a, \(-4x+5+2x-1=3\Leftrightarrow-2x=-1\Leftrightarrow x=\dfrac{1}{2}\)

b, \(-2x+2=2\Leftrightarrow x=0\)

c, \(-2x-6=-8\Leftrightarrow x=1\)

a: \(x^3-4x^2-x+4=0\)

=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)

=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)

=>\(\left(x-4\right)\left(x^2-1\right)=0\)

=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)

b: Sửa đề: \(x^3+3x^2+3x+1=0\)

=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)

=>\(\left(x+1\right)^3=0\)

=>x+1=0

=>x=-1

c: \(x^3+3x^2-4x-12=0\)

=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)

=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)

=>\(\left(x+3\right)\left(x^2-4\right)=0\)

=>\(\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\)

=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)

d: \(\left(x-2\right)^2-4x+8=0\)

=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)

=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)

=>\(\left(x-2\right)\left(x-2-4\right)=0\)

=>(x-2)(x-6)=0

=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

1: P(x)=M(x)+N(x)

=-2x^3+x^2+4x-3+2x^3+x^2-4x-5

=2x^2-8

2: P(x)=0

=>x^2-4=0

=>x=2 hoặc x=-2

3: Q(x)=M(x)-N(x)

=-2x^3+x^2+4x-3-2x^3-x^2+4x+5

=-4x^3+8x+2

\(4x^2+4x-3=0\)

\(\left[\left(2x\right)^2+2.2x.1+1\right]-4=0\)

\(\left(2x+1\right)^2-2^2=0\)

\(\left(2x+1-2\right).\left(2x+1+2\right)=0\) 

\(\left(2x-1\right).\left(2x+3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-1=0\\2x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{3}{2}\end{cases}}}\)

Vậy \(\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{3}{2}\end{cases}}\)

\(x^4-3x^3-x+3=0\)

\(x^3.\left(x-3\right)-\left(x-3\right)=0\)

\(\left(x-3\right).\left(x^3-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x^3-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)

Vậy \(\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

\(x^2.\left(x-1\right)-4x^2+8x-4=0\)

\(x^2.\left(x-1\right)-\left[\left(2x\right)^2-2.2x.2+2^2\right]=0\)

\(x^2.\left(x-1\right)-\left(2x-2\right)^2=0\)

\(x^2.\left(x-1\right)-4.\left(x-1\right)^2=0\)

\(\left(x-1\right).\left[x^2-4.\left(x-1\right)\right]=0\)

\(\left(x-1\right).\left[x^2-2.x.2+2^2\right]=0\)

\(\left(x-1\right).\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)

Vậy \(\begin{cases}x=1\\x=2\end{cases}\)

Tham khảo nhé~

a) \(4x^3-36x=0\)

\(\Leftrightarrow4x\left(x^2-9\right)=0\)

\(\Leftrightarrow4x\left(x+3\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=0\\x+3=0\\x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=3\end{matrix}\right.\)

b) \(\left(x-2\right)^2-4x+8=0\)

\(\Leftrightarrow\left(x-2\right)^2-\left(4x-8\right)=0\)

\(\Leftrightarrow\left(x-2\right)^2-4\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-2-4\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

c) \(x^3+\left(x+3\right)\left(x-9\right)=-27\)

\(\Leftrightarrow\left(x^3+27\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)

\(\Leftrightarrow x\left(x+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)

a) \(x^3+4x=0\)

\(\Rightarrow x\left(x^2+4\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x^2+4=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x^2=-4\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x\in\phi\end{array}\right.\)

Vậy: \(x=0\)

b) \(2\left(5-x\right)=4x-3\)

\(\Rightarrow10-2x=4x-3\)

\(\Rightarrow10+3=4x+2x\)

\(\Rightarrow13=6x\)

\(\Rightarrow x=\frac{13}{6}\)

x³+ 4x=0

<=> x(x²+4)=0

=> x=0 hoặc x²+4=0

Mà: x²+4 >4

=>x=0

b) (5-x).2=4x-3

<=>10-2x=4x-3

<=>10+3= 2x+4x

<=>13=6x

=>x=\(\frac{13}{6}\)

\(\left(4x-3\right)-\left(x+5\right)=3\left(10-x\right)\)

<=> \(4x-3-x-5=30-3x\)

<=>\(3x-8-30+3x=0\)

<=> \(0x=38\)

<=>x thuộc rỗng, không tìm được giá trị x trong trường hợp này