Whakaoti i te 4(x-3)^2-(2x-5)^2geq2

$$4 ( x - 3 ) ^ { 2 } - ( 2 x - 5 ) ^ { 2 } \geq 2$$

Whakaoti mō x

$$x\leq \frac{9}{4}$$

Ngā Upane Otinga

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

$$4\left(x^{2}-6x+9\right)-\left(2x-5\right)^{2}\geq 2$$

Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te x^{2}-6x+9.

$$4x^{2}-24x+36-\left(2x-5\right)^{2}\geq 2$$

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

$$4x^{2}-24x+36-\left(4x^{2}-20x+25\right)\geq 2$$

Hei kimi i te tauaro o 4x^{2}-20x+25, kimihia te tauaro o ia taurangi.

$$4x^{2}-24x+36-4x^{2}+20x-25\geq 2$$

Pahekotia te 4x^{2} me -4x^{2}, ka 0.

$$-24x+36+20x-25\geq 2$$

Pahekotia te -24x me 20x, ka -4x.

$$-4x+36-25\geq 2$$

Tangohia te 25 i te 36, ka 11.

$$-4x+11\geq 2$$

Tangohia te 11 mai i ngā taha e rua.

$$-4x\geq 2-11$$

Tangohia te 11 i te 2, ka -9.

$$-4x\geq -9$$

Whakawehea ngā taha e rua ki te -4. I te mea he tōraro a -4, ka huri te ahunga koreōrite.

$$x\leq \frac{-9}{-4}$$

Ka taea te hautanga \frac{-9}{-4} te whakamāmā ki te \frac{9}{4} mā te tango tahi i te tohu tōraro i te taurunga me te tauraro.

$$x\leq \frac{9}{4}$$