Whakaoti i te 4(x^2+1)+8(3x-4)>4x^2

$$4 ( x ^ { 2 } + 1 ) + 8 ( 3 x - 4 ) > 4 x ^ { 2 }$$

Whakaoti mō x

$$x>\frac{7}{6}$$

Ngā Upane Otinga

Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te x^{2}+1.

$$4x^{2}+4+8\left(3x-4\right)>4x^{2}$$

Whakamahia te āhuatanga tohatoha hei whakarea te 8 ki te 3x-4.

$$4x^{2}+4+24x-32>4x^{2}$$

Tangohia te 32 i te 4, ka -28.

$$4x^{2}-28+24x>4x^{2}$$

Tangohia te 4x^{2} mai i ngā taha e rua.

$$4x^{2}-28+24x-4x^{2}>0$$

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

$$-28+24x>0$$

Me tāpiri te 28 ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

$$24x>28$$

Whakawehea ngā taha e rua ki te 24. I te mea he tōrunga te 24, kāore e huri te ahunga koreōrite.

$$x>\frac{28}{24}$$

Whakahekea te hautanga \frac{28}{24} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$x>\frac{7}{6}$$