Whakaoti i te sqrt{2/3-5x}-sqrt{3x+1/2}=0.x>0

$$\sqrt { \frac { 2 } { 3 } - 5 x } - \sqrt { 3 x + \frac { 1 } { 2 } } = 0 . x > 0$$

Whakaoti mō x

$$x=\frac{1}{48}\approx 0.020833333$$

Ngā Upane Otinga

Me tango -\sqrt{3x+\frac{1}{2}} mai i ngā taha e rua o te whārite.

$$\sqrt{\frac{2}{3}-5x}=\sqrt{3x+\frac{1}{2}}$$

Pūruatia ngā taha e rua o te whārite.

$$\left(\sqrt{\frac{2}{3}-5x}\right)^{2}=\left(\sqrt{3x+\frac{1}{2}}\right)^{2}$$

Tātaihia te \sqrt{\frac{2}{3}-5x} mā te pū o 2, kia riro ko \frac{2}{3}-5x.

$$\frac{2}{3}-5x=\left(\sqrt{3x+\frac{1}{2}}\right)^{2}$$

Tātaihia te \sqrt{3x+\frac{1}{2}} mā te pū o 2, kia riro ko 3x+\frac{1}{2}.

$$\frac{2}{3}-5x=3x+\frac{1}{2}$$

Tangohia te 3x mai i ngā taha e rua.

$$\frac{2}{3}-5x-3x=\frac{1}{2}$$

Pahekotia te -5x me -3x, ka -8x.

$$\frac{2}{3}-8x=\frac{1}{2}$$

Tangohia te \frac{2}{3} mai i ngā taha e rua.

$$-8x=\frac{1}{2}-\frac{2}{3}$$

Ko te maha noa iti rawa atu o 2 me 3 ko 6. Me tahuri \frac{1}{2} me \frac{2}{3} ki te hautau me te tautūnga 6.

$$-8x=\frac{3}{6}-\frac{4}{6}$$

Tā te mea he rite te tauraro o \frac{3}{6} me \frac{4}{6}, me tango rāua mā te tango i ō raua taurunga.

$$-8x=\frac{3-4}{6}$$

Tangohia te 4 i te 3, ka -1.

$$-8x=-\frac{1}{6}$$

Whakawehea ngā taha e rua ki te -8.

$$x=\frac{-\frac{1}{6}}{-8}$$

Tuhia te \frac{-\frac{1}{6}}{-8} hei hautanga kotahi.

$$x=\frac{-1}{6\left(-8\right)}$$

Whakareatia te 6 ki te -8, ka -48.

$$x=\frac{-1}{-48}$$

Ka taea te hautanga \frac{-1}{-48} te whakamāmā ki te \frac{1}{48} mā te tango tahi i te tohu tōraro i te taurunga me te tauraro.

$$x=\frac{1}{48}$$

Whakakapia te \frac{1}{48} mō te x i te whārite \sqrt{\frac{2}{3}-5x}-\sqrt{3x+\frac{1}{2}}=0.

$$\sqrt{\frac{2}{3}-5\times \frac{1}{48}}-\sqrt{3\times \frac{1}{48}+\frac{1}{2}}=0$$

Whakarūnātia. Ko te uara x=\frac{1}{48} kua ngata te whārite.

$$0=0$$

Ko te whārite \sqrt{\frac{2}{3}-5x}=\sqrt{3x+\frac{1}{2}} he rongoā ahurei.

$$x=\frac{1}{48}$$