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

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

a: \(x\in\left\{0;25\right\}\)

c: \(x\in\left\{0;5\right\}\)

B1: Đk: 5x ≥ 0 => x ≥ 0

Vì |x + 1| ≥ 0 => |x + 1| = x + 1

     |x + 2| ≥ 0 => |x + 2| = x + 2

     |x + 3| ≥ 0 => |x + 3| = x + 3

     |x + 4| ≥ 0 => |x + 4| = x + 4

=> |x + 1| + |x + 2| + |x + 3| + |x + 4| = 5x

 => x + 1 + x + 2 + x + 3 + x + 4 = 5x

=> 4x + 10 = 5x

=> x = 10

B2: Ta có: |x - 2018| = |2018 - x|

=> A=|x + 2000| + |2018 - x| ≥ |x + 2000 + 2018 - x| = |4018| = 4018

Dấu " = " xảy ra <=> (x + 2000)(x - 2018) ≥ 0

Th1: \(\hept{\begin{cases}x+2000\ge0\\x-2018\ge0\end{cases}\Rightarrow}\hept{\begin{cases}x\ge-2018\\x\le2018\end{cases}}\Rightarrow-2018\le x\le2018\)

Th2: \(\hept{\begin{cases}x+2000\le0\\x-2018\le0\end{cases}\Rightarrow}\hept{\begin{cases}x\le-2018\\x\ge2018\end{cases}}\)(vô lý)

Vậy GTNN của A = 4018 khi -2018 ≤ x ≤ 2018

B3:

a, Vì |x + 1| ≥ 0 ; |2y - 4| ≥ 0

=> |x + 1| + |2y - 4| ≥ 0

Dấu " = " xảy ra <=> \(\hept{\begin{cases}x+1=0\\2y-4=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-1\\y=2\end{cases}}\)

Vậy...

b, Vì |x - y + 1| ≥ 0 ; (y - 3)² ≥ 0

 => |x - y + 1| + (y - 3)² ≥ 0 

Dấu " = " xảy ra <=> \(\hept{\begin{cases}x-y+1=0\\y-3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x-y=-1\\y=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x-3=-1\\y=3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=3\end{cases}}\)

Vậy...

c, Vì |x + y| ≥ 0 ; |x - z| ≥ 0  ; |2x - 1| ≥ 0 

=> |x + y| + |x - z| + |2x - 1| ≥ 0 

Dấu " = " xảy ra <=> \(\hept{\begin{cases}x+y=0\\x-z=0\\2x-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+y=0\\x=z\\x=\frac{1}{2}\end{cases}\Leftrightarrow}}\hept{\begin{cases}\frac{1}{2}+y=0\\x=z=\frac{1}{2}\end{cases}\Leftrightarrow}\hept{\begin{cases}y=\frac{-1}{2}\\x=z=\frac{1}{2}\end{cases}}\)

coi lại mới thấy trình bày ngờ-u :)) 

B1: Đk: 5x ≥ 0 => x ≥ 0

=> x + 1 > 0 => |x + 1| = x + 1

=> x + 2 > 0 => |x + 2| = x + 2 

=> x + 3 > 0 => |x + 3| = x + 3 

=> x + 4 > 0 => |x + 4| = x + 4 

Ta có:  |x + 1| + |x + 2| + |x + 3| + |x + 4| = 5x

=> .... Làm tiếp như dưới

\(a,x\cdot\left(x+2\right)=0\)

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

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

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

a)x.(x+2)=0=>x=0 hoặc x+2=0

                  =>x=0 hoặc x=-2

b)7/12 .(x-3)=0

=>x-3=0

=>x=3

Ta có :

a) \(x\left(x+2\right)=0\)

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

Vậy \(x=0\)hoặc \(x=-2\)

b) \(\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)\left(x-3\right)=0\)

Vì \(\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)\ne0\)

\(\Rightarrow\)\(x-3=0\)

\(\Rightarrow\)\(x=3\)

Vậy \(x=3\)

a: \(x^3-4x^2-x+4=0\)

