Question

Решите уравнение,пожалуйста

Answer 1

Ответ: 6

$3\sqrt{x+3}-\sqrt{x-2}=7$

$3\sqrt{x+3}=7+\sqrt{x-2}\ \ \ |()^2$

$9(x+3)=49+14\sqrt{x-2}+x-2$

$9x+27-49-x+2=14\sqrt{x-2}$

$8x-20=14\sqrt{x-2}\ \ \ |()^2$

$64x^2 - 320x + 400=196(x-2)$

$64x^2 - 320x + 400=196x-392$

$64x^2 - 516x + 792=0\ \ \ |:4$

$16x^2 - 129x + 198=0$

$D=(-129)^2-4\cdot16\cdot 198=16641-12672=3969$

$\sqrt{D}=\sqrt{3969}=63$

$x_1=\frac{129-63}{2\cdot16}=\frac{66}{32}=\frac{33}{16}$

$x_2=\frac{129+63}{2\cdot16}=\frac{192}{32}=6$

$3\sqrt{\frac{33}{16}+3}-\sqrt{\frac{33}{16}-2}=3\sqrt{\frac{81}{16}}-\sqrt{\frac{1}{16}}=3\cdot\frac{9}{4}-\frac{1}{4}=\frac{27}{4}-\frac{1}{4}=\frac{26}{4}=6,5 \neq 7$

$3\sqrt{6+3}-\sqrt{6-2}=3\sqrt{9}-\sqrt{4}=3\cdot3-2=9-2=7$