\(a,\Leftrightarrow\left(4x-8\right)\left(x+1\right)=0\\ \Leftrightarrow4\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ b,\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x^2=-1\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=-1\\ c,\Leftrightarrow x^2-2x-4x+8=0\\ \Leftrightarrow\left(x-2\right)\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\\ d,\Leftrightarrow x^3-3x^2+3x-9x+2x-6=0\\ \Leftrightarrow\left(x-3\right)\left(x^2+3x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x^2+x+2x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+1\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\\x=-2\end{matrix}\right.\)

a) \(\Rightarrow4\left(x+1\right)\left(x-2\right)=0\)

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

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

\(\Rightarrow\left(x+1\right)\left(x^2+1\right)=0\)

\(\Rightarrow x=-1\left(do.x^2+1\ge1>0\right)\)

c) \(\Rightarrow x\left(x-4\right)-2\left(x-4\right)=0\)

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

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

d) \(\Rightarrow x^2\left(x-3\right)+3x\left(x-3\right)+2\left(x-3\right)\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\\x=-1\end{matrix}\right.\)

\(a,\Leftrightarrow2x^2+10x-2x^2=12\Leftrightarrow x=\dfrac{12}{10}=\dfrac{6}{5}\\ b,\Leftrightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\\ \Leftrightarrow\left(1-2x\right)\left(9-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\\ d,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ e,\Leftrightarrow4x^2-4x+1-4x^2+196=0\\ \Leftrightarrow-4x=-197\Leftrightarrow x=\dfrac{197}{4}\)

\(f,\Leftrightarrow x^2+8x+16-x^2+1=16\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\\ g,Sửa:\left(3x+1\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(3x+1-x-1\right)\left(3x+1+x+1\right)=0\\ \Leftrightarrow2x\left(4x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\\ h,\Leftrightarrow x^2+8x-x-8=0\\ \Leftrightarrow\left(x+8\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-8\end{matrix}\right.\\ i,\Leftrightarrow2x^2-13x+15=0\\ \Leftrightarrow2x^2+2x-15x-15=0\\ \Leftrightarrow\left(x+1\right)\left(2x-15\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{15}{2}\end{matrix}\right.\)

a: \(3\left(x-3\right)-6x=0\)

=>\(3x-9-6x=0\)

=>-3x-9=0

=>3x+9=0

=>3x=-9

=>\(x=-\dfrac{9}{3}=-3\)

b: Đề thiếu vế phải rồi bạn

c: \(2\left(x-3\right)+3x=9\)

=>2x-6+3x=9

=>5x-6=9

=>5x=6+9=15

=>x=15/5=3

d: \(x\left(x-11\right)+2\left(x-11\right)=0\)

=>\(\left(x-11\right)\left(x+2\right)=0\)

=>\(\left[{}\begin{matrix}x-11=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-2\end{matrix}\right.\)

e: \(x\left(x+2\right)+8=x^2\)

=>\(x^2+2x+8=x^2\)

=>2x+8=0

=>2x=-8

=>x=-8/2=-4

f: \(8\left(x+1\right)+2x=-2\)

=>\(8x+8+2x=-2\)

=>10x=-2-8=-10

=>\(x=-\dfrac{10}{10}=-1\)

g: 12-3(x+2)=0

=>3(x+2)=12

=>x+2=12/3=4

=>x=4-2=2

a) Ta có: 70 = 2.5.7; 84 =  2 2 . 3 . 7 => ƯCLN(70,84) = 2.7 = 14

=> ƯC(70,84) = Ư(14) = {1;2;7;14}

Mà x ∈ ƯC(70, 84) và x > 8.Vậy x = 14

b) Ta có: 64 =  2 6 ; 48 =  2 4 . 3 ; 88 =  2 3 . 11 => ƯCLN(64,48,88) =  2 3 = 8

=> ƯC(64,48,88) = Ư(8) = {1;2;4;8}

Mà x ∈ ƯC(64,48,88) và x > 4 . Vậy x = 8

c) Vì 126 ⋮ x; 210 ⋮ x nên x ∈ ƯC(126,210)

Ta có: 126 =  2 . 3 2 . 7 ; 210 = 2.3.5.7 => ƯCLN(126,210) = 2.3.7 = 42

=> ƯC(126,210) = Ư(42) = {1;2;3;6;7;14;21;42}

Mặt khác: 15 < x < 30. Vậy x = 21

d) Vì 150 ⋮ x; 84 ⋮ x; 30 ⋮ x nên x ∈ ƯC(150,84,30)

Ta có: 150 =  2 . 3 . 5 2 ; 84 =  2 2 . 3 . 7 ; 30 = 2.3.5 => ƯCLN(150,84,30) = 2.3 = 6

=> ƯC(150,84,30) = Ư(6) = {1;2;3;6}

Mặt khác: 2 < x < 6. Vậy x = 3

a)\(2x+3\left(x+4\right)-14=8\)

\(2x+3x+12-14=8\)

\(2x+3x-2=8\)

\(5x-2=8\)

\(5x=8+2\)

\(5x=10\)

\(x=10:5\)

\(x=2\)

b)\(x+2x+3x=4x+16\)

\(x\left(1+2+3\right)=4x+16\)

\(6x-4x=16\)

\(2x=16\)

\(x=16:2\)

\(x=8\)

c)\(2\left(x+1\right)+3\left(x+5\right)=0\)

\(2x+2+3x+15=0\)

\(5x+17=0\)

\(5x=0-17\)

\(5x=-17\)

\(x=-17:5\)

\(x=-\frac{17}{5}\)

a) Ta có: B(12) = {0;12;24;36;48;60;...}

x ∈ B(12) và 20 ≤ x ≤ 50 nên x = 24;36;48.

b) x ∈ Ư(20) và x > 8.

Ta có: x ∈ Ư(20) = {1;2;3;4;5;10;20;...}

x ∈ Ư(20) và x > 8 nên x = 10; 20.

c) Ta có: x ⋮ 5 nên x là bội của 15

B(15) = {0;15;30;45;60...} vì 0 < x ≤ 40 nên x = 15; 30.

d) Ta có: 16 ⋮ x nên x là ước của 16.

Ư(16) = {1;2;4;8;16}. Vậy x = 1,2,4,8,16.

e) Ta có: B(18) = {0;18;36;54;72;90;108}

Vì 9 < x < 120 nên x ∈ {18;36;54;72;90;108}

f) Vì 6 ⋮ (x – 1) nên (x – 1) là ước của 6.

=> (x – 1) ∈ {1;2;3;6} => x ∈ {2;3;4;7}