Whakaoti i te (2a-5)^2/2-(a-3)^2+1/2geqa^2

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

Whakaoti mō a

$$a\leq 1$$

Ngā Upane Otinga

Whakareatia ngā taha e rua o te whārite ki te 2. I te mea he tōrunga te 2, kāore e huri te ahunga koreōrite.

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

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(2a-5\right)^{2}.

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

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(a-3\right)^{2}.

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

Hei kimi i te tauaro o a^{2}-6a+9, kimihia te tauaro o ia taurangi.

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

Whakamahia te āhuatanga tohatoha hei whakarea te 2 ki te \frac{4a^{2}-20a+25}{2}-a^{2}+6a-9.

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

Tuhia te 2\times \frac{4a^{2}-20a+25}{2} hei hautanga kotahi.

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

Me whakakore te 2 me te 2.

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

Pahekotia te 4a^{2} me -2a^{2}, ka 2a^{2}.

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

Pahekotia te -20a me 12a, ka -8a.

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

Tangohia te 18 i te 25, ka 7.

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

Tāpirihia te 7 ki te 1, ka 8.

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

Tangohia te 2a^{2} mai i ngā taha e rua.

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

Pahekotia te 2a^{2} me -2a^{2}, ka 0.

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

Tangohia te 8 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$-8a\geq -8$$

Whakawehea ngā taha e rua ki te -8. I te mea he tōraro a -8, ka huri te ahunga koreōrite.

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

Whakawehea te -8 ki te -8, kia riro ko 1.

$$a\leq 1$$