=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)

=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)

=>\(\left(x-4\right)\left(x^2-1\right)=0\)

=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)

b: Sửa đề: \(x^3+3x^2+3x+1=0\)

=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)

=>\(\left(x+1\right)^3=0\)

=>x+1=0

=>x=-1

c: \(x^3+3x^2-4x-12=0\)

=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)

=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)

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

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

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

d: \(\left(x-2\right)^2-4x+8=0\)

=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)

=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)

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

=>(x-2)(x-6)=0

=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

Bài 2: 

a) Ta có: \(\left|x-2\right|=\left|4-x\right|\)

\(\Leftrightarrow x-2=4-x\)

\(\Leftrightarrow2x=6\)

hay x=3

b) Ta có: \(\left(\left|2x-1\right|-3\right)\cdot\left(-2\right)+\left(-5\right)=6\)

\(\Leftrightarrow\left(\left|2x-1\right|-3\right)\cdot\left(-2\right)=11\)

\(\Leftrightarrow\left|2x-1\right|-3=\dfrac{-11}{2}\)

\(\Leftrightarrow\left|2x-1\right|=\dfrac{-11}{2}+\dfrac{6}{2}=\dfrac{-5}{2}\)(Vô lý)

a) Tìm được x = -4.        

b) Tìm được x = 3.

c) Tìm được x = ±1.

bài 2: (x-3).(y+2) = -5

    Vì x, y \(\in\)Z   => x-3 \(\in\)Ư(-5) = {5;-5;1;-1}

Ta có bảng: 

bài 3: a(a+2)<0

TH1 : \(\orbr{\begin{cases}a< 0\\a+2>0\end{cases}}\)=>\(\orbr{\begin{cases}a< 0\\a>-2\end{cases}}\)=> -2<a<0 ( TM)

TH2: \(\orbr{\begin{cases}a>0\\a+2< 0\end{cases}}\Rightarrow\orbr{\begin{cases}a>0\\a< -2\end{cases}}\Rightarrow loại\)
 

           Vậy -2<a<0

Bài 5: \(\left(x^2-1\right)\left(x^2-4\right)< 0\)

TH 1 : \(\hept{\begin{cases}x^2-1>0\\x^2-4< 0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x^2>1\\x^2< 4\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x>1\\x< 2\end{cases}}\)\(\Rightarrow\)1 < a < 2

TH 2: \(\hept{\begin{cases}x^2-1< 0\\x^2-4>0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x^2< 1\\x^2>4\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x< 1\\x>2\end{cases}}\)\(\Rightarrow\)loại

                         Vậy 1<a<2

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

\(\Rightarrow\orbr{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}\Rightarrow}\orbr{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)

b. \(\left(x^2+1\right)\left(x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^2+1=0\\x-4=0\end{cases}\Rightarrow}\orbr{\begin{cases}x^2=-1\left(Voly\right)\\x=4\end{cases}\Rightarrow x=4}\)

c, \(2x^2-\frac{1}{3}x=0\)

\(\Leftrightarrow x\left(2x-\frac{1}{3}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\2x-\frac{1}{3}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{6}\end{cases}}\)

d, \(\left(\frac{4}{5}\right)^{5x}=\left(\frac{4}{5}\right)^7\)

\(\Rightarrow5x=7\)

\(\Rightarrow x=\frac{7}{5}\)

e, Ta có: \(A=\frac{x+5}{x-2}=\frac{\left(x-2\right)+7}{x-2}=1+\frac{7}{x-2}\)

Để A ∈ Z <=> (x - 2) ∈ Ư(7) = { ±1; ±7 }

 Vậy....

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

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

\(\Leftrightarrow\orbr{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)

Vậy : ....

b) \(\left(x^2+1\right)\left(x-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x-4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=-1\left(loại\right)\\x=4\end{cases}}\)

c) \(2x^2-\frac{1}{3}x=0\)

\(\Leftrightarrow x\left(2x-\frac{1}{3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x-\frac{1}{3}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{6}\end{cases}}\)

Vậy :...