Tìm x bt:
a) x3+5x2−4x−20=0
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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:
a) x3 +3x2 - 10x = 0
b) x3 - 5x2 - 14x =0
c) x3 + 5x2- 24x =0
Giải giúp mình với ạ !
Mình cảm ơn !
x3+3x2-10x=0
=>x(3+3.2-10)=0
=>x=0
x3-5x2-14x=0
=>x(3-5.2-14)=0
=>x=0
x3+5x2-24x=0
=>x(3+5.2-24)=0
=>x=0
Câu a)
\(x^3+3x^2-10=0\Rightarrow x\left(x^2+3x-10\right)=0\Rightarrow x\left(x^2-2x+5x-10\right)=0\Rightarrow x\left(x\left(x-2\right)+5\left(x-2\right)\right)=0\Rightarrow x\left(x+5\right)\left(x-2\right)=0\)
\(\Rightarrow x=0;x=5;x=2\)
\(x^3-9x^2+26x-24\)
\(=x^3-4x^2-5x^2+20x+6x-24\)
\(=\left(x-4\right)\left(x^2-5x+6\right)\)
\(=\left(x-4\right)\left(x-2\right)\left(x-3\right)\)
a)7.2x=56⇔2x=8⇔2x=23⇔x=3
b)(2x+1)3=9.81⇔(2x+1)3=93⇔2x+1=9⇔2x=8⇔x=4
c)x3=82⇔x3=26⇔x=22⇔x=4
d)4.2x-3=1⇔4.2x=4⇔2x=1⇔2x=20⇔x=0
e)2.3x=162⇔3x=81⇔3x=34⇔x=4
a) x = -1. b) x = 4 hoặc x = 5.
c) x = ± 2 . d) x = 1 hoặc x = 2.
\(\Leftrightarrow\left(x-5\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\\x=-1\end{matrix}\right.\)
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: \(\Leftrightarrow\left(x-1\right)\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=5\\x=-5\end{matrix}\right.\)
a) \(5x\left(x-2000\right)-x+2000=0\)
\(\Leftrightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\)
\(\Leftrightarrow\left(5x-1\right)\left(x-2000\right)=0\)
\(\Leftrightarrow x\in\left\{\frac{1}{5};2000\right\}\)
a: 5x-20xy
\(=5x\cdot1-5x\cdot4y=5x\left(1-4y\right)\)
b: \(x^2-9=\left(x-3\right)\left(x+3\right)\)
c: \(x^2-2xy+y^2-z^2\)
\(=\left(x-y\right)^2-z^2\)
=(x-y-z)(x-y+z)
d: \(5x\left(x-1\right)-2\left(x-1\right)=\left(x-1\right)\left(5x-2\right)\)
e; \(x^2+4x+3=x^2+x+3x+3\)
=x(x+1)+3(x+1)
=(x+1)(x+3)
f: \(x^3-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left\lbrack\left(x+y\right)^2-1\right\rbrack\)
=(x+y)(x+y-1)(x+y+1)
g: \(x^2-x-y^2-y\)
\(=\left(x^2-y^2\right)-\left(x+y\right)\)
=(x-y)(x+y)-(x+y)
=(x+y)(x-y-1)
h: \(16x-5x^2-3\)
\(=-5x^2+15x+x-3\)
=-5x(x-3)+(x-3)
=(x-3)(-5x+1)
i: \(x^3-4x=x\left(x^2-4\right)=x\left(x-2\right)\left(x+2\right)\)
j: \(2x^2-6x=2x\cdot x-2x\cdot3=2x\left(x-3\right)\)
k: \(x^3-3x^2-4x+12\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\cdot\left(x-2\right)\left(x+2\right)\)
l: \(x^2-y^2-5x+5y\)
=(x-y)(x+y)-5(x-y)
=(x-y)(x+y-5)
a) x3+5x2−4x−20=0
x2.x+5.x2-4x-4.5=0
x2.(x+5)-4.(x+5)=0
(x+5).(x2-4)=0
Ta có:x2-4=(x+2)(x-2)
Do đó: (x+5).(x2-4)=0
(x+5)(x+2)(x-2)=0
\(\Rightarrow\left[\begin{array}{nghiempt}x+5=0\\x+2=0\\x-2=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=-5\\x=-2\\x=2\end{array}\right.\)
Vậy x=-5;-2;2
a)x3+5x2-4x-20=0
=>x2(x+5)-4(x+5)=0
=>(x2-4)(x+5)=0
=>(x-2)(x+2)(x+5)=0
=>x-2=0 hoặc x+2=0 hoặc x+5=0
\(\Rightarrow\begin{cases}x=-5\\x=-2\\x=2\end{cases}\)