a)  \(x^3+3x^2+3x+2=0\)

<=>  \(x^3+x^2+x+2x^2+2x+2=0\)

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

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

tự làm

b) \(x^4-2x^3+2x-1=0\)

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

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

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

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

tự làm

c)   \(x^4-3x^3-6x^2+8x=0\)

<=>   \(x\left(x^3-3x^2-6x+8\right)=0\)

<=>  \(x\left[\left(x^3+x^2-2x\right)-\left(4x^2+4x-8\right)\right]=0\)

<=>\(x\left[x\left(x^2+x-2\right)-4\left(x^2+x-2\right)\right]=0\)

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

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

tự làm

\(\left(x+2\right)^3-x^2\left(x-6\right)-4=0\\ \Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2-4=0\\ \Leftrightarrow12x-12=0\\ \Leftrightarrow12x=12\\ \Leftrightarrow x=1\)

\(6x^2-\left(2x-3\right)\left(3x+2\right)=1\\ \Leftrightarrow6x^2-\left[3x.\left(2x-3\right)+2.\left(2x-3\right)\right]=1\\ \Leftrightarrow6x^2-\left(6x^2-9x+4x-6\right)=1\\ \Leftrightarrow6x^2-\left(6x^2-5x-6\right)=1\\ \Leftrightarrow6x^2-6x^2+5x+6=1\\ \Leftrightarrow5x=-5\\ \Leftrightarrow x=-1\)

1) \(\Rightarrow10x-16-12x+15=12x-16+11\)

\(\Rightarrow14x=4\Rightarrow x=\dfrac{2}{7}\)

2) \(\Rightarrow4x^2+4x+1-4x^2+13x-3-15=0\)

\(\Rightarrow17x=17\Rightarrow x=1\)

3) \(\Rightarrow\left(3x-1\right)\left(2x-7+6x-5\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

2: Ta có: \(\left(2x+1\right)^2-\left(4x-1\right)\left(x-3\right)-15=0\)

\(\Leftrightarrow4x^2+4x+1-4x^2+12x+x-3-15=0\)

\(\Leftrightarrow17x=17\)

hay x=1

a. 3x(x-2)-x+2=0

3x(x-2)-(x-2)=0

(3x-1)(x-2)=0

=>\(\hept{\begin{cases}3x-1=0\\x-2=0\end{cases}}\)

=> \(\hept{\begin{cases}3x=1\\x=2\end{cases}}\)

=>\(\hept{\begin{cases}x=\frac{1}{3}\\x=2\end{cases}}\)

vậy x thuộc (1/3;2)

b. 4x(x-3)-2x+6=0

4x(x-3) -2(x-3)=0

(4x-2)(x-3)

=>*4x-2=0

4x=2

x=1/2

*x-3=0

x=3

vậy x thuộc (1/2;3)

a. 3x(x-2)-x+2=0

=>(x-2)(3x-1)=0

=>x-2=0 hoặc 3x-1=0

*x-2=0=>x=2

*3x-1=0=>x=\(\frac{1}{3}\)

Vậy x=2;x=\(\frac{1}{3}\)

b. 4x(x-3)-2x+6=0

=>(x-3)(4x-2)=0

=>(x-3)=0 hoặc 4x-2=0

*x-3=0=>x=3

*4x-2=0=>\(\frac{1}{2}\)

Vậy x=3; x=\(\frac{1}{2}\)

câu c, d,e thì chút nx mk gửi cho nha

\(a,-2x\left(2-3x\right)+3\left(-5+7x-6x^2\right)=-4\)

\(\Rightarrow-4x+6x^2-15+21x-18x^2=-4\)

\(\Rightarrow-12x^2+17x-11=0\)

\(\Rightarrow12x^2-17x+11=0\)

\(\Rightarrow9x^2-2.3.\frac{17}{6}x+\left(\frac{17}{6}\right)^2-\left(\frac{17}{6}\right)^2+11=0\)

\(\Rightarrow\left(3x-\frac{17}{6}\right)^2+\frac{107}{36}=0VN\)

Không có gt x thỏa mãn 

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

\(\Rightarrow3x-9x^2+12x^3-12x^3+18x^2=0\)

\(\Rightarrow9x^2+3x=0\)

\(\Rightarrow3x\left(3x+1\right)=0\)

