\(\left|1-3x\right|=x-8\)

\(\Leftrightarrow\orbr{\begin{cases}1-3x=x-8\\1-3x=-x+8\end{cases}\Leftrightarrow\orbr{\begin{cases}9-4x=0\\-7-2x=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{9}{4}\\x=-\frac{7}{2}\end{cases}}}\)

ĐKXĐ : \(x-8\ge0\Rightarrow x\ge8\)

Khi đó |1 - 3x| = x - 8

<=> \(\orbr{\begin{cases}1-3x=x-8\\1-3x=-x+8\end{cases}}\Rightarrow\orbr{\begin{cases}-4x=-9\\-2x=7\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{9}{4}\text{(loại)}\\x=-3,5\left(\text{loại}\right)\end{cases}}\)

=> \(x\in\varnothing\)

\(\left(3x-2\right)\left(3x+8\right)\left(x+1\right)^2+16=0\)

\(\Leftrightarrow\left(9x^2+18x-16\right)\left(x^2+2x+1\right)+16=0\)

\(\Leftrightarrow\left[9\left(x^2+2x+1\right)-25\right]\left(x^2+2x+1\right)+16=0\)

Đặt \(x^2+2x+1=a\ge0\)

\(\left(9a-25\right)a+16=0\)

\(\Leftrightarrow9a^2-25a+16=0\)

\(\Rightarrow\left[{}\begin{matrix}a=1\\a=\frac{16}{9}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2+2x+1=1\\x^2+2x+1=\frac{16}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\left(x+2\right)=0\\\left(x+1\right)^2=\left(\frac{4}{3}\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x+1=\frac{4}{3}\\x+1=-\frac{4}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=\frac{1}{3}\\x=-\frac{7}{3}\end{matrix}\right.\)

(x - 1)(5x + 3) = (3x - 8)(x - 1)

<=> (x - 1)(5x + 3) - (3x - 8)(x - 1)= 0

<=> (x - 1)(5x + 3 - 3x + 8) = 0

<=> (x - 1)(2x + 11) = 0

\(\Leftrightarrow \begin{bmatrix} x - 1 = 0 & & \\ 2x + 11 = 0 & & \end{bmatrix}\)pn bỏ dấu ngoặc bên phải nha

\(\Leftrightarrow \begin{bmatrix} x = 1 & & \\ x = \frac{-11}{2} & & \end{bmatrix}\)

Vậy ............

\(\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)

\(\Rightarrow\left(x-1\right)\left(5x+3\right)-\left(3x-8\right)\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(5x+3-3x+8\right)=0\)

\(\Rightarrow\left(x-1\right)\left(2x+11\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2x+11=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{11}{2}\end{matrix}\right.\)

Chúc bạn học tốt!

(x-1)(5x+3)=(3x-8)(x-1)

<=>(x-1)(5x+3)-(x-1)(3x-8)=0

<=>(x-1)(5x+3-3x+8)=0

<=>(x-1)(2x+11)=0

=> x-1= hoặc 2x+11=0

=> x=1 x = \(-\dfrac{11}{2}\)

Vậy S={1;\(-\dfrac{11}{2}\)}

1.a)|−7x|=3x+16

Vì |-7x| ≥ 0  nên 3x+16 ≥ 0 ⇔ x ≥ \(\dfrac{-16}{3}\)    (*)

Với đk (*), ta có: |-7x|=3x+16

\(\left[\begin{array}{} -7x=3x+16\\ -7x=-3x-16 \end{array} \right.\) ⇔  \(\left[\begin{array}{} -7x-3x=16\\ -7x+3x=-16 \end{array} \right.\)

⇔ \(\left[\begin{array}{} x=-1,6 (t/m)\\ x= 4 (t/m) \end{array} \right.\)

b) \(\dfrac{x-1}{x+2}\) - \(\dfrac{x}{x-2}\) = \(\dfrac{5x-8}{x^2-4}\)

⇔ \(\dfrac{(x-1)(x-2)}{x^2-4}\) - \(\dfrac{x(x+2)}{x^2-4}\) = \(\dfrac{5x-8}{x^2-4}\)

⇒ x² - 2x - x + 2 - x² - 2x = 5x - 8  

⇔ -5x - 5x = -8 - 2

⇔ -10x = -10

⇔ x=1

2.7x+5 < 3x−11

⇔ 7x - 3x < -11 - 5

⇔ 4x < -16

⇔ x < -4

bạn tự biểu diễn trên trục số nha !

Bạn nên gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để được hỗ trợ tốt hơn. Viết đề như thế này gây khó đọc.

`x^2+3x+8=0`

`<=>x^2+2*x*3/2+9/4-9/4+8=0`

`<=>(x+3/2)^2-2-1/4+8=0`

`<=>(x+3/2)^2+5+3/4=0`

`<=>(x+3/2)^2=-23/4` vô lý

Vì `VT>=0,VP<0`

`=>` PT vô nghiệm

đề thế này ak \(\left(2x+4\right)\sqrt{x+8}=3x^2+7x+8\)

viet^2 thế nào đấy