\(a,\Leftrightarrow\left(x+2\right)\left(x+2-x+3\right)=0\\ \Leftrightarrow5\left(x+2\right)=0\Leftrightarrow x=-2\\ b,\Leftrightarrow2x\left(x-1\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\ c,\Leftrightarrow\left(x-1-2x-1\right)\left(x-1+2x+1\right)=0\\ \Leftrightarrow3x\left(-x-2\right)=0\Leftrightarrow-3x\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

Bài 15:

a: 2x+x=45

=>3x=45

=>\(x=\frac{45}{3}=15\)

b: 2x+7x=918

=>\(x\cdot\left(7+2\right)=918\)

=>9x=918

=>\(x=\frac{918}{9}=102\)

c: 2x+3x=60+5

=>5x=65

=>\(x=\frac{65}{5}=13\)

d: \(11x+22x=33\cdot2\)

=>33x=66

=>\(x=\frac{66}{33}=2\)

Bài 14:

a: (12-x)(2-x)=0

=>\(\left[\begin{array}{l}12-x=0\\ 2-x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=12\\ x=2\end{array}\right.\)

b: (x-33)(11-x)=0

=>\(\left[\begin{array}{l}x-33=0\\ 11-x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=33\\ x=11\end{array}\right.\)

c: (21-x)(12-x)=0

=>\(\left[\begin{array}{l}21-x=0\\ 12-x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=21\\ x=12\end{array}\right.\)

d: (50-x)(x-150)=0

=>\(\left[\begin{array}{l}50-x=0\\ x-150=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=50\\ x=150\end{array}\right.\)

Bài 13:

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

=>\(\left[\begin{array}{l}x-2=0\\ x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=3\end{array}\right.\)

b: (x-3)(x-4)=0

=>\(\left[\begin{array}{l}x-3=0\\ x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=4\end{array}\right.\)

c: (x-7)(6-x)=0

=>\(\left[\begin{array}{l}x-7=0\\ 6-x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=7\\ x=6\end{array}\right.\)

d: (x-3)(x-13)=0

=>\(\left[\begin{array}{l}x-3=0\\ x-13=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=13\end{array}\right.\)

mình cần gấp

a: Ta có: \(2x\left(x-1\right)-2x^2=-6\)

\(\Leftrightarrow2x^2-2x-2x^2=-6\)

\(\Leftrightarrow x=3\)

b: Ta có: \(2x\left(x-3\right)+5\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)

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

Đặt x² + 3x + 3 = a ;  x² - x - 1 = b ; -2x² - 2x - 1 = c ; -1 = d

Ta nhận thấy a³ + b³ + c³ + d³ = 0 (1) 

và a + b + c + d = 0

Khi đó ta có (1) <=>  (a + b)³ + (c + d)³ - 3ab(a + b) - 3cd(c + d) = 0

<=> ab(a + b) + cd(c + d) = 0

<=> (a + b)(ab - cd) = 0   

<=> \(\left[{}\begin{matrix}a=-b\\ab=cd\end{matrix}\right.\)

Với a = -b ta được x² + 3x + 3 = -x² + x + 1

<=> x² + x + 1 = 0 

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

=> Phương trình vô nghiệm

Với ab = cd 

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

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

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

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

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

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

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

x = -1

Bài 1:
$2x(x+3)+(2x+3)(5-x)=2$

$\Leftrightarrow 2x^2+6x+(10x-2x^2+15-3x)=2$

$\Leftrightarrow 2x^2+6x+7x-2x^2+15=2$

$\Leftrightarrow 13x+15=2$

$\Leftrightarrow 13x=2-15=-13$

$\Leftrightarrow x=-13:13=-1$

Bài 2:

$x-y=4\Rightarrow x=y+4$. Thay vào $xy=5$ thì:

$(y+4)y=5$

$\Leftrightarrow y^2+4y-5=0$

$\Leftrightarrow (y-1)(y+5)=0$

$\Leftrightarrow y=1$ hoặc $y=-5$

Nếu $y=1$ thì $x=y+4=5$. Khi đó $x^3+y^3=5^3+1^3=126$

Nếu $y=-5$ thì $x=y+4=-1$. Khi đó: $x^3+y^3=(-1)^3+(-5)^3=-126$

3:

a: 3^x*3=243

=>3^x=81

=>x=4

b; 2^x*16^2=1024

=>2^x=4

=>x=2

c: 64*4^x=16^8

=>4^x=4^16/4^3=4^13

=>x=13

d: 2^x=16

=>2^x=2^4

=>x=4

b: \(\Leftrightarrow4x+13=5\)

hay x=-2

a) 2x(3x+1) – (2x+3)(3x-2) = 12

\(\Leftrightarrow6x^2+2x-\left(6x^2-4x+9x-6\right)=12\)

\(\Leftrightarrow6x^2+2x-6x^2+4x-9x+6=12\)

\(\Leftrightarrow-3x+6=12\) 

\(\Leftrightarrow-3x=6\)  

\(\Leftrightarrow x=-2\)  

vậy x = -2

 b)  (x+2)² – (x-3)(x+3) = 5

\(\Leftrightarrow\left(x+2\right)^2-\left(x^2-9\right)=5\)

\(\Leftrightarrow x^2+4x+4-x^2+9-5=0\)  

\(\Leftrightarrow4x+8=0\)

\(\Leftrightarrow4x=-8\) 

\(\Leftrightarrow x=-2\)

Vậy  x = -2