a) \(\left(\frac{1}{7}x-\frac{2}{3}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\frac{1}{7}x-\frac{2}{3}=0\\-\frac{1}{5}x+\frac{3}{5}=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\frac{1}{7}x=\frac{2}{3}\\-\frac{1}{5}x=-\frac{3}{5}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{14}{3}\\x=3\end{cases}}\)

b)\(\frac{1}{10}x-\frac{4}{5}x+1=0\)

\(\Leftrightarrow x.\left(\frac{1}{10}-\frac{4}{5}\right)+1=0\)

\(\Rightarrow-\frac{7}{10}x=-1\)

\(\Rightarrow x=\frac{10}{7}\)

c)\(\left(2x-\frac{1}{3}\right).\left(5x+\frac{2}{7}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{3}=0\\5x+\frac{2}{7}=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=\frac{1}{3}\\5x=-\frac{2}{7}\end{cases}}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{6}\\x=-\frac{2}{35}\end{cases}}\)

a, (1/7 . x - 2/3) . (-1/5 . x + 3/5) = 0

Suy ra : 1/7 .x -2/3 = 0 hoặc -1/5 .x + 3/5 =0

Vậy : 1/7 .x = 2/3 hoặc -1/5 .x = 3/5

         x =2/3 : 1/7 hoặc x = 3/5 : (-1/5)

        x = 14/3 hoặc x = -3

b, 1/10 .x - 4/5 .x + 1 =0

   x . (1/10 - 4/5) + 1 = 0

   x . (-7/10) + 1 = 0

   x . -7/10 =0 +1 = 1

   x = 1 : (-7/10)

   x = -10/7

c, (2x - 1/3 ) . (5x +2/7) = 0

Suy ra : 2x - 1/3 = 0 hoặc 5x + 2/7 = 0

Vậy : 2x = 1/3 hoặc 5x = 2/7

         x = 1/3 : 2 hoặc x = 2/7 : 5

         x = 1/6 hoặc x = 2/35

cái chỗ x = -3 bạn sửa lại là 3 nha , mk nhấn nhầm

a) (3x + 2)(x^2 - 1) = (9x^2 - 4)(x + 1)

<=> 3x^3 - 3x + 2x^2 - 2 = 9x^3 + 9x^2 - 4x - 4

<=> 3x^3 - 3x + 2x^2 - 2 - 9x^3 - 9x^2 + 4x + 4 = 0

<=> 6x^3 + 7x^2 - x - 2 = 0 (doi dau)

<=> (x + 1)(2x - 1)(3x + 2) = 0

<=> x + 1 = 0 hoặc 2x - 1 = 0 hoặc 3x + 2 = 0

<=> x = -1 hoặc x = 1/2 hoặc x = -2/3

Bài 1: 

a) Ta có: \(x\left(x^2-4\right)=0\)

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

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

Vậy: \(x\in\left\{0;2;-2\right\}\)

b) Ta có: \(\left(2x-3\right)+\left(-3x\right)-\left(x-5\right)=40\)

\(\Leftrightarrow2x-3-3x-x+5=40\)

\(\Leftrightarrow-2x+2=40\)

\(\Leftrightarrow-2x=38\)

hay x=-19

Vậy: x=-19

Bài 2: 

a) Ta có: \(-45\cdot12+34\cdot\left(-45\right)-45\cdot54\)

\(=-45\cdot\left(12+34+54\right)\)

\(=-45\cdot100\)

\(=-4500\)

b) Ta có: \(43\cdot\left(57-33\right)+33\cdot\left(43-57\right)\)

\(=43\cdot57-43\cdot33+43\cdot33-33\cdot57\)

\(=43\cdot57-33\cdot57\)

\(=57\cdot\left(43-33\right)\)

\(=57\cdot10=570\)

+)   (5x-1). (2x+3)-3. (3x-1)=0

10x^2+15x-2x-3 - 9x+3=0

10x^2 +8x=0

2x(5x+4)=0

=> x=0 hoặc x= -4/5

+)    x^3 (2x-3)-x^2 (4x^2-6x+2)=0

2x^4 -3x^3 -4x^4 + 6x^3 - 2x^2=0

-2x^4 + 3x^3-2x^2=0

x^2(-2x^2+x-2)=0

-2x^2(x-1)^2=0

=> x=0 hoặc x=1

+)   x (x-1)-x^2+2x=5

x^2 -x -x^2+2x=5

x=5

+)     8 (x-2)-2 (3x-4)=25

8x - 16-6x+8=25

2x=33

x=33/2