<=> 75x+23 = 3x-1  or  75x+23 = -(3x-1)

<=> x = -1/3  or  x = -11/39

*vn:vô nghiệm.

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vn\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\pm\sqrt{2}\)

-Vậy \(S=\left\{\pm\sqrt{2}\right\}\).

b. \(16x^2-8x+5=0\)

\(\Leftrightarrow16x^2-8x+1+4=0\)

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

-Vậy S=∅.

c. \(2x^3-x^2-8x+4=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\pm2\end{matrix}\right.\)

-Vậy \(S=\left\{\dfrac{1}{2};\pm2\right\}\).

d. \(3x^3+6x^2-75x-150=0\)

\(\Leftrightarrow3x^2\left(x+2\right)-75\left(x+2\right)=0\)

\(\Leftrightarrow3\left(x+2\right)\left(x^2-25\right)=0\)

\(\Leftrightarrow3\left(x+2\right)\left(x+5\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\pm5\end{matrix}\right.\)

-Vậy \(S=\left\{-2;\pm5\right\}\)

1,=\(x^2-3x-2x^2+6x=-x^2+3x\)

2,=\(3x^2-x-5+15x=3x^2+14x-5\)

3,=\(5x+15-6x^2-6x=-6x^2-x+15\)

4,=\(4x^2+12x-x-3=4x^2+11x-3\)

5: =>(x+5)^3=0

=>x+5=0

=>x=-5

6: =>(2x-3)^2=0

=>2x-3=0

=>x=3/2

7: =>(x-6)(x-10)=0

=>x=10 hoặc x=6

8: \(\Leftrightarrow x^3-12x^2+48x-64=0\)

=>(x-4)^3=0

=>x-4=0

=>x=4

\(b,4x^2-x-5=0\)

\(\Leftrightarrow4x^2-5x+4x-5=0\)

\(\Leftrightarrow x\left(4x-5\right)+4x-5=0\)

\(\Leftrightarrow\left(4x-5\right)\left(x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{5}{4}\end{cases}}\)

Bài 2

\(a,x^3+5x^2+3x-9\)

\(\Leftrightarrow x^3-x^2+6x^2-6x+9x-9\)

\(\Leftrightarrow x^2\left(x-1\right)+6x\left(x-1\right)+9\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+6x+9\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x+3\right)^2\)

b,\(x^3-7x-6\)

\(\Leftrightarrow x^3-3x^2+3x^2-9x+2x-6\)

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

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

\(\Leftrightarrow\left(x-3\right)\left(x+1\right)\left(x+2\right)\)

c,\(3x^3-7x^2+17x-5\)

\(\Leftrightarrow3x^3-x^2-6x^2+2x+15x-5\)

\(\Leftrightarrow x^2\left(3x-1\right)-2x\left(3x-1\right)+5\left(3x-1\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-2x+5\right)\)

\(4x^2-x-5=0\)

<=>  \(4x^2+4x-5x-5=0\)

<=>   \(4x\left(x+1\right)-5\left(x+1\right)=0\)

<=>  \(\left(x+1\right)\left(4x-5\right)=0\)

tự lm tiếp

Bài 1:

a)\(28x^3+15x^2+75x+125=0\)

\(\Leftrightarrow\left(4x+5\right)\left(7x^2-5x+25\right)=0\)

Dễ thấy: \(7x^2-5x+25=7\left(x-\frac{5}{14}\right)^2+\frac{675}{28}>0\)

\(\Rightarrow4x+5=0\Rightarrow x=-\frac{5}{4}\)

b)\(4x^2-x-5=0\)

\(\Leftrightarrow\left(x+1\right)\left(4x-5\right)=0\)

\(\Rightarrow x=-1;x=\frac{5}{4}\)

Bài 2:

a)\(x^3+5x^2+3x-9\)

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

b)\(x^3-7x-6\)

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

c)\(3x^3-7x^2+17x-5\)

\(=\left(3x-1\right)\left(x^2-2x+5\right)\)

\(4x^2-x-5=0\)

<=> \(4x^2+4x-5x-5=0\)

<=>  \(4x\left(x+1\right)-5\left(x+1\right)=0\)

<=>  \(\left(x+1\right)\left(4x-5\right)=0\)

tự giải nốt

b)\(3x^3+6x^2-75x-150=0\Leftrightarrow3\left(x^3+2x^2-25x-50\right)=0\Leftrightarrow x^3+2x^2-25x-50=0\)

<=>\(x^2\left(x+2\right)-25\left(x+2\right)=0\Leftrightarrow\left(x^2-25\right)\left(x+2\right)=0\Leftrightarrow\left(x-5\right)\left(x+5\right)\left(x+2\right)=0\)

<=>x-5=0 hoặc x+5=0 hoặc x+2=0<=>x=5 hoặc x=-5 hoặc x=-2

c)\(2x^5-3x^4+6x^3-8x^2+3=0\Leftrightarrow2x^5+x^4-4x^4-2x^3+8x^3+4x^2-12x^2+3=0\)

<=>\(x^4\left(2x+1\right)-2x^3\left(2x+1\right)+4x^2\left(2x+1\right)-3\left(4x^2-1\right)=0\)

<=>\(x^4\left(2x+1\right)-2x^3\left(2x+1\right)+4x^2\left(2x+1\right)-3\left(2x-1\right)\left(2x+1\right)=0\)

<=>\(\left(2x+1\right)\left(x^4-2x^3+4x^2-6x+3\right)=0\)

<=>\(\left(2x+1\right)\left(x^4-2x^3+x^2+3x^2-6x+3\right)=0\)

<=>\(\left(2x+1\right)\left[x^2\left(x^2-2x+1\right)+3\left(x^2-2x+1\right)\right]=0\)

<=>\(\left(2x+1\right)\left(x^2+3\right)\left(x^2-2x+1\right)=0\Leftrightarrow\left(2x+1\right)\left(x^2+3\right)\left(x-1\right)^2=0\)

Vì \(x^2\ge0\Rightarrow x^2+3\ge3>0\Rightarrow\orbr{\begin{cases}2x+1=0\\\left(x-1\right)^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}\)

a) 2x³ - x² - 8x + 4 = 0

x².(2x - 1) - 4.(2x - 1) = 0

(x² - 4)(2x - 1) = 0

\(\Rightarrow\orbr{\begin{cases}x^2-4=0\\2x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x^2=4\\x=\frac{1}{2}\end{cases}}\)

Với x² = 4

=> x = 2 hoặc x = -2

=> x = {-2 ; 2 ; \(\frac{1}{2}\))

a) x²(2x-1) - 4(2x-1) = 0 <=> (2x-1)(x²- 4)=0 <=> x=\(\frac{1}{2}\)hay x=-2 hay x= 2