a) (x + 5)(x - 4) = 0

x + 5 = 0 hoặc  x - 4=  0

x thuộc {-5 ; 4}

b) (x - 10)(x-  3) = 0

x - 10  = 0 hoặc x - 3 = 0

x thuộc {3;10}

c) (3 - x)(x - 3)  = 0

3 - x = 0 ; x - 3 = 0 

< = . x=  3 (thõa mãn cả 2 ĐK)

d) x(x + 1) = 0

x = 0 hoặc x+  1 = 0

=> x = -1

Vậy x thuộc {-1 ; 0} 

a)(x+5)(x-4)=0

nên x+5=0 hoặc x-4=0

x=0-5                 x=0+4

x=-5                   x=4

b)(x-10)(x-3)=0

nên x-10=0 hoặc x-3=0

x=0+10               x=0+3

x=10                   x=3

c)(3-x)(x-3)=0

nên 3-x=0 hoặc x-3=0

x=3-0               x=0+3

x=3

d)x(x+1)=0

nên x=0 hoặc x+1=0

                     x=0-1

                     x=-1

a/(x + 5).(x - 4) = 0

=> x + 5 = 0 => x = -5

hoặc x - 4 = 0 => x = 4

Vậy x = -5 , x = 4

 b/(x - 10).(x - 3) = 0 

=> x - 10 = 0 => x = 10

hoặc x - 3 = 0 => x = 3

Vậy x = 10, x = 3

c/(3 - x).(x - 3) = 0 

=> (3 - x)² = 0 => 3 - x = 0 => x = 3

Vậy x = 3

d/x.(x + 1) = 0

=> x = 0 

hoặc x + 1 = 0 => x = -1

Vậy x = 0 , x = -1

mình không biets khó quá

\(3x\left(x+\frac{1}{5}\right)=0\)

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

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

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

\(\left(x-3\right)\left(x+9\right)>0\)

Th1 : \(\hept{\begin{cases}x-3>0\\x+9>0\end{cases}\Rightarrow\hept{\begin{cases}x>3\\x>-9\end{cases}\Rightarrow}x>3}\)

Th2 : \(\hept{\begin{cases}x-3< 0\\x+9< 0\end{cases}\Rightarrow\hept{\begin{cases}x< 3\\x< -9\end{cases}\Rightarrow}x< -9}\)

\(x^3-2x^2+x-2=0\\ \Leftrightarrow x^2\left(x-2\right)+\left(x-2\right)=0\\ \Leftrightarrow\left(x^2+1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=2\end{matrix}\right.\\ Vậy:x=2\\ ---\\ 2x\left(3x-5\right)=10-6x\\ \Leftrightarrow6x^2-10x-10+6x=0\\ \Leftrightarrow6x^2-4x-10=0\\ \Leftrightarrow6x^2+6x-10x-10=0\\ \Leftrightarrow6x\left(x+1\right)-10\left(x+1\right)=0\\ \Leftrightarrow\left(6x-10\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}6x-10=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)

\(4-x=2\left(x-4\right)^2\\ \Leftrightarrow4-x=2\left(x^2-8x+16\right)\\ \Leftrightarrow2x^2-16x+32+x-4=0\\ \Leftrightarrow2x^2-15x+28=0\\ \Leftrightarrow2x^2-8x-7x+28=0\\ \Leftrightarrow2x\left(x-4\right)-7\left(x-4\right)=0\\ \Leftrightarrow\left(2x-7\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\\ ---\\ 4-6x+x\left(3x-2\right)=0\\ \Leftrightarrow4-6x+3x^2-2x=0\\ \Leftrightarrow3x^2-8x+4=0\\ \Leftrightarrow3x^2-6x-2x+4=0\\ \Leftrightarrow3x\left(x-2\right)-2\left(x-2\right)=0\\ \Leftrightarrow\left(3x-2\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)

a)

\(3x\left(x+\frac{1}{5}\right)=0\)

=>_3x=0

   |_x+1/5=0

=> _x=0

     |_x=-15

b)(x-2)(3+x)=0

=> _x-2=0

     |_ 3+x=0

=> _x=2

     |_x=-3

c) x khác -9 và 3

4>x>-9

c) để (x-3)(x+9)>0

thì x≠3; x≠-9

để (x-3)(x+9)>0

=> (x-3)  và (x+9) phải cùng dấu 

TH1:để (x-3) và (x+9) cùng lớn hơn 0 thì x>4

TH2:để(x-3) và (x+9) cùng nhỏ hơn 0 thì x< -9

vậy để (x-3)(x+9) >0 thì -9<x<4 và x≠3; x≠-9 

1/ Ta có \(\left(x-2\right)\left(x+\frac{2}{3}\right)>0\)

