Issolvi (2a-5)^2/2-(a-3)^2+1/2geqa^2

$$\frac { ( 2 a - 5 ) ^ { 2 } } { 2 } - ( a - 3 ) ^ { 2 } + \frac { 1 } { 2 } \geq a ^ { 2 }$$

Solvi għal a

$$a\leq 1$$

Passi tas-Soluzzjonijiet

Immultiplika ż-żewġ naħat tal-ekwazzjoni b'2. Peress li 2 huwa pożittiv, id-direzzjoni tal-inugwaljanza inbidlet.

$$2\left(\frac{\left(2a-5\right)^{2}}{2}-\left(a-3\right)^{2}\right)+1\geq 2a^{2}$$

Uża teorema binomjali \left(a-b\right)^{2}=a^{2}-2ab+b^{2} biex tespandi \left(2a-5\right)^{2}.

$$2\left(\frac{4a^{2}-20a+25}{2}-\left(a-3\right)^{2}\right)+1\geq 2a^{2}$$

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

$$2\left(\frac{4a^{2}-20a+25}{2}-\left(a^{2}-6a+9\right)\right)+1\geq 2a^{2}$$

Biex issib l-oppost ta' a^{2}-6a+9, sib l-oppost ta' kull terminu.

$$2\left(\frac{4a^{2}-20a+25}{2}-a^{2}+6a-9\right)+1\geq 2a^{2}$$

Uża l-propjetà distributtiva biex timmultiplika 2 b'\frac{4a^{2}-20a+25}{2}-a^{2}+6a-9.

$$2\times \frac{4a^{2}-20a+25}{2}-2a^{2}+12a-18+1\geq 2a^{2}$$

Esprimi 2\times \frac{4a^{2}-20a+25}{2} bħala frazzjoni waħda.

$$\frac{2\left(4a^{2}-20a+25\right)}{2}-2a^{2}+12a-18+1\geq 2a^{2}$$

Annulla 2 u 2.

$$4a^{2}-20a+25-2a^{2}+12a-18+1\geq 2a^{2}$$

Ikkombina 4a^{2} u -2a^{2} biex tikseb 2a^{2}.

$$2a^{2}-20a+25+12a-18+1\geq 2a^{2}$$

Ikkombina -20a u 12a biex tikseb -8a.

$$2a^{2}-8a+25-18+1\geq 2a^{2}$$

Naqqas 18 minn 25 biex tikseb 7.

$$2a^{2}-8a+7+1\geq 2a^{2}$$

Żid 7 u 1 biex tikseb 8.

$$2a^{2}-8a+8\geq 2a^{2}$$

Naqqas 2a^{2} miż-żewġ naħat.

$$2a^{2}-8a+8-2a^{2}\geq 0$$

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

$$-8a+8\geq 0$$

Naqqas 8 miż-żewġ naħat. Xi ħaġa mnaqqsa minn żero tagħti numru negattiv.

$$-8a\geq -8$$

Iddividi ż-żewġ naħat b'-8. Peress li -8 huwa negattiv, id-direzzjoni tal-inugwaljanza inbidlet.

$$a\leq \frac{-8}{-8}$$

Iddividi -8 b'-8 biex tikseb1.

$$a\leq 1$$