Question

Bài 1:

a) x (\(x^2\) + 2) + 2x\((1-\dfrac{1}{2}x^2)=4\) 
b) (2x)\(^2\) (x – 1) + x(\(x^2\) + 4x) = 40
c) 3x(x – 2) – 3(\(x^2\) – 3) = 8
d) 2\(x^2\)(4\(x^3\) + 2x) + (\(x^2\) – 2)(- 2x)\(^3\) = 20

Bài 2: 

P = 3x(\(\dfrac{2}{3}\)\(x^2\) − \(3x^4)\) + (3x)\(^2\) (\(x^3\) – 1) + (- 2x + 9)\(x^2\) - 12

 

Answer 1

Bài 2: 

Ta có: \(P=3x\left(\dfrac{2}{3}x^2-3x^4\right)+9x^2\left(x^3-1\right)+x^2\left(-2x+9\right)-12\)

\(=2x^3-9x^5+9x^5-9x^2-2x^3+9x^2-12\)

=-12

Answer 2

Bài 1: 

a: Ta có: \(x\left(x^2+2\right)+2x\left(1-\dfrac{1}{2}x^2\right)=4\)

\(\Leftrightarrow x^3+2x+2x-x^3=4\)

hay x=1

b: Ta có: \(4x^2\left(x-1\right)+x\left(x^2+4x\right)=40\)

\(\Leftrightarrow4x^3-4x^2+x^3+4x^2=40\)

\(\Leftrightarrow5x^3=40\)

hay x=2

c: Ta có: \(3x\left(x-2\right)-3\left(x^2-3\right)=8\)

\(\Leftrightarrow3x^2-6x-3x^2+9=8\)

\(\Leftrightarrow-6x=-1\)

hay \(x=\dfrac{1}{6}\)