a: ĐKXĐ: \(x\in R\)

\(\sqrt{x^2-4x+4}=7\)

=>\(\sqrt{\left(x-2\right)^2}=7\)

=>|x-2|=7

=>\(\left[{}\begin{matrix}x-2=7\\x-2=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-5\end{matrix}\right.\)

b: ĐKXĐ: x>=-3

\(\sqrt{4x+12}-3\sqrt{x+3}+\dfrac{4}{3}\cdot\sqrt{9x+27}=6\)

=>\(2\sqrt{x+3}-3\sqrt{x+3}+\dfrac{4}{3}\cdot3\sqrt{x+3}=6\)

=>\(3\sqrt{x+3}=6\)

=>\(\sqrt{x+3}=2\)

=>x+3=4

=>x=1(nhận)

a: \(x^3-4x^2-x+4=0\)

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

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

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

=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)

b: Sửa đề: \(x^3+3x^2+3x+1=0\)

=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)

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

=>x+1=0

=>x=-1

c: \(x^3+3x^2-4x-12=0\)

=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)

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

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

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

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

d: \(\left(x-2\right)^2-4x+8=0\)

=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)

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

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

=>(x-2)(x-6)=0

=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

Ta có: 5x(x-3)+4x-12=0

=> 5x(x-3)+4(x-3)=0

=> (5x+4).(x-3)=0

\(\Rightarrow\orbr{\begin{cases}5x+4=0\Rightarrow5x=-4\Rightarrow x=-\frac{4}{5}\\x-3=0\Rightarrow x=3\end{cases}.}\)

Vậy x=3; -4/5

5x(x-3)+4(x-3)=0

(x-3)(5x+4)=0

=>\(\hept{\begin{cases}x-3=0\\5x+4=0\end{cases}}\)=>\(\hept{\begin{cases}x=3\\x=-\frac{4}{5}\end{cases}}\)

a, Áp dụng tính chất dãy tỉ số bằng nhau ta có : 

\(\frac{x}{3}=\frac{y}{-2}=\frac{2x+5y}{2.3+5.\left(-2\right)}=-\frac{12}{-4}=3\)

\(x=-3;y=6\)

b, Theo bài ra ta có : \(x:y=4:5\Leftrightarrow\frac{x}{4}=\frac{y}{5}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có : 

\(\frac{x}{4}=\frac{y}{5}=\frac{x-y}{4-5}=\frac{13}{-1}=-13\)

\(x=-52;y=-65\)

c, Theo bài ra ta có: \(4x=7y\Leftrightarrow\frac{x}{7}=\frac{y}{4}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{x}{7}=\frac{y}{4}=\frac{x-y}{7-4}=\frac{12}{3}=4\)

\(x=28;y=16\)

\(2x\left(x-3\right)+x\left(2x+5\right)=4x\left(x-3\right)+12\)

\(2x^2-6x+2x^2+5x=4x^2-12x+12\)

\(11x=12\)

\(x=\frac{12}{11}\)

trả lời 

11x=12

x=12:11

x=12/11

11x=12

x=12:11

x=12/11

hc tốt

4x+12=3x-7

4x-3x=-12-7

x=-19

x²(x - 3) - 4x + 12 = 0

x³ - 3x² - 4x + 12 = 0

x²(x - 3) - 4(x - 3) = 0

(x - 3)(x² - 4) = 0

x - 3 = 0 hoặc x² - 4 = 0

x = 0 + 3         x² = 0 + 4

x = 3               x² = 4

                       x = 2; -2

=> x = 3 hoặc x = 2; -2

-2x +3(x-2) +9= 12x+4

= -2x+3x -6 +9 = 12+4x

= x+3= 12+4x

= -3x=9

=) x= -3

4x + 3 = x - 12

4x - x = -12 - 3

3x = -15

x = ( -15 ) : 3

x = -5

~ Hok tốt ~

4x + 3 = x - 12

4x - x = (-12) - 3

3x      = -15

 x       = (-15) : 3

 x       = -5

#Hok tốt