\(\Leftrightarrow6x^2+4x+27x+18-6x^2-12x-x-2=x^2-x-6x-6\)

\(\Leftrightarrow18x+16=x^2-7x-6\)

\(\Leftrightarrow x^2-7x-18x=16+6\)

\(\Leftrightarrow x^2-15x=22\)

\(\Leftrightarrow x^2-15x-22=0\)

......

\(\Leftrightarrow\left(6x^2+27x+4x+18\right)-\left(6x^2+x+12x+2\right)=x-1-x+6\)

\(\Leftrightarrow6x^2+31x+18-6x^2-x-12x-2=7\)

\(\Leftrightarrow18x+16=7\)

\(\Leftrightarrow18x=-9\)

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

                       \(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x-1\right)-\left(x-6\right)\)

\(3x\left(2x+9\right)+2\left(2x+9\right)-x\left(6x+1\right)-2\left(6x+1\right)=x-1-x+6\)

                 \(6x^2+27x+4x+18-6x^2-x-12x-2=5\)

        \(6x^2+\left(27x+4x\right)+18-6x^2-\left(12x+x\right)-2=5\)

                                 \(6x^2+31x+18-6x^2-13x-2=5\)

                    \(\left(6x^2-6x^2\right)+\left(31x-13x\right)+\left(18-2\right)=5\)

                                                                           \(18x+16=5\)

                                                                                     \(18x=5+16\)

                                                                                     \(18x=21\)

                                                                                          \(x=21:18\)

                                                                                          \(x=\frac{7}{6}\)

                                                Vậy \(x=\frac{7}{6}\)

P/s: Mình mới lớp 6 nên hi vọng bn xem bài của mik thật kĩ xem có sai sót không,cảm ơn.

a, |x+1| +2x \(=\)-2

\(\Leftrightarrow\) |x+1|\(=\)-2-2x (*)

TH1: Nếu x+1\(\ge0\)\(\Leftrightarrow x\ge-1\)thì \(\left|x+1\right|=x+1\)

Thay vào (*) ta có:

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

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

\(\Leftrightarrow3x=-3\)

\(\Leftrightarrow x=-1\)(TMĐK)

TH2: Nếu \(x+1< 0\Leftrightarrow x< -1\Rightarrow\left|x+1\right|=-\left(x+1\right)\)

Thay vào (*), ta có:

\(-x-1=\)\(-2-2x\)

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

\(\Leftrightarrow x=-1\)(kTMĐK)

Vậy S\(=\){-1}

b, \(x^2-6x+9=1\)

\(\Leftrightarrow x^2-2.3.x+3^2\)

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

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

\(\Leftrightarrow\left(x-3-1\right)\left(x-3+1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

Vậy S\(=\){4;2}

a) |x + 1| + 2x = 2

\(\Rightarrow\) |3x + 1| = 2

\(\Rightarrow\) |3x| = 1

\(\Rightarrow\) |x| = 1 : 3 = \(\dfrac{1}{3}\)

\(\Rightarrow\) x = \(\dfrac{1}{3}\) hoặc \(-\dfrac{1}{3}\)

a: \(\Leftrightarrow\left|x+1\right|=-2x-2\)

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

b: \(\Leftrightarrow\left(x-3\right)^2=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\Leftrightarrow x\in\left\{4;2\right\}\)

a) (3x+2)(2x+9) - (x+2)(6x+1) = (x+1) - (x-6)

<=> 6x² + 27x + 4x + 18 - 6x² - x - 12x - 2 = x+1 - x+6

<=> 18x + 16 = 7

<=> 18x = -9

<=> x = \(-\dfrac{1}{2}\)

b) 3(2x-1)(3x-1) - (2x-3)(9x-1) = 0

<=> 3.(6x²-2x-3x+1) - (18x²-2x-27x+3) = 0

<=> 3.(6x²-5x+1) - 18x²+29x-3 = 0

<=> 18x²-15x+3 - 18x²+29x - 3 = 0

<=> 14x = 0

<=> x = 0

\(x^2-6x+9=\left(2x+1\right)^2\)

