Question

Розв'язати рiвняння: √3х+1-√8x-7=1.

Answer 1

ОДЗ :

$\displaystyle\bf\\\left \{ {{3x+1\geq 0} \atop {8x-7\geq 0}} \right. \ \ \ \Rightarrow \ \ \ \left \{ {{x\geq -\dfrac{1}{3} } \atop {x\geq \dfrac{7}{8} }} \right. \ \ \ \Rightarrow \ \ \ x\geq \frac{7}{8} \\\\\\\sqrt{3x+1} -\sqrt{8x-7} =1\\\\\Big(\sqrt{3x+1}\Big)^{2} =\Big(1+\sqrt{8x-7}\Big)^{2} \\\\3x+1=1+2\sqrt{8x-7} +8x-7\\\\2\sqrt{8x-7} =8x-6-3x-1\\\\\Big(2\sqrt{8x-7}\Big)^{2} =\Big(7-5x\Big)^{2} \\\\4\cdot\Big(8x-7\Big)=25x^{2} -70x+49\\\\32x-28=25x^{2} -70x+49$

$\displaystyle\bf\\25x^{2} -102x+77=0\\\\D=(-102)^{2}-4\cdot 25\cdot 77= 10404-7700=2704=52^{2} \\\\\\x_{1} =\frac{102+52}{50} =\frac{154}{50} =3,08\\\\\\x_{2} =\frac{102-52}{50} =1\\\\\\Otvet \ : \ 3,08 \ ; \ 1$