b^{2} biex tespandi \left(\sqrt{2}-1\right)^{2}.

$$3+2\sqrt{2}+\left(\sqrt{2}\right)^{2}-2\sqrt{2}+1$$

Il-kwadrat ta' \sqrt{2} huwa 2.

$$3+2\sqrt{2}+2-2\sqrt{2}+1$$

Żid 2 u 1 biex tikseb 3.

$$3+2\sqrt{2}+3-2\sqrt{2}$$

Żid 3 u 3 biex tikseb 6.

$$6+2\sqrt{2}-2\sqrt{2}$$

Ikkombina 2\sqrt{2} u -2\sqrt{2} biex tikseb 0.

$$6$$

Fattur

$$2\times 3$$