Tìm x:
x3+27+(x+3)(x-9)=0
Tìm x:
x3 + 9x = 0
Lời giải:
$x^3+9x=0$
$\Leftrightarrow x(x^2+9)=0$
$\Leftrightarrow x=0$ hoặc $x^2+9=0$
Xét TH $x^2+9=0\Leftrightarrow x^2=-9<0$ (vô lý)
Vậy $x=0$
Tìm x biết x^3+27+(x+3)(x-9)=0
=x3+33+(x+3)(x-9)
=(x+3)(x2-3x+9)+(x+3)(x-9)
=(x+3)(x2-3x+9+x-9)
=(x+3)(x2-2x)
=(x+3)(x-2)x
Tìm x biết: 4x².(x-2)-x+2=0 x³+27+(x+3).(x-9)
\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)
a: Ta có: \(4x^2\left(x-2\right)-x+2=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
b: Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=0\\x=2\end{matrix}\right.\)
Tìm x biết: x3 + 27 + (x + 3)(x - 9) = 0
\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-2=0\\x+3=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=2\\x=-3\end{array}\right.\)
<=> (x+3)(x2+3x+9)+(x+3)(x - 9)
<=> (x+3)(x2-3x+9+x - 9)=0
<=> (x+3)(x2-2x)=0
<=> (x+3)x(x-2)=0
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=-3\\x=2\end{array}\right.\)
\(x^3+27+\left(x+3\right)\left(x-9\right)\)=0
\(\Leftrightarrow x^3+3^3+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x+3=0\\x-2=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-3\\x=2\end{array}\right.\)
Vậy x=0 hoặc x=-3 hoặc x=2
Tìm x:
X^3 + 27 + (x+3)×(x-9)=0
(x+3)(x^2-3x+9)+(x+3)(x-9)=0
(x+3)(x^2-3x+9+x-9)=0
(x+3)(x^2-2x)=0
rồi bạn tự làm đi
(x+3)x(x-2)=0
=> x=-3;0;2
OK. NHỚ K CHO MK NHÉ
tìm x, biết: x3+27+(x+3)(x-9)=0
Đáp án của bạn Nguyễn Tiến Hải còn thiếu trường hợp:
(x + 1/2)2 = 25/4
TH1: x + 1/2 = 5/2 và giải như bạn Hải
TH2: x + 1/2 = -5/2
x = -3
=> (x+3)(x2-3x+9) + (x+3)(x-9) =0
=> (x+3)(x2-2x)=0 => (x+3)(x-2)x=0
=> x=-3 hoặc x=2 hoặc x=0
tìm số nguyên x , biết :
a) x + 546 = 46
b) 2x - 19 x 3 = 27
c) x + 12= 23 + 3 x 3^4
d) x- 12 = 3 - 3 x 2^4
e) ( 27 - x ) nhân ( x +9 ) = 0
f) ( -x) x ( x-43) = 0
a) \(x+546=46\\ x=46-546\\ x=-500\)
b) \(2x-19\times3=27\\ 2x-57=27\\ 2x=27+57\\ 2x=84\\ x=84:2\\ x=42\)
c) \(x+12=23+3\times3^4\\ x+12=23+3\times81\\ x=23+243-12\\ x=254\)
d) \(x-12=3-3\times2^4\\ x-12=3-3\times16\\ x=3-48+12\\ x=-33\)
e) \(\left(27-x\right)\left(x+9\right)=0\\ \Rightarrow\left[{}\begin{matrix}27-x=0\\x+9=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=27\\x=-9\end{matrix}\right.\)
f) \(\left(-x\right)\left(x-43\right)=0\\ \Rightarrow\left[{}\begin{matrix}-x=0\\x-43=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=43\end{matrix}\right.\)
a) x + 546 = 46
x = 46 - 546
x = -500
b) 2x - 19 x 3 = 27
2x - 57 = 27
2x = 27 + 57
2x = 84
x = 84 : 2
x = 42
c) x + 12 = 23 + 3 x 34
x + 12 = 23 + 3 x 81
x + 12 = 23 + 243
x + 12 = 266
x = 266 - 12
x = 254
d) x - 12 = 3 - 3 x 24
x - 12 = 3 - 3 x 16
x - 12 = 3 - 48
x - 12 = -45
x = -45 + 12
x = -33
e) (27 - x) x (x + 9) = 0
TH1: 27 - x = 0
x = 27 - 0
x = 27
TH2: x + 9 = 0
x = 0 + 9
x = 9
⇒ x = 27 hoặc x = 9
f) (-x) x (x - 43) = 0
TH1: -x = 0
⇒ x = 0
TH2: x - 43 = 0
x = 0 + 43
x = 43
⇒ x = 0 hoặc x = 43.
tìm x :x^3-6x^2+12x-8=0 16x^2-9(x+1)^2=0 -27+27*x-ax^2+x^3=0
Tìm x
a, (x-2)^2-3x^2+6x=0
b, x^3+27+(x+3)(x-9)=0
\(a,\Leftrightarrow\left(x-2\right)^3-3x\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-2-3x\right)=0\\ \Leftrightarrow\left(x-2\right)\left(-2x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\\ b,\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\\ \Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)