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

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

\(\Rightarrow\left(x-3-2x-1\right)\left(x-3+2x+1\right)=0\)

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

\(\Rightarrow\left[\begin{array}{nghiempt}x+4=0\\3x-2=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=-4\\x=\frac{2}{3}\end{array}\right.\)

Vậy x = -4 ; \(\frac{2}{3}\)

x² + 6x + 9 = (2x-1) ²

sử dụng công thức nhân tóm tắt (ab) ² = a²-2ab + b² và chúng tôi mang lại phương trình để các hình thức:

x² + 6x + 9 = 4x²-4x + 1

 đặt phương trình:

-3x² + 10x + 8 = 0

đã nhận được bình đẳng vuông, mà bây giờ được giải quyết:

-3x² + 10x + 8 = 0

Δ = b²-4ac = 100- (4 * (- 3) * 8) = 100 - (- 96) = 100 + 96 = 196

√Δ = 14

x₁ = (- b-√Δ): 2a = (- 10-14): (- 6) = (- 24): (- 6) = 4

x₂ = (- b + √Δ): 2a = (- 10 + 14) (- 6) = 4 (- 6) = -4/6 = -2/3

x∈ {-2/3; 4}

Giải:

\(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x-1\right)-\left(x-6\right)\)

\(\Leftrightarrow6x^2+4x+27x+18-\left(6x^2+12x+x+2\right)=x-1-x+6\)

\(\Leftrightarrow6x^2+4x+27x+18-6x^2-12x-x-2=5\)

\(\Leftrightarrow16+18x=5\)

\(\Leftrightarrow18x=-11\)

\(\Leftrightarrow x=-\dfrac{11}{18}\)

Vậy ...

b) Ta có: \(\left(x-2\right)\left(x^2-2x+4\right)\left(x+2\right)\left(x^2+2x+4\right)-x^6+2x=1\)

\(\Leftrightarrow\left(x^3-8\right)\left(x^3+8\right)-x^6+2x-1=0\)

\(\Leftrightarrow x^6-64-x^6+2x-1=0\)

\(\Leftrightarrow2x-65=0\)

\(\Leftrightarrow2x=65\)

hay \(x=\frac{65}{2}\)

Vậy: \(x=\frac{65}{2}\)

c) Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)

\(\Leftrightarrow x^3-27-x\left(x+2\right)\left(x-2\right)-1=0\)

\(\Leftrightarrow x^3-27-x\left(x^2-4\right)-1=0\)

\(\Leftrightarrow x^3-27-x^3+4x-1=0\)

\(\Leftrightarrow4x-28=0\)

\(\Leftrightarrow4x=28\)

hay x=7

Vậy: x=7

a) Để phương trình \(\left(2x+1\right)^2\cdot\left(9x+2k\right)-5\left(x+2\right)=40\) có nghiệm là x=2 thì Thay x=2 vào phương trình \(\left(2x+1\right)^2\cdot\left(9x+2k\right)-5\left(x+2\right)=40\), ta được:

\(\left(2\cdot2+1\right)^2\cdot\left(9\cdot2+2k\right)-5\left(2+2\right)=40\)

\(\Leftrightarrow25\cdot\left(2k+18\right)-20=40\)

\(\Leftrightarrow25\left(2k+18\right)=60\)

\(\Leftrightarrow2k+18=\dfrac{12}{5}\)

\(\Leftrightarrow2k=-\dfrac{78}{5}\)

hay \(k=\dfrac{-39}{5}\)

Vậy: \(k=\dfrac{-39}{5}\)

(9x+2k) - 5(x+2)=40 có nghiệm là x=2

=>(2*2+1)²(9*2+2k)-5(2+2)=40

=>25(18+5k)-20=40

=>25(18+5k)=60

=>18+5k=2.4

=>5k=-15.6 =>k=-0.624

b) 2(2x-1)+18=9(x+2)(2x+k) có nghiệm là x=1

=>2(2*1-1)+18=9(1+2)(2*1+k)

=>2+18=27(2+k)

=>2+k=20/27

=>k=-34/27