\(\text{1) }3x^3-48x=0\\ \Leftrightarrow x\left(3x^2-48\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x^2-48=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\3x^2=48\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm4\end{matrix}\right.\\ \text{Vậy }x=0\text{ hoặc }x=\pm4\)

\(\text{2) }x^3+x^2-4x=4\\ \Leftrightarrow x^3+x^2-4x-4=0\\ \Leftrightarrow\left(x^3+x^2\right)-\left(4x+4\right)=0\\ \Leftrightarrow x^2\left(x+1\right)-4\left(x+1\right)=0\\ \Leftrightarrow\left(x^2-4\right)\left(x+1\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=1\end{matrix}\right.\\ \text{Vậy }x=2\text{ hoặc }x=-2\text{ hoặc }x=1\)

1) \(3x^3-48x=0\)

\(\Leftrightarrow3x\left(x^2-16\right)=0\)

\(\Leftrightarrow3x\left(x^2-4^2\right)=0\)

\(\Leftrightarrow3x\left(x-4\right)\left(x+4\right)=0\)

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

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

Vậy x=0 ; x=4 ; x=-4

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

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

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\\x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\\x=-2\end{matrix}\right.\)

Vậy x=-1 ; x=2 ; x=-2

x⁴+4x³-4x²-48x-48=0

=> x⁴+4(x³-x²) - 48x = 48

=> x⁴ + 4[x²(x-1)] - 48x = 48 

     \(x^4+4x^3-4x^2-48x-48=0\)

\(\Leftrightarrow\)\(x^4-2x^3-4x^2+6x^3-12x^2-24x+12x^2-24x-48=0\)

\(\Leftrightarrow\)\(x^2\left(x^2-2x-4\right)+6x\left(x^2-2x-4\right)+12\left(x^2-2x-4\right)=0\)

\(\Leftrightarrow\)\(\left(x^2-2x-4\right)\left(x^2+6x+12\right)\)

\(\Leftrightarrow\)\(\left[\left(x-1\right)^2-5\right]\left(x^2+6x+12\right)=0\)

\(\Leftrightarrow\)\(\left(x-1-\sqrt{5}\right)\left(x-1+\sqrt{5}\right)\left(x^2+6x+12\right)=0\)

Ta có:   \(x^2+6x+12=\left(x+3\right)^2+3>0\)

\(\Rightarrow\)\(\orbr{\begin{cases}x-1-\sqrt{5}=0\\x-1+\sqrt{5}=0\end{cases}}\)      

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=1+\sqrt{5}\\x=1-\sqrt{5}\end{cases}}\)

Vậy...

c: =>\(\dfrac{2x-1}{\left(x+5\right)\left(x-1\right)}+\dfrac{x-2}{\left(x-1\right)\left(x-9\right)}=\dfrac{3x-12}{\left(x-9\right)\left(x+5\right)}\)

=>(2x-1)(x-9)+(x-2)(x+5)=(3x-12)(x-1)

=>2x^2-19x+9+x^2+3x-10=3x^2-15x+12

=>-16x-1=-15x+12

=>-x=13

=>x=-13

\(x^3+12x^2+48x+64=x^3+3.x^2.4+3.x.4^2+4^3=\left(x+4\right)^3\)

\(4x^3+32x^2+64x=4x\left(x^2+8x+16\right)=4x\left(x+4\right)^2\)

\(\frac{4x}{\left(x+4\right)^3}=\frac{16x^2}{4x\left(x+4\right)^3},\frac{x-4}{4x\left(x+4\right)^2}=\frac{x^2-16}{4x\left(x+4\right)^3}\)

a) (4x-5)^2 -4 (x-2)^2 =0
⇔(4x-5)²-[2(x-2)]²=0
​⇔(4x-5)²-(2x-4)²=0
​⇔(4x-5-2x+4)(4x-5+2x-4)=0
​⇔(2x-1)(6x-9)=0
​⇔\(\left[{}\begin{matrix}2x-1=0\\6x-9=0\end{matrix}\right.\)
​⇔\(\left[{}\begin{matrix}2x=1\\6x=9\end{matrix}\right.\text{​⇔}\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\)

Bài 2: 

a: \(\Leftrightarrow2x^2-10x-3x-2x^2=26\)

=>-13x=26

hay x=-2

b: \(\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\)

hay \(x\in\left\{1;\dfrac{1}{5}\right\}\)

c: \(\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\)

hay \(x\in\left\{-5;2\right\}\)

\(x^2+4x+4\)

\(=\left(x+2\right)^2\)

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

\(=x^3-27\)

\(x^2-4x+4\)

\(=\left(x-2\right)^2\)