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

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

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

\(\Rightarrow x+2=0\) (x² +2 loại)

\(\Rightarrow x=-2\)

2) \(x^3+4x^2-2x-8=0\)

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

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

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

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x^2+x+4=0\\x-1=0\end{matrix}\right.\Rightarrow x=1\)

4) \(x^3+x-30=0\)

\(\Rightarrow x^2\left(x-3\right)+3x\left(x-3\right)+10\left(x-3\right)=0\)

\(\Rightarrow\left(x^2+3x+10\right)\left(x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2+3x+10=0\\x-3=0\end{matrix}\right.\Rightarrow x=3.\)

P/S: mấy bạn đừng giải lại nếu như có cách làm khác.

Câu 1+ 2 :

Câu 3:

Câu 4:

1) x³ + 2x² + 2x + 4 = 0

x² ( x + 2 ) + 2 ( x + 2 ) = 0

( x² + 2 ) ( x + 2 ) = 0

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

\(\Leftrightarrow x=-2\)

2) x³ + 4x² - 2x - 8 = 0

x² ( x + 4 ) - 2 ( x + 4 ) = 0

( x² - 2 ) ( x + 4 ) = 0

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\sqrt{2}\\x=-4\end{matrix}\right.\)

a) \(7x^2-28=0\Leftrightarrow7\left(x^2-4\right)=0\Leftrightarrow x^2-4=0\)

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

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

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

c) \(2x^3-50x=0\Leftrightarrow2x\left(x^2-25\right)=0\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\x-5=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\) vậy \(x=0;x=5;x=-5\)

d) \(9\left(3x-2\right)=x\left(2-3x\right)\Leftrightarrow9\left(3x-2\right)=-x\left(3x-2\right)\)

\(\Leftrightarrow9\left(3x-2\right)+x\left(3x-2\right)=0\Leftrightarrow\left(9+x\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}9+x=0\\3x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-9\\3x=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-9\\x=\dfrac{2}{3}\end{matrix}\right.\) vậy \(x=-9;x=\dfrac{2}{3}\)

e) \(5x\left(x-3\right)-2x+6=0\Leftrightarrow5x\left(x-3\right)-2\left(x-3\right)=0\)

\(\Leftrightarrow\left(5x-2\right)\left(x-3\right)=0\) \(\Leftrightarrow\left\{{}\begin{matrix}5x-2=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5x=2\\x=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\x=3\end{matrix}\right.\) vậy \(x=\dfrac{2}{5};x=3\)

Sao bạn không tự làm bớt đi , bài dễ mà

a: \(x^2-4x+3=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

=>x=1 hoặc x=3

b: \(x^2+x-12=0\)

=>(x+4)(x-3)=0

=>x=3 hoặc x=-4

c: \(3x^2+2x-5=0\)

\(\Leftrightarrow3x^2+5x-3x-5=0\)

=>(3x+5)(x-1)=0

=>x=1 hoặc x=-5/3

d: \(x^4-2x^2-3=0\)

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

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

hay \(x\in\left\{\sqrt{3};-\sqrt{3}\right\}\)

16-2x=40-3x

40-16=3x-2x

24=x

x=24

2x-16=40+x

  2x-x=40+16

      x=56

tick nhé cảm ơn

(x-7)(x+2005) = 0

\(\Rightarrow\) x-7 = 0

Vậy x = 7

2x-16=40+x=>x=56

16-2x=40-3x=>x=24

2(4x-2x)-7x=16=>x=-5,(3)

-2(-3-4x)-3(3x+7)=31=>x=-46

9(x+4)-4(2x+15)=3=>x=27

-2[x+(-7)]+(x-3)=12=>x=-1

(x-7)(x+2005)=0 =>x=7;-2005

Bài 1: Tìm x

a) 2x - 15 = -27
b) 2 (x + 1) – 3 = 7

c) 14 – (40 – x) = -27
d) 96 – 2(4 – 5x) = -12

e) – 40 – (– 3 – 33) + (40 – x) = – (– 47)
f) x(3x – 9). (121 – x2) = 0

g) – 62 – (38 + x) + 2x = – 100
h) (x + 1)2.(x2 + 1) = 0

i) (x – 12) – (2x + 31) = 6
k) 17/ (x + 3)3 : 3 – 1 = – 10

Bài 1: a) 2x - 15 = 27 => 2x = 27 + 15 = 42 => x = 42 : 2 = 21. b) 2(x + 1) - 3 = 7 => 2(x + 1) = 7 + 3 = 10 => x + 1 = 10 : 2 = 5 => x = 5 - 1 = 4. c) 14 - (40 - x) = -27 => 40 - x = 14 - (-27) = 41 => x = 40 - 41 = -1. d) 96 - 2(4 - 5x) = -12 => 2(4 - 5x) = 96 - (-12) = 108 => 4 - 5x = 108 : 2 = 54 => 5x = 4 - 54 = -50 => x = (-50) : 5 = -10. 

e) (-40) - [(-3) - 33] + (40 - x) = -(-47) => (-40) - (-36) + 40 - x = 47 => (-40) + 36 + 40 - x = 47 => 36 - x = 47 => x = 36 - 47 = -11. f) x(3x - 9).(121 - 2x) = 0 => x hoặc (3x - 9) hoặc (121 - 2x) bằng 0 => ta có 3 TH: TH1: x = 0 ; TH2: 3x - 9 = 0 => 3x = 0 + 9 = 9 => x = 9 : 3 = 3 ; TH3: 121 - 2x = 0 => 2x = 121 - 0 = 121 (vô lý)(loại). Vậy x ∊ {0;3} 

a)\(2x+3\left(x+4\right)-14=8\)

\(2x+3x+12-14=8\)

\(2x+3x-2=8\)

\(5x-2=8\)

\(5x=8+2\)

\(5x=10\)

\(x=10:5\)

\(x=2\)

b)\(x+2x+3x=4x+16\)

\(x\left(1+2+3\right)=4x+16\)

\(6x-4x=16\)

\(2x=16\)

\(x=16:2\)

\(x=8\)

c)\(2\left(x+1\right)+3\left(x+5\right)=0\)

\(2x+2+3x+15=0\)

\(5x+17=0\)

\(5x=0-17\)

\(5x=-17\)

\(x=-17:5\)

\(x=-\frac{17}{5}\)

a) (2x-5)(x-1)<0 nên âm

=> 2x-5>0 ; x-1<0 hoặc 2x-5<0 ; x-1>0

=> 2x>5 ; x<1 hoặc 2x<5 ; x>1

=> x>=2 ; x<1 hoặc x<=2 ; x>1

chúc bnaj học tốt!!!