Đáp án:x=3
Giải thích các bước giải:$$ đặt u = \sqrt {2{x^2} + 3x + 10} $$
$$v = \sqrt {2{x^2} + x + 4} $$
suy ra $${u^2} - {v^2} = 6x + 6 = > 3(x + 1) = \frac{{{u^2} - {v^2}}}{2}$$
pttt: $$\eqalign{
& u + v = \frac{{{u^2} - {v^2}}}{2} \cr
& < = > (u - v)(u + v) = 2(u + v) \cr
& < = > (u + v)(u - v - 2) = 0 \cr
& < = > [_{u - v - 2 = 0}^{u + v = 0} \cr
& = > [_{\sqrt {2{x^2} + 7x + 10} - \sqrt {2{x^2} + x + 4} - 2 = 0}^{\sqrt {2{x^2} + 7x + 10} + \sqrt {2{x^2} + x + 4} = 0(loại)} \cr} $$
ta có $$\eqalign{
& \sqrt {2{x^2} + 7x + 10} - \sqrt {2{x^2} + x + 4} - 2 = 0 \cr
& < = > \sqrt {2{x^2} + 7x + 10} = \sqrt {2{x^2} + x + 4} + 2 \cr
& < = > 2{x^2} + 7x + 10 = 2{x^2} + x + 4 + 4 + 4\sqrt {2{x^2} + x + 4} \cr
& < = > 3x + 1 = 2\sqrt {2{x^2} + x + 4} \cr
& < = > \{ _{9{x^2} + 6x + 1 = 8{x^2} + 4x + 16}^{x \geqslant \frac{{ - 1}}{3}} \cr
& < = > \{ _{{x^2} + 2x - 15 = 0}^{x \geqslant \frac{{ - 1}}{3}} \cr
& < = > \{ _{[_{x = - 5}^{x = 3}}^{x \geqslant \frac{{ - 1}}{3}} < = > x = 3 \cr} $$
