Bài 1:a) x (\(x^2\) + 2) + 2x\((1-\dfrac{1}{2}x^2)=4\) b) (2x)\(^2\) (x – 1) + x(\(x^2\) + 4x) = 40c) 3x(x – 2) – 3(\(x^2\) – 3) = 8d) 2\(x^2\)(4\(x^3\) + 2x) + (\(x^2\) – 2)(- 2x)\(^3\) = 20Bài 2: P = 3x(\(\dfrac{2}{3}\)\(x^2\) − \(3x^4)\) + (3x)\(^2\) (\(x^3\) – 1) + (- 2x + 9)\(x^2\) -...
Đọc tiếp
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
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
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 \(x=\dfrac{1}{6}\)