=> \(\hept{\begin{cases}x-2>0\\x+\frac{2}{3}>0\end{cases}}\)hoặc \(\hept{\begin{cases}x-2< 0\\x+\frac{2}{3}< 0\end{cases}}\)

=> \(\hept{\begin{cases}x>2\\x>-\frac{2}{3}\end{cases}}\)hoặc \(\hept{\begin{cases}x< 2\\x< -\frac{2}{3}\end{cases}}\)

=> \(\orbr{\begin{cases}x>2\\x< -\frac{2}{3}\end{cases}}\)

Vậy \(\orbr{\begin{cases}x>2\\x< -\frac{2}{3}\end{cases}}\)thì \(\left(x-2\right)\left(x+\frac{2}{3}\right)>0\)

2    \(xy=\frac{x}{y}\Rightarrow y=\frac{x}{xy}=\frac{1}{y}\Rightarrow y^2=1\Rightarrow y=+-1\)

nếu \(y=1\Rightarrow x+y=xy=x+1=x\Rightarrow x-x=-1\Rightarrow0=-1\)vô lí (loại)

\(\Rightarrow y=-1\Rightarrow x+y=xy=x-1=-x\Rightarrow2x=1\Rightarrow x=\frac{1}{2}\)(thỏa mãn)

vậy \(x=\frac{1}{2};y=-1\)

a) \(\left|x-9\right|=2x+5\)

khi \(x\ge-\frac{5}{2}\), ta có:

\(\orbr{\begin{cases}x-9=2x+5\\x-9=-2x-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}-x=14\\3x=4\end{cases}}\Rightarrow\orbr{\begin{cases}x=-14\\x=\frac{4}{3}\end{cases}}\)

\(x=-14\)không thỏa mãn

\(x=\frac{4}{3}\)thỏa mãn

vậy x=4/3

Lm câu b, câu c cho mik luôn bn ơi

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

\(\Leftrightarrow\left(x+1\right)\left[5x-5\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left(x+1\right)\left(5x-5x+10\right)=0\)

\(\Leftrightarrow10\left(x+1\right)=0\)

mà \(10\ne0\)

nên x+1=0

hay x=-1

Vậy: x=-1

b) Ta có: \(\left(4x+1\right)\left(x-2\right)-\left(2x-3\right)=4\)

\(\Leftrightarrow4x^2-8x+x-2-2x+3-4=0\)

\(\Leftrightarrow4x^2-9x-3=0\)

\(\Leftrightarrow\left(2x\right)^2-2\cdot2x\cdot\frac{9}{4}+\frac{81}{16}-\frac{129}{16}=0\)

\(\Leftrightarrow\left(2x-\frac{9}{4}\right)^2=\frac{129}{16}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{9}{4}=\frac{\sqrt{129}}{4}\\2x-\frac{9}{4}=-\frac{\sqrt{129}}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\frac{9+\sqrt{129}}{4}\\2x=\frac{9-\sqrt{129}}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{9+\sqrt{129}}{8}\\x=\frac{9-\sqrt{129}}{8}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{9+\sqrt{129}}{8};\frac{9-\sqrt{129}}{8}\right\}\)

c) Ta có: \(2x^3-18x=0\)

\(\Leftrightarrow2x\left(x^2-9\right)=0\)

\(\Leftrightarrow2x\left(x-3\right)\left(x+3\right)=0\)

mà \(2\ne0\)

nên \(\left[{}\begin{matrix}x=0\\x+3=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=3\end{matrix}\right.\)

Vậy: \(x\in\left\{0;-3;3\right\}\)

d) Ta có: \(\left(3x-2\right)\left(2x+1\right)-6x\left(x+2\right)=11\)

\(\Leftrightarrow6x^2+3x-4x-2-6x^2-12x=11\)

\(\Leftrightarrow-13x-2=11\)

\(\Leftrightarrow-13x=13\)

hay x=-1

Vậy: x=-1

e) Ta có: \(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)=3\left(1-x^2\right)\)

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

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

\(\Leftrightarrow3x-12=0\)

\(\Leftrightarrow3x=12\)

hay x=4

Vậy: x=4

f) Ta có: \(6x^2-\left(2x+5\right)\left(3x-2\right)=-1\)

\(\Leftrightarrow6x^2-\left(6x^2-4x+15x-10\right)+1=0\)

\(\Leftrightarrow6x^2-6x^2+4x-15x+10+1=0\)

\(\Leftrightarrow-11x+11=0\)

\(\Leftrightarrow-11x=-11\)

hay x=1

Vậy: x=1