3) \(x\left(x-4\right)+\left(x-4\right)^2=0\Leftrightarrow\left(x-4\right)\left(x+x-4\right)=0\Leftrightarrow2\left(x-4\right)\left(x-2\right)=0\)

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

4x.(x+1)-8(x+1)=0

(4x-8)(x+1)=0

suy ra x=2 hoặc x=-1

1) \(4x\left(x+1\right)=8\left(x+1\right)\Leftrightarrow4x^2+4x=8x+8\Leftrightarrow4x^2-4x-8=0\Leftrightarrow x^2-x-2=0\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)

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

Ta có : \(\dfrac{\left(x-3\right)\left(x+2\right)\left(x+1\right)}{\left(x+3\right)\left(x-4\right)}>0\)

- Đặt \(f\left(x\right)=\dfrac{\left(x-3\right)\left(x+2\right)\left(x+1\right)}{\left(x+3\right)\left(x-4\right)}\)

- Lập bảng xét dấu :

- Từ bảng xét dấu : - Để f(x) > 0

\(\Leftrightarrow\left[{}\begin{matrix}-3< x< -2\\-1< x< 3\\x>4\end{matrix}\right.\)

Vậy ...

\(a,x+5x^2=0\\ \Rightarrow a,x\left(1+5x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{5}\end{matrix}\right.\\ b,\left(x+3\right)^2+\left(4+x\right)\left(4-x\right)=0\\ \Rightarrow x^2+6x+9+16-x^2=0\\ \Rightarrow6x+25=0\\ \Rightarrow6x=-25\\ \Rightarrow x=-\dfrac{25}{6}\)

\(c,5x\left(x-1\right)=x-1\\ \Rightarrow c,5x\left(x-1\right)-\left(x-1\right)\\ \Rightarrow\left(x-1\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ d,x^2-2x-3=0\\ \Rightarrow\left(x^2-3x\right)+\left(x-3\right)=0\\ \Rightarrow x\left(x-3\right)+\left(x-3\right)=0\\ \Rightarrow\left(x+1\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)

1) Do x ∈ Z và 0 < x < 3

⇒ x ∈ {1; 2}

2) Do x ∈ Z và 0 < x ≤ 3

⇒ x ∈ {1; 2; 3}

3) Do x ∈ Z và -1 < x ≤ 4

⇒ x ∈ {0; 1; 2; 3; 4}

a: \(\Leftrightarrow\left|\dfrac{5}{3}x\right|=\dfrac{1}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{5}{3}=\dfrac{1}{6}\\x\cdot\dfrac{5}{3}=-\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}:\dfrac{5}{3}=\dfrac{3}{30}=\dfrac{1}{10}\\x=-\dfrac{1}{10}\end{matrix}\right.\)

b: \(\Leftrightarrow\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|=\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{3}{2}\)

\(\Leftrightarrow\left|x-1\right|=\dfrac{3}{2}:\dfrac{3}{4}=2\)

=>x-1=2 hoặc x-1=-2

=>x=3 hoặc x=-1

c: \(\Leftrightarrow\left|x+\dfrac{3}{5}\right|=\left|x-\dfrac{7}{3}\right|\)

\(\Leftrightarrow x+\dfrac{3}{5}=\dfrac{7}{3}-x\)

=>2x=44/15

hay x=22/15

a,2x-1=3

     2x = 3+1

     2x =4

      x = 4:2

      x=2

b,(x-4)-x+4=0

x-4-x+4=0

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

0+0=0

x thuộc Z

c,tương tự

x=3

Ta có : x⁵ + x⁴ + x + 1 = 0

<=> x⁴(x + 1) + (x + 1) = 0

<=> (x + 1)(x⁴ + 1) = 0

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^4+1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x^4=-1\end{cases}}\)

Vậy x = -1

Ta có : x⁴ + 3x³ - x - 3 = 0

<=> x³(x + 3) - (x + 3) = 0

<=> (x + 3) (x³ - 1) = 0

 \(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x^3-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-3\\x^3=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-3\\x=1\end{cases}}\)

Vậy x thuộc {-3;1}

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

<=> x+5=0 hoặc x-4=0

<=> x=-5 hoặc x=4

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

<=> x-1=0 hoặc x-3=0

<=> x=1 hoặc x=3

a) (x+5).(9x-4)=0

=> x+5=0 hoặc 9x-4=0

Nếu x+5=0: x=0-5=-5

Nếu 9x-4=0: 9x=0+4=4

                     x=4/9

b) (x-1).(x-3)=0

=> x-1=0 hoặc x-3=0

Nếu x-1=0: x=0+1=1

Nếu x-3=0: x=0+3=3

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

=> 3-x=0 hoặc x-3=0

Nếu 3-x=0: x=3-0=0

Nếu x-3=0: x=0+3=3

d) x.(x+1)=0

=> x=0 hoặc x+1=0

Nếu x+1=0: x=0-1=-1

\(a,\left(x+5\right)\left(9x-4\right)=0\)

<=>\(\orbr{\begin{cases}x+5=0\\9x-4=0\end{cases}}\)

<=>\(\orbr{\begin{cases}x=-5\\x=\frac{4}{9}\end{cases}}\)

\(b,\left(x-1\right)\left(x-3\right)=0\)

<=>\(\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}}\)

<=>\(\orbr{\begin{cases}x=1\\x=3\end{cases}}\)

\(c,\left(3-x\right)\left(x-3\right)=0\)

<=>\(\orbr{\begin{cases}3-x=0\\x-3=0\end{cases}}\)

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

\(d,x\left(x+1\right)=0\)

<=>\(\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\)

<=>\(\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)