c: Ta có: \(x^3+3x^2+3x-7=0\)

\(\Leftrightarrow x+1=2\)

hay x=1

b: Ta có: \(x\left(x-3\right)-4x+12=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

=>4x+12=3x-21

=>4x-3x=-12-21

=>x=-33

Vậy x=-33

4x + 12 = 3x - 21 => x = - 33

4x+12=3(x-7)

=>4x+12=3x-21

=>3x-4x=21+12

=>-x=33

=>x=-33

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)^2 - 4x - 12 = 0

<=> (x + 3)^2 - 4(x + 3) = 0

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

<=> x = -3 hoặc x = 1

b) x(x + 5)(x- 5) - (x - 3)(x^2 + 3x + 9) = 7

<=> x^3 - 25x - x^3 + 27 = 7

<=> -25x + 27 = 7

<=> x = 4/5

a/ \(\left(x+3\right)^2-4x-12=0\)

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

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

\(\left[{}\begin{matrix}x+3=0\Rightarrow x=-3\\x+3-4=0\Rightarrow x=1\end{matrix}\right.\)

---

b/ \(x\left(x+5\right)\left(x-5\right)-\left(x-3\right)\left(x^2+3x+9\right)=7\)

\(x\left(x^2-25\right)-\left(x^3-27\right)=7\)

\(x^3-25x-x^3+27=7\)

\(-25x=-20\)

\(x=\dfrac{20}{25}=\dfrac{4}{5}\)

a, <=>x² +6x+9-4x-12=0

<=> x² +2x -3=0

<=> x² +3x -x-3=0

<=> x.(x+3) - (x+3) =0

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

<=> x=1 hoặc x=-3

b, <=> x(x² -25) - (x-3)(x+3)² -7=0 

<=> x³ -25x + (9-x²) (x+3) -7=0

<=> x³ -25x+ 9x+27-x³ -3x² -7=0

<=> -3x² -16x +20=0

<=>(3x-10)(x-2) =0 (đoạn này tự phân tích nha ^ ^)

<=> x= 10/3 hoặc x=2

Chúc bạn học tốt nha!

-5(3x-7) - ( -15x+3) - ( 12 -x ) =-4

=>-15x+35+15x-3-12+x=-4

=>20+x=-4

=>x=-4-20

=>x=-24

-3(4x-2) - ( -12+8) - (-x)=0

=>-12x+6+12-8+x=0

=>-11x+10=0

=>-11x=0-10

=>-11x=-10

=>11x=10

=>x=10:11=\(\frac{10}{11}\)

Ta có : \(\frac{4x+7}{3}=\frac{12}{4x+7}\)

\(\Rightarrow\left(4x+7\right)^2=12.3=36\)

\(\Rightarrow4x+7=\orbr{\begin{cases}-6\\6\end{cases}}\)

\(\Rightarrow4x=\orbr{\begin{cases}-13\\-1\end{cases}}\)

\(\Rightarrow x=\orbr{\begin{cases}\frac{-13}{4}\\\frac{-1}{4}\end{cases}}\)

4x+12=3x-7

4x-3x=-12-7

x=-19

5x-16=40+x

=> 5x-16-x = 40

=> 5x-x -16=40

4x-16=40

4x= 40+16

4x=56

x= 56:4

x=14

Vậy...

4x-10=15-x

=> 4x-10+x= 15

4x+x -10=15

5x= 15+10

5x= 25

x= 25:5

x=5

Vậy....

5x -16=40+x

=> 5x-x=40+16

=>4x=56

=>x=56:4

x=14

\(d,=24x^2-38x+3\\ e,=x^2-12x+35\\ f,=\left(x^2-144\right)\left(4x-1\right)=4x^3-x^2-576x+144\)

Tính giá trị biểu thức

ý d, e, f

d: \(\left(12x-1\right)\left(2x-3\right)\)

\(=24x^2-36x-2x+3\)

\(=24x^2-38x+3\)