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

3x^2 + 6x - 3x^2 = 36

6x = 36

x= 6

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

\(\Leftrightarrow x^2-x=12\)

\(\Leftrightarrow x^2-x-12=0\)

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

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

Dễ rồi tự làm

\(3x^2-3x\left(-2+x\right)=36\)

\(=3x^2+6x-3x=36\)

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

\(=3x\left(x-3\right)=36\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}3x=36\\x-3=36\end{cases}\Leftrightarrow\orbr{\begin{cases}x=36:3=12\\x=36+3=39\end{cases}}}\)

Vậy S = { 12 ; 39 }

sai rồi nhs, bn giải pt cho 3x²-3x=36 =3x(x-3)=36 chỗ đó sai rồi

\(d,\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=15\\ \Leftrightarrow24x=-10\Leftrightarrow x=-\dfrac{5}{12}\\ e,\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\\ \Leftrightarrow9x=10\Leftrightarrow x=\dfrac{10}{9}\\ f,\Leftrightarrow9x^2+18x+9-18x=36+x^3-27\\ \Leftrightarrow x^3-9x^2=0\Leftrightarrow x^2\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\)

1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)

=-27x^3-18x^2+4x+10

2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27

=7x^3+37x^2+46x+33

5:

\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)

\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)

=7x^3-48x^2+8x-35

a)3x+10-2x>-11
3x - 2x > -10-11
1x > -21
x > -21
b) 3x2 - 6x + 3x2 < 36
-6x < 36
x < -6

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

\(\Rightarrow3x+10-2x=0\)

\(\Rightarrow x+10=0\)

\(\Rightarrow x=-10\)

\(b,x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5\)

\(\Rightarrow\left(2x^2-x\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5\)

\(\Rightarrow2x^3+9x^2-5x-2x^3-9x^2-4,5=3,5\)

\(\Rightarrow-5x-4,5=3,5\)

\(\Rightarrow-5x=8\)

\(\Rightarrow x=-\dfrac{8}{5}\)

\(c,3x^2-3x\left(x-2\right)=36\)

\(\Rightarrow3x^2-3x^2+6x=36\)

\(\Rightarrow6x=36\)

\(\Rightarrow x=6\)

\(d,\left(3x^2-x+1\right)\left(x-1\right)=x^2\left(4-3x\right)=\dfrac{5}{2}\)

\(\Rightarrow3x^3-3x^2-x^2+x+x-1+4x^2-3x^3=\dfrac{5}{2}\)

\(\Rightarrow2x-1=\dfrac{5}{2}\)

\(\Rightarrow2x=\dfrac{7}{2}\)

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

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

\(3x+10-2x=0\)

\(x+10=0\)

\(x=-10\)

c,\(3x^2-3x\left(x-2\right)=36\)

\(3x^2-3x^2+6x=36\)

\(6x=36\)

\(x=6\)

a, Ta có : \(\left(x-1\right)\left(x^2+5x-2\right)-x^3+1=0\)

=> \(\left(x-1\right)\left(x^2+5x-2\right)-\left(x^3-1\right)=0\)

=> \(\left(x-1\right)\left(x^2+5x-2\right)-\left(x-1\right)\left(x^2+x+1\right)=0\)

=> \(\left(x-1\right)\left(x^2+5x-2-x^2-x-1\right)=0\)

=> \(\left(x-1\right)\left(4x-3\right)=0\)

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

=> \(\left[{}\begin{matrix}x=1\\x=\frac{3}{4}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là \(S=\left\{1,\frac{3}{4}\right\}\)

b, Ta có : \(5\left(x^2+3x\right)-9\left(3x+3\right)=x^2-36\)

=> \(5x^2+15x-27x-27=x^2-36\)

=> \(5x^2+15x-27x-27-x^2+36=0\)

=> \(4x^2-12x+9=0\)

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

=> \(x=\frac{3}{2}\)

Vậy phương trình có tập nghiệm là \(S=\left\{\frac{3}{2}\right\}\)

\(a.\left(x-1\right)\left(x^2+5x-2\right)-x^3+1=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+5x-2\right)-\left(x^3-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+5x-2\right)-\left(x-1\right)\left(x^2+x+1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+5x-2-x^2-x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{3}{4}\end{matrix}\right.\)

Vậy pt có tập nghiệm \(S=\left\{1;\frac{3}{4}\right\}\)

\(b.5\left(x^2+3x\right)-9\left(3x+3\right)=x^2-36\\ \Leftrightarrow5x^2+15x-27x-27=x^2-36\\ \Leftrightarrow5x^2+15x-27x-27-x^2+36=0\\ \Leftrightarrow4x^2-12x+9=0\\ \Leftrightarrow\left(2x-3\right)^2=0\\ \Leftrightarrow x=\frac{3}{2}\)

Vậy pt có tập nghiệm \(S=\left\{\frac{3}{2}\right\}\)

Chúc bạn học tốt!!!!!!!!!!!