b) \(27x^3-54x^2+36x=8\)

\(\Rightarrow27x^3-54x^2+36x-8=0\)

\(\Rightarrow\left(3x\right)^3-3.\left(3x\right)^2.2+3.3x.2^2-2^3=0\)

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

\(\Rightarrow3x-2=0\)

\(\Rightarrow3x=2\)

\(\Rightarrow x=\dfrac{2}{3}\)

(2x-5)^2-(5+2x)^2=0
<=>(2x-5-5-2x)(2x-5+5+2x)=0
<=>(-10).(4x)=0
<=>(-40x)=0
<=>x =0
27x^3-54x^2+36x=8
<=>27x^3-54x^2+36x-8=0
<=>(3x-2)^3=0
<=>3x-2=0
<=>3x=2
<=>x=2/3

a,\(\left(2x-5\right)^2-\left(5+2x\right)^2=0\)

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

Do \(\left(2x-5\right)^2\ge0\)

\(\Rightarrow\left(2x-5\right)^2=0\)

\(\Rightarrow2x-5=0\)

\(\Rightarrow x=\dfrac{5}{2}\)

@Yukru giúp câu b với!

(3x + 1)² - (3x - 1)²

= (3x + 1 + 3x - 1)(3x + 1 - 3x + 1)

= 6x.2

= 12x

các bạn giúp mih nốt mấy bài kia với mih đang cần gấp

Bài 2:

a: \(=x\left(x^2-4\right)=x\left(x-2\right)\left(x+2\right)\)

b: \(=2xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(2xy-1\right)\)

Bài 3:

=>x^2=5

hay \(x=\pm\sqrt{5}\)

phân tích đa thức sau thành nhân tử:

a) x²+2x-y²+1

=x\(^2\)+2x+1-y\(^2\)

=(x+1)\(^2\)-y\(^2\)

=(x+1-y)(x+1+y)

b) x²+3x-y²+3y

=x\(^2\)-y\(^2\)+3x+3y

=(x-y)(x+y)+3(x+y)

=(x+y)(x-y+3)

c) 3(x+3)-x²+9

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

=3(x+3)-(x-3)(x+3)

=(x+3)[3-(x-3)]

=(x+3)(3-x+3)

a) x2+2x-y2+1

=x+2x+1-y

=(x+1)-y

=(x+1-y)(x+1+y)

b) x2+3x-y2+3y

=x-y+3x+3y

=(x-y)(x+y)+3(x+y)

=(x+y)(x-y+3)

c) 3(x+3)-x2+9

=3(x+3)-(x-3)

=3(x+3)-(x-3)(x+3)

=(x+3)[3-(x-3)]

=(x+3)(3-x+3)

=(x+3)(6-x)

a)

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

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

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

\(\Leftrightarrow (2x+3)(2x+2)=0\Rightarrow \left[\begin{matrix} 2x+3=0\\ 2x+2=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=-\frac{3}{2}\\ x=-1\end{matrix}\right.\)

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

\(\Leftrightarrow (x-5)^2+(x-5)=0\)

\(\Leftrightarrow (x-5)(x-5+1)=0\)

\(\Leftrightarrow (x-5)(x-4)=0\)

\(\Rightarrow \left[\begin{matrix} x-5=0\\ x-4=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=5\\ x=4\end{matrix}\right.\)

c) \((x+2)^3=x+2\)

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

\(\Leftrightarrow (x+2)[(x+2)^2-1]=0\)

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

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

\(\Rightarrow \left[\begin{matrix} x+2=0\\ x+1=0\\ x+3=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=-2\\ x=-1\\ x=-3\end{matrix}\right.\)

d)

\(|3x-1|=(1-3x)^2\)

\(\Leftrightarrow |3x-1|=|3x-1|^2\)

\(\Leftrightarrow |3x-1|^2-|3x-1|=0\)

\(\Leftrightarrow |3x-1|(|3x-1|-1)=0\)

\(\Rightarrow \left[\begin{matrix} |3x-1|=0\\ |3x-1|-1=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} |3x-1|=0\\ |3x-1|=1\end{matrix}\right.\)

\(\Rightarrow \left[\begin{matrix} 3x-1=0\\ 3x-1=\pm 1\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{1}{3}\\ x=\frac{2}{3}\\ x=0\end{matrix}\right.\)

e)

\(2x+(x+3)(3-x)+(x+1)(x-1)=7\)

\(\Leftrightarrow 2x+(3^2-x^2)+(x^2-1^2)=7\)

\(\Leftrightarrow 2x=-1\Rightarrow x=-\frac{1}{2}\)

4(2x + 7) - 3(3x - 2) = 24

8x + 28 - 9x + 6 = 24

-x + 34 = 24

-x = 24 - 34

-x = -10

x = 10

=>  8x +28 -9x+6=24

=>-x=24-6-28

=>-x=-10

=> x=10

  study well

 k nha

 ai k cho mk mk trả lại gấp đôi

=>8x+28-9x+6=24

-x=24-6-28

-x=-10

=>x=10

a: =>x+38+2x=-3-8+2x

=>3x+38=2x-11

=>x=-49

b: \(\Leftrightarrow65+x-15-5x=12-5x\)

=>-4x+50=-5x+12

=>x=-38

c: \(\Leftrightarrow3x+12-7x+21=-3-5x-2=-5x-5\)

=>-4x+33=-5x-5

=>x=-38

d: \(\Leftrightarrow-123+2x+23=x-120\)

=>2x-100=x-120

=>x=-20

e: =>-45+25+5x=16-x

=>5x-20=-x+16

=>6x=36

=>x=6