Toán Lớp 9: giai phuong trinh
√2x²+11x+19+√2x²+5x+7=3(x+2)
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Toán Lớp 9: giai phuong trinh
√2x²+11x+19+√2x²+5x+7=3(x+2)
Comments ( 2 )
$\sqrt{2x^2+11x+19}+\sqrt{2x^2+5x+7}=3\left(x+2\right)$
$⇔\sqrt{2x^2+11x+19}+\sqrt{2x^2+5x+7}=3x+6$
$⇔2x^2+5x+7=\frac{9x^2}{4}+6x+4$
$⇔11x^2+41x-6\sqrt{2x^2+5x+7}x+43-12\sqrt{2x^2+5x+7}=2x^2+11x+19$
⇔(x + 2) (-3 x + 2 sqrt(x (2 x + 5) + 7) – 4) = 0
⇔x (11 x – 6 sqrt(x (2 x + 5) + 7) + 41) – 12 sqrt(x (2 x + 5) + 7) + 43 = x (2 x + 11) + 19
⇔2 sqrt(x (2 x + 5) + 7) = 3 x + 4
$⇔8x^2+20x+28=9x^2+24x+16$
\(⇔\left[ \begin{array}{l}x=-6(ktm)\\x=2(tm)\end{array} \right.\)
Vậy pt có nghiệm S= {2}
\sqrt{2x^2 + 11x + 19} + \sqrt{2x^2 + 5x + 7} = 3(x + 2)
ĐKXĐ: x > – 2
Đặt: \sqrt{2x^2 + 11x + 19} = a
\sqrt{2x^2 + 5x + 7} = b
⇒ a^2 – b^2 = (2x^2 + 11x + 19) – (2x^2 + 5x + 7) = 6x + 12 = 6(x + 2)
Ta có hệ PT sau:
$\begin{cases} a + b = 3(x + 2)\ (1)\\a^2 – b^2 = 6(x + 2) = 2.3.(x – 2)\ (2)\\ \end{cases}$
(2) ⇔ a^2 – b^2 = 2.(a + b)
⇔ (a + b)(a – b) = 2(a + b)
⇔ a – b = 2
⇔ a = b + 2
Thay vào PT (1), ta có:
(b + 2) + b = 3(x + 2)
⇔ 2b = 3x + 4
⇒ 2\sqrt{2x^2 + 5x + 7} = 3x + 4
⇔ (2\sqrt{2x^2 + 5x + 7})^2 = (3x + 4)^2 $\text{(ĐK: x >}$ -4/3)
⇔ 4(2x^2 + 5x + 7) = 9x^2 + 24x + 16
⇔ x^2 + 4x – 12 = 0
⇔ x^2 – 2x + 6x – 12 = 0
⇔ x(x – 2) + 6(x – 2) = 0
⇔ (x – 2)(x + 6) = 0
⇔ $\left[\begin{matrix} x = 2\ \text{(TM)}\\ x = -6\ \text{(Loại)}\end{matrix}\right.$
Vậy PT có nghiệm duy nhất là: x = 2