Tìm x biết:
(x - 3).(x2 + 3x +9) - (3x - 17) = x3 - 12
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a: \(\Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
a: Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x^2-3\right)=0\)
\(\Leftrightarrow x^3-27-x^3+3x=0\)
\(\Leftrightarrow x=9\)
b: Ta có: \(8x^4+x=0\)
\(\Leftrightarrow x\left(8x^3+1\right)=0\)
\(\Leftrightarrow x\left(2x+1\right)\left(4x^2-2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)
a) (x - 140) : 7 = 33 - 23 . 3
(x - 140) : 7 = 27 - 8 . 3 = 27 - 24 = 3
x - 140 = 3 x 7 = 21
x = 21 + 140 = 161
b) x3 . x2 = 28 : 23
x5 = 25
=> x = 2
c) (x + 2) . ( x - 4) = 0
x = -2 hoặc 4
d) 3x-3 - 32 = 2 . 32 =
3x-3 - 9 = 2 . 9 = 18
3x-3 = 18 + 9 = 27
3x-3 = 33
=> x - 3 = 3
x = 3 + 3 = 6
(x−3)(x2+3x+9)−(3x−17)=x3−12(x−3)(x2+3x+9)−(3x−17)=x3−12
⇔x3−27−3x+17−x3+12=0⇔x3−27−3x+17−x3+12=0
⇔2−3x=0⇔2−3x=0
⇔3x=2⇒x=23
a) x = 1; x = - 1 3 b) x = 2.
c) x = 3; x = -2. d) x = -3; x = 0; x = 2.
e: =>-40+3+33+40-x=47
=>36-x=47
=>x=-11
f: =>x(x-3)(11-x)(11+x)=0
hay \(x\in\left\{0;3;11;-11\right\}\)
g: =>-62-38-x+2x=-100
=>x-100=-100
hay x=0
i: =>x-12-2x-31=6
=>-x-43=6
=>x+43=-6
hay x=-49
h: =>(x+1)=0
=>x=-1
f: =>x(x-3)(x+11)(x-11)=0
hay \(x\in\left\{0;3;-11;11\right\}\)
\(\left(x-3\right)\left(x^2+3x+9\right)-\left(3x-17\right)=x^3-12\)
\(x^3+3x^2+9x-3x^2-9x-37-3x+17=x^3-12\)
\(x^3+3x^2+9x-3x^2-9x-3x-x^3=37-17+12\)
\(-3x=32\)
\(x=\frac{32}{-3}=-\frac{32}{3}\)
Vậy x = \(-\frac{32}{3}\)
(x-3)(x\(^2\)+3x+9)-(3x-17)=x\(^3\)-12
(x\(^3\)-3\(^3\))-(3x-17)=x\(^3\)-12
3\(^3\)-3\(^3\)-3x+17=x\(^3\)-12
x\(^3\)-27-3x+17=x\(^3\)-12
x\(^3\)-27-3x+17-x\(^3\)=-12
-27-3x+17=-12
-27-3x=-29
3x=2
x=\(\frac{2}{3}\)