a)\(2x\left(x-2016\right)-2x+4032=0\)

\(\Leftrightarrow2x\left(x-2016\right)-2\left(x-2016\right)=0\)

\(\Leftrightarrow\left(2x-2\right)\left(x-2016\right)=0\)

\(\Leftrightarrow2\left(x-1\right)\left(x-2016\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x-2016=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=2016\end{array}\right.\)

b)\(5x\left(x-3\right)=x-3\)

\(\Leftrightarrow5x\left(x-3\right)-\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-3=0\\5x-1=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=3\\x=\frac{1}{5}\end{array}\right.\)

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

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

\(\Leftrightarrow\left(3x-1+x+2\right)\left[\left(3x-1\right)-\left(x+2\right)\right]=0\)

\(\Leftrightarrow\left(4x+1\right)\left(2x-3\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}4x+1=0\\2x-3=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{4}\\x=\frac{3}{2}\end{array}\right.\)

\(5x\left(x-3\right)=x-3\)

\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)

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

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

A= (x^2+2x)^2+9x^2+18x+20

=x^4+4x^4+4x^2+9x^2+18x+20

=5x^4+13x^2+18x+20

Cái bài này bạn yêu cầu không rõ nên mình chỉ giúp bạn được bấy nhiêu thôi. Nếu bạn yêu cầu rút gọn thì như trên còn yêu câu khác thì mình chưa chắc nên bạn phải ghi cụ thể nha.

`a)2x^2+3(x-1)(x+1)=5x(x+1)`

`<=>2x^2+3x^2-3=5x^2+5x`

`<=>5x=-3`

`<=>x=-3/5`

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`b)(x-3)^3+3-x=0` nhỉ?

`<=>(x-3)^3-(x-3)=0`

`<=>(x-3)(x^2-1)=0`

`<=>[(x=3),(x^2=1<=>x=+-1):}`

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`c)5x(x-2000)-x+2000=0`

`<=>5x(x-2000)-(x-2000)=0`

`<=>(x-2000)(5x-1)=0`

`<=>[(x=2000),(x=1/5):}`

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`d)3(2x-3)+2(2-x)=-3`

`<=>6x-9+4-2x=-3`

`<=>4x=2`

`<=>x=1/2`

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`e)x+6x^2=0`

`<=>x(1+6x)=0`

`<=>[(x=0),(x=-1/6):}`

Với \(2x-1\ge0\Leftrightarrow x\ge\frac{1}{2}\)

\(\Rightarrow D=2\left(2x-1\right)-3\left(2x+3\right)\)

\(\Rightarrow D=4x-2-6x-9\)

\(\Rightarrow D=-2x-11\)

Với \(2x+3< 0\Rightarrow x< -\frac{3}{2}\)

\(\Rightarrow D=2\left(1-2x\right)-3\left(2x+3\right)\)

\(\Rightarrow D=2-4x-6x-9\)

\(\Rightarrow D=-10x-7\)

Thanks nha!

48∈B(2x+1)

⇒48⋮2x+1

⇒2x+1∈Ư(48)

Ư(48)={1;2;3;4;6;8;12;16;24;48}

⇒x∈{0;1}

\(2x+1\inƯ\left(48\right)\)

\(\Leftrightarrow2x+1\in\left\{1;3\right\}\)

hay \(x\in\left\{0;1\right\}\)

48 là bội của 2x + 1

\(\Rightarrow\) 2x + 1 \(\in\) Ư (48)

Ta có:

Ư (48) = {1; 2; 3; 4; 6; 8; 12; 24; 48}

\(\Rightarrow x\in\left\{0;1\right\}\)

\(A=27x^3-8-27x^3+3=-5\\ B=8x^3+y^3+8x^3-y^3-16x^3=0\)

2x + 5 ⋮ x + 2 <=> 2( x + 2 ) + 1 ⋮ x + 2

Vì 2( x + 2 ) ⋮ x + 2 Để 2( x + 2 ) + 1 ⋮ x + 2 <=> 1 ⋮ x + 2 

=> x + 2 là ước của 1 => Ư(1) = { -1 ; 1 }

Ta có : x + 2 = 1 => x = -1

           x + 2 = - 1 => x = -3

Vậy x = { -1;'-3 }