8) \(\left(x+4\right)\left(6x-12\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+4=0\\6x-12=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-4\\6x=12\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-4\\x=2\end{cases}}}\)

Vậy \(x\in\left\{-4;2\right\}\)

11) \(\left(\frac{7}{8}-2x\right)\left(3x+\frac{1}{3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{7}{8}-2x=0\\3x+\frac{1}{3}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=\frac{7}{8}-0\\3x=-\frac{1}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=\frac{7}{8}\\x=-\frac{1}{9}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{16}\\x=-\frac{1}{9}\end{cases}}}\)

Vậy \(x\in\left\{\frac{7}{16};-\frac{1}{9}\right\}\)

12) \(3x-2x^2=0\)

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

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

Vậy \(x\in\left\{0;\frac{3}{2}\right\}\)

13) \(5x+10x^2=0\)

\(\Leftrightarrow5x\left(1+2x\right)=0\)

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

Vậy \(x\in\left\{0;-\frac{1}{2}\right\}\)

8) (x+4)(6x-12)=0

\(\Leftrightarrow\orbr{\begin{cases}x+4=0\\6x-12=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=2\end{cases}}}\)

Vậy x={-4;2}

11) \(\left(\frac{7}{8}-2x\right)\left(3x+\frac{1}{3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{7}{8}-2x=0\\3x+\frac{1}{3}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=\frac{7}{8}\\3x=\frac{-1}{3}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{16}\\x=\frac{-1}{9}\end{cases}}}\)

( x - 1 )( x + 2 ) - x - 2 = 0

<=> ( x - 1 )( x + 2 ) - ( x + 2 ) = 0

<=> ( x + 2 )( x - 2 ) = 0

<=> x = ±2

( 2x - 7 )³ = 8( 7 - 2x )²

<=> ( 2x - 7 )³ - 8( 2x - 7 )² = 0

<=> ( 2x - 7 )²( 2x - 15 ) = 0

<=> x = 7/2 hoặc x = 15/2

Bài 1: 

\(2x^2+8x+30\)

\(=2\left(x^2+4x+15\right)\)

\(=2\left(x^2+4x+4+11\right)\)

\(=2\left(x+2\right)^2+22>0\forall x\)

Bài 2: 

\(-x^2-2x-12\)

\(=-\left(x^2+2x+12\right)\)

\(=-\left(x^2+2x+1+11\right)\)

\(=-\left(x+1\right)^2-11< 0\forall x\)

\(1,x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\\ 2,-2x^2-x-1=-2\left(x^2+2\cdot\dfrac{1}{4}x+\dfrac{1}{16}+\dfrac{7}{16}\right)\\ =-2\left(x+\dfrac{1}{4}\right)^2-\dfrac{7}{8}\le-\dfrac{7}{8}< 0\\ 3,\dfrac{1}{2}x^2-2x+2=\dfrac{1}{2}\left(x^2-4x+4\right)=\dfrac{1}{2}\left(x-2\right)^2\ge0\)

1: \(x^2-x+1\)

\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\)

Bài 5: 

a: \(8A=8+8^2+...+8^8\)

\(\Leftrightarrow7A=8^8-1\)

hay \(A=\dfrac{8^8-1}{7}\)

b: \(8B=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)

\(\Leftrightarrow8B=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\)

\(\Leftrightarrow8B=3^{16}-1\)

hay \(B=\dfrac{3^{16}-1}{8}\)

Ta xet 3 truong hop

TH1 : x la so nguyen duong 

Co 2x^8 + 2x^7 + 1 = duong + duong + duong = duong

Ma so duong luon lon hon 0

=> 2x^8 + 2x^7 + 1 > 0 voi x la so nguyen duong

TH2 : x la so nguyen am

Co 2x^8 + 2x^7 + 1 = duong + am + duong .

Do 2x^8 > 2x^7 nen tong tren mang dau duong

Ma so duong luon lon hon 0

=> 2x^8 + 2x^7 + 1 > 0 voi x la so nguyen am

TH3 : x = 0 

Voi x = 0 ta co 2x^8 + 2x^7 + 1 = 0 + 0 + 1 = 1

Ma 1 > 0 

=> 2x^8 + 2x^7 + 1 > 0 voi x = 0 

Vay 2x^8 + 2x^7 + 1 > 0 voi moi x