8: =>6x^2-9x+2x-3-6x^2-12x=16

=>-19x=19

=>x=-1

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

bài 5:

1: \(\dfrac{12x^3y^2}{18xy^5}=\dfrac{12x^3y^2:6xy^2}{18xy^5:6xy^2}=\dfrac{2x^2}{3y^3}\)

2: \(\dfrac{10xy-5x^2}{2x^2-8y^2}=\dfrac{5x\cdot2y-5x\cdot x}{2\left(x^2-4y^2\right)}\)

\(=\dfrac{5x\left(2y-x\right)}{-2\left(x+2y\right)\left(2y-x\right)}=\dfrac{-5x}{2\left(x+2y\right)}\)

3: \(\dfrac{x^2-xy-x+y}{x^2+xy-x-y}\)

\(=\dfrac{\left(x^2-xy\right)-\left(x-y\right)}{\left(x^2+xy\right)-\left(x+y\right)}\)

\(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)

4: \(\dfrac{\left(x+1\right)\left(x^2-2x+1\right)}{\left(6x^2-6\right)\left(x^3-1\right)}\)

\(=\dfrac{\left(x+1\right)\left(x-1\right)^2}{6\left(x^2-1\right)\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{\left(x+1\right)\left(x-1\right)}{6\left(x-1\right)\left(x+1\right)\cdot\left(x^2+x+1\right)}\)

\(=\dfrac{1}{6\left(x^2+x+1\right)}\)

5: \(\dfrac{2x^2-7x+3}{1-4x^2}\)

\(=-\dfrac{2x^2-7x+3}{4x^2-1}\)

\(=-\dfrac{2x^2-6x-x+3}{\left(2x-1\right)\left(2x+1\right)}\)

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

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

Bài 3:

1: \(9x^3-xy^2\)

\(=x\cdot9x^2-x\cdot y^2\)

\(=x\left(9x^2-y^2\right)\)

\(=x\left(3x-y\right)\left(3x+y\right)\)

2: \(x^2-3xy-6x+18y\)

\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)

\(=x\left(x-3y\right)-6\left(x-3y\right)\)

\(=\left(x-3y\right)\left(x-6\right)\)

3: \(x^2-3xy-6x+18y\)

\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)

\(=x\left(x-3y\right)-6\left(x-3y\right)\)

\(=\left(x-3y\right)\left(x-6\right)\)

4: \(6xy-x^2+36-9y^2\)

\(=36-\left(x^2-6xy+9y^2\right)\)

\(=36-\left(x-3y\right)^2\)

\(=\left(6-x+3y\right)\left(6+x-3y\right)\)

5: \(x^4-6x^2+5\)

\(=x^4-x^2-5x^2+5\)

\(=x^2\left(x^2-1\right)-5\left(x^2-1\right)\)

\(=\left(x^2-5\right)\left(x^2-1\right)\)

\(=\left(x^2-5\right)\left(x-1\right)\left(x+1\right)\)

6: \(9x^2-6x-y^2+2y\)

\(=\left(9x^2-y^2\right)-\left(6x-2y\right)\)

\(=\left(3x-y\right)\left(3x+y\right)-2\left(3x-y\right)\)

\(=\left(3x-y\right)\left(3x+y-2\right)\)

a ) ( 6x + 1 )² + ( 6x - 1 )²  - 2 . ( 6x + 1 )( 6x - 1 )

= ( 6x + 1 )² - 2 . ( 6x + 1 )( 6x - 1 ) + ( 6x - 1 )²

= ( 6x + 1 - 6x + 1 )²

= 2² = 4

b ) x . ( 2x² - 3 ) - x² . ( 5x + 1 ) + x²

= 2x³ - 3x - 5x³ - x² + x²

= ( 2x³ - 5x³ ) - 3x - ( x² - x² )

= - 3x³ - 3x

= - 3x . ( x² + 1)

a) \(x^2-64=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)

b) \(4x^2-4x+1=0\)

\(\Leftrightarrow\left(2x-1\right)^2=0\Leftrightarrow2x-1=0\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

c) \(9-6x+x^2=0\)

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

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

a: Ta có: \(x^2-64=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)

b: Ta có: \(4x^2-4x+1=0\)

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

hay \(x=\dfrac{1}{2}\)

c: ta có: \(x^2-6x+9=0\)

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

hay x=3

1: \(=3x^2-6x-5x+5x^2-8x^2+24=-11x+24\)

2: \(=8x^2+12x-10x-15-4\left(2x^2-x+4x-2\right)+10x+7\)

\(=8x^2+12x-8-8x^2+4x-16x+8\)

\(=0\)

3: \(=\left(6x+1-6x+1\right)^2=4\)

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

\(=3x^3+6x-3x^3+3x=9x\)

a: ĐKXĐ: \(3x^2+6x\ne0\)

=>\(x^2+2x\ne0\)

=>\(x\cdot\left(x+2\right)\ne0\)

=>\(x\notin\left\{0;-2\right\}\)

b: ĐKXĐ: \(x^3+64\ne0\)

=>\(x^3\ne-64\)

=>\(x\ne-4\)

c: ĐKXĐ: \(x^2-1\ne0\)

=>\(x^2\ne1\)

=>\(x\notin\left\{1;-1\right\}\)