Tìm x biết:
4x3-9x=0
x3+8x=0
-x3+9x=0
Tìm x:
a) 64x5-4x3=0
b) x3-81x=0
c) x(5-3x)-5+3x=0
d) 4x3+12x2-9x-27=0
e) x4-6x2+7=0
Tìm x:
a) 64x5-4x3=0
b) x3-81x=0
c) x(5-3x)-5+3x=0
d) 4x3+12x2-9x-27=0
e) x4-6x2+7=0
b) x3 – 5x2 – x + 5 = 0.
c) x3 – x2 – 25x + 25 = 0
d) 4x3 – 8x2 – 9x + 18 = 0.
b: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\\x=-1\end{matrix}\right.\)
c: \(\Leftrightarrow\left(x-1\right)\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=5\\x=-5\end{matrix}\right.\)
Bài 5. Tìm x, biết:
a) x (2x - 7) + 4x -14 = 0
b) x3 - 9x = 0
c) 4x2 -1 - 2(2x -1)2 = 0
d) (x3 - x2 ) - 4x2 + 8x - 4 = 0
\(a,\Leftrightarrow x\left(2x-7\right)+2\left(2x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{7}{2}\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-2\left(2x-1\right)^2=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1-4x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(-2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Tìm x biết:
a) 5x(x – 2) + 3x – 6 = 0
b) x 3 – 9 x = 0
Tìm x :
x3 + 9x=0
\(x^3+9x=0\)
\(x\left(x^2+9\right)=0\)
\(x^2+9=0\)
\(x^2=-9\)
mà \(x^2\text{≥}\text{∀}x\)
⇒ Không tồn tại x thỏa mãn
Lời giải:
$x^3+9x=0$
$\Leftrightarrow x(x^2+9)=0$
\(\Rightarrow \left[\begin{matrix} x=0\\ x^2+9=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=0(\text{chọn})\\ x^2=-9<0(\text{vô lý})\end{matrix}\right.\)
Bài 5:
1) a) Cho hai đa thức:
P (x) = 5x2 + 3x3 - 5x2 + 2x3 – 2 +4x – 4x2 + x3
Q(x) = 6x – x3 + 5 – 4x3 + 6 – 3x2 – 7x2
Tính M(x) = P(x) + Q(x)
b) Tìm C(x) biết: (5x2 + 9x – 3x4 + 7x3 -12) + C(x) = -2x3 + 9 – 6x + 7x4 -2x3
2) Tìm nghiệm của các đa thức sau
a) 4x - b) x2 – 4x +3
a: P(x)=6x^3-4x^2+4x-2
Q(x)=-5x^3-10x^2+6x+11
M(x)=x^3-14x^2+10x+9
b: \(C\left(x\right)=7x^4-4x^3-6x+9+3x^4-7x^3-5x^2-9x+12\)
=10x^4-11x^3-5x^2-15x+21
Tìm x biết:
a) x3 - 7x2 - 9x + 63 = 0
b) x3 - 3x2 + 3x - 1 + 2.(x2 - x) = 0
a) (x – 2)(x2 + 2x + 4) – x( x2 +2) = 12 b) (x – 3)2 – (x+2)(x–2) = 16
c) x3 – 9x = 0 d) x3 – 6x2 + 9x – 54 = 0
giúp e vs ạ
\(a,\Leftrightarrow x^3-8-x^3-2x=12\Leftrightarrow-2x=20\Leftrightarrow x=-10\\ b,\Leftrightarrow x^2-6x+9-x^2+4=16\Leftrightarrow=-6x=3\Leftrightarrow x=-\dfrac{1}{2}\\ c,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-6\right)+9\left(x-6\right)=0\\ \Leftrightarrow\left(x^2+9\right)\left(x-6\right)=0\\ \Leftrightarrow x=6\left(x^2+9>0\right)\)