\(\Leftrightarrow12x^2-12x+20-10x+17=0\)

\(\Leftrightarrow12x^2-22x+37=0\)

\(\text{Δ}=\left(-22\right)^2-4\cdot12\cdot37=-1292< 0\)

Do đó: Phương trình vô nghiệm

\(12x^2-4\left(3x-5\right)=10x-17\)

\(\Leftrightarrow12x^2-12x+20=10x-17\)

\(\Leftrightarrow12x^2-12x-10x=-17-20\)

\(\Leftrightarrow-12x^2-22x=-37\)

Ta có:  3 x 2 + 2 x + 4 = 8 x 3 + 12 x 2 + 8 x + 1 3 x 2 + 2 x + 5 = ( 2 x + 1 ) 3 + 2 x + 1 3 x 2 + 2 x + 5  (1)

Dễ thấy  3 x 2 + 2 x + 4 > 0  với mọi x. Đặt  u = 3 x 2 + 2 x + 4 v = 2 x + 1 .

 Ta có:   ( 1 ) ⇔ u = v 3 + v u 2 + 1 ⇔ u 3 + u = v 3 + v ⇔ ( u − v ) ( u 2 + u v + v 2 + 1 ) = 0 ⇔ u = v

(Vì  u 2 + u v + v 2 + 1 = u + v 2 2 + 3 4 v 2 + 1 > 0 )

  u = v ⇔ 3 x 2 + 2 x + 4 = 2 x + 1 ⇒ 3 x 2 + 2 x + 4 = 4 x 2 + 4 x + 1 x 2 − 2 x − 3 = 0 ⇒ x = 3   h o a c   x = − 1.

Thử lại, ta nhận x= 3

A

Chọn D

\(\dfrac{-5}{12}\times\dfrac{2}{11}+\dfrac{-5}{11}\times\dfrac{9}{11}+\dfrac{5}{12}\\ =\dfrac{5}{12}\times\left(1-\dfrac{2}{11}\right)+\dfrac{-5}{11}\times\dfrac{9}{11}\\ =\dfrac{5}{12}\times\dfrac{9}{11}+\dfrac{-5}{11}\times\dfrac{9}{11}\\ =\dfrac{9}{11}\times\left(\dfrac{5}{12}-\dfrac{5}{11}\right)\\ =\dfrac{9}{11}\times\dfrac{-5}{132}=-\dfrac{15}{484}\)

c: Ta có: \(x^3-12x^2+48x-64=0\)

\(\Leftrightarrow x-4=0\)

hay x=4

c: Ta có: \(x^3-12x^2+48x-64=0\)

\(\Leftrightarrow x-4=0\)

hay x=4