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

hjhuhh

hfjjfjfjfrtrf

Rút gọn hết ta được :

a/ 41x - 17 = -21

=> 41x = -4 => x = 4/41

b/ 34x - 17 = 0 

=> 34x = 17

=> x = 17/34 = 1/2

c/ 19x + 56 = 52 

=> 19x = -4

=> x = -4/19

d/ 20x² - 16x - 34 = 10x² + 3x - 34

=> 10x² - 19x = 0

=> x(10x - 19) = 0

=> x = 0 

hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10

Vậy x = 0 ; x = 19/10

Rút gọn hết ta được :

a/ 41x - 17 = -21

=> 41x = -4 => x = 4/41

b/ 34x - 17 = 0

=> 34x = 17

=> x = 17/34 = 1/2

c/ 19x + 56 = 52

=> 19x = -4

=> x = -4/19

d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34

=> 10x 2 - 19x = 0

=> x(10x - 19) = 0

=> x = 0 hoặc 10x - 19 = 0

=> 10x = 19

=> x = 19/10

Vậy x = 0 ; x = 19/10 

a) ( 6x - 3 ) ( 2x + 4 ) + ( 4x - 1 ) ( 5 - 3x ) = -21

<=> 12x² + 24x - 6x - 12 + 20x - 12x² - 5 + 3x = -21

<=> 41x = -21 + 12 + 5 

<=> 41x = -4

<=> x = -4/41

\(x^3-2x^2+x-2=0\\ \Leftrightarrow x^2\left(x-2\right)+\left(x-2\right)=0\\ \Leftrightarrow\left(x^2+1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=2\end{matrix}\right.\\ Vậy:x=2\\ ---\\ 2x\left(3x-5\right)=10-6x\\ \Leftrightarrow6x^2-10x-10+6x=0\\ \Leftrightarrow6x^2-4x-10=0\\ \Leftrightarrow6x^2+6x-10x-10=0\\ \Leftrightarrow6x\left(x+1\right)-10\left(x+1\right)=0\\ \Leftrightarrow\left(6x-10\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}6x-10=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)

\(4-x=2\left(x-4\right)^2\\ \Leftrightarrow4-x=2\left(x^2-8x+16\right)\\ \Leftrightarrow2x^2-16x+32+x-4=0\\ \Leftrightarrow2x^2-15x+28=0\\ \Leftrightarrow2x^2-8x-7x+28=0\\ \Leftrightarrow2x\left(x-4\right)-7\left(x-4\right)=0\\ \Leftrightarrow\left(2x-7\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\\ ---\\ 4-6x+x\left(3x-2\right)=0\\ \Leftrightarrow4-6x+3x^2-2x=0\\ \Leftrightarrow3x^2-8x+4=0\\ \Leftrightarrow3x^2-6x-2x+4=0\\ \Leftrightarrow3x\left(x-2\right)-2\left(x-2\right)=0\\ \Leftrightarrow\left(3x-2\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)

a ) \(x^3+3x^2+3x+2=0\)

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

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

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

\(\Leftrightarrow x+1=-1\)

\(\Leftrightarrow x=-2\)

Vậy \(x=-2\)

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

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

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

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

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

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

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

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

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

a, \(x^3+3x^2+3x+2=0\)

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

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

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

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

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

\(\Leftrightarrow\left(x-1\right)\left[\left(x^3-x^2\right)\left(x-1\right)\right]=0\)

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

\(\Leftrightarrow x=1\)

a ) x^3+3x^2+3x+2=0x3+3x2+3x+2=0

\Leftrightarrow x^3+3x^2+3x+1+1=0⇔x3+3x2+3x+1+1=0

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

\Leftrightarrow\left(x+1\right)^3=-1⇔(x+1)3=−1

\Leftrightarrow x+1=-1⇔x+1=−1

\Leftrightarrow x=-2⇔x=−2

Vậy x=-2x=−2

b ) x^4-2x^3+2x-1=0x4−2x3+2x−1=0

\Leftrightarrow x^4-1-2x\left(x^2-1\right)=0⇔x4−1−2x(x2−1)=0

\Leftrightarrow\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)=0⇔(x2−1)(x2+1)−2x(x2−1)=0

\Leftrightarrow\left(x^2-1\right)\left(x^2+1-2x\right)=0⇔(x2−1)(x2+1−2x)=0

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

\Leftrightarrow\left(x-1\right)^3\left(x+1\right)=0⇔(x−1)3(x+1)=0

\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x=-1\end{matrix}\right.⇔[(x−1)3=0x+1=0​⇔[x−1=0x=−1​

\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.⇔[x=1x=−1​

Vậy \left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.[x=1x=−1​  

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^