3(5−x)2+1​=13−2x

Bình phương:

\(9 \left[\right. \left(\right. 5 - x \left.\right)^{2} + 1 \left]\right. = \left(\right. 13 - 2 x \left.\right)^{2}\)

Khai triển:

\(9 \left(\right. x^{2} - 10 x + 25 + 1 \left.\right) = 169 - 52 x + 4 x^{2}\) \(9 x^{2} - 90 x + 234 = 169 - 52 x + 4 x^{2}\)

Chuyển vế:

\(5 x^{2} - 38 x + 65 = 0\)

\(\Delta = \left(\right. - 38 \left.\right)^{2} - 4 \cdot 5 \cdot 65 = 1444 - 1300 = 144\) \(\sqrt{\Delta} = 12\)

cosα=a12​+b12​​a22​+b22​​∣a1​a2​+b1​b2​∣​

Với:

\(\left(\right. a_{1} , b_{1} \left.\right) = \left(\right. 3 , - 4 \left.\right) , \left(\right. a_{2} , b_{2} \left.\right) = \left(\right. 12 , - 5 \left.\right)\)

Tính:

\(a_{1} a_{2} + b_{1} b_{2} = 3 \cdot 12 + \left(\right. - 4 \left.\right) \left(\right. - 5 \left.\right) = 36 + 20 = 56\) \(\sqrt{a_{1}^{2} + b_{1}^{2}} = \sqrt{9 + 16} = 5\) \(\sqrt{a_{2}^{2} + b_{2}^{2}} = \sqrt{144 + 25} = 13\) \(cos ⁡ \alpha = \frac{56}{5 \cdot 13} = \frac{56}{65}\)

\(cos ⁡ \alpha = \frac{56}{65}\)

Δ=(−9)2−4⋅2⋅6=81−48=33 \(x = \frac{9 \pm \sqrt{33}}{4}\)

\(x = \frac{9 + \sqrt{33}}{4} \text{ho}ặ\text{c} x = \frac{9 - \sqrt{33}}{4}\)a,x2−9x−10=0 \(\Delta = \left(\right. - 9 \left.\right)^{2} - 4 \cdot 1 \cdot \left(\right. - 10 \left.\right) = 81 + 40 = 121\) \(\sqrt{\Delta} = 11\)