a) (2x - 5)² - (5 + 2x) = 0

<=> 4x² - 22x + 20 = 0 

\(\Leftrightarrow\left(2x-\dfrac{11}{2}\right)^2=\dfrac{41}{4}\)

\(\Leftrightarrow x=\dfrac{\pm\sqrt{41}+11}{4}\)

b) \(27x^3-54x^2+36x=0\)

\(\Leftrightarrow x\left(3x^2-6x+4\right)=0\)

\(\Leftrightarrow x=0\) (Vì \(3x^2-6x+4=3\left(x-1\right)^2+1>0\forall x\))

c) x³ + 8 - (x + 2).(x - 4) = 0

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

\(\Leftrightarrow\left(x+2\right)\left(x^2-3x+8\right)=0\)

\(\Leftrightarrow x=-2\) (Vì \(x^2-3x+8=\left(x-\dfrac{3}{2}\right)^2+\dfrac{23}{4}>0\))

d) \(x^6-1=0\)

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

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

\(\Leftrightarrow x^2-1=0\) (Vì \(x^4+x^2+1>0\))

\(\Leftrightarrow x=\pm1\)

\(d,x^6-1=0\\ \Leftrightarrow\left(x^2\right)^3-1^3=0\\ \Leftrightarrow\left(x^2-1\right)\left(x^4+x^2+1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^4+x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x^4+x^2+1=0\left(Vô.lí,vì:x^4\ge0;x^2\ge0,\forall x\in R\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\\ c,\left(x^3+8\right)-\left(x+2\right)\left(x-4\right)=0\\ \Leftrightarrow\left(x^3+8\right)-\left(x^2-2x-8\right)=0\\ \Leftrightarrow x^3-x^2+2x+16=0\\ \Leftrightarrow x^3+2x^2-3x^2-6x+8x+16=0\\ \Leftrightarrow x^2\left(x+2\right)-3x\left(x+2\right)+8\left(x+2\right)=0\\ \Leftrightarrow\left(x^2-3x+8\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2-3x+8=0\left(Vô.lí\right)\\x+2=0\end{matrix}\right.\Leftrightarrow x=-2\)

c)(x^3+ 8) - (x + 2)(x - 4) = 0
<=> x^3 -x^2 + 2x +8 + 8 = 0
<=> x^3 -x^2 + 2x + 16 = 0
<=> (x+2)(x^2-3x+8) = 0
=> x = -2 

2: \(=\dfrac{-2}{75}+\dfrac{5}{39}=\dfrac{33}{325}\)

3: \(=\dfrac{6}{11}\left(\dfrac{4}{9}+\dfrac{5}{9}\right)=\dfrac{6}{11}\)

4: \(=\dfrac{7}{19}\left(\dfrac{5}{13}+\dfrac{8}{13}-1\right)=-2\cdot\dfrac{7}{19}=-\dfrac{14}{19}\)

5: \(=\dfrac{2}{7}\left(\dfrac{4}{23}-\dfrac{27}{23}+1\right)=0\)

6: \(=\dfrac{3}{8}\left(\dfrac{3}{7}+\dfrac{4}{7}\right)+\dfrac{11}{8}=\dfrac{3}{8}+\dfrac{11}{8}=\dfrac{14}{8}=\dfrac{7}{4}\)

a.

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

b.

\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)

c.

\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)

\(=\left(x+3\right)^3\)

d.

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

e.

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

\(=\left(x-y\right)\left(x+y-5\right)\)

f.

\(x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)

g. 10x(x-y)-6y(y-x)

=10x(x-y)+6y(x-y)

=(x-y)(10x+6y)

h.x²-4x-5

=(x-5)(x+1)

i.x⁴-y⁴ = (x²-y²)(x²+y²)

B2.

a.5(x-2)=x-2

⇔5(x-2)-(x-2)=0

⇔4(x-2)=0

⇔x=2

b.3(x-5)=5-x

⇔3(x-5)+(x-5)=0

⇔4(x-5)=0

⇔x=5

c.(x+2)²-(x+2)(x-2)=0

⇔(x+2)[(x+2)-(x-2)]=0

⇔4(x+2)=0

⇔x=-2

a) = x^2 - 9 - (x^2 + 3x - 10)

= -3x + 1

b) = 3x + 1 - 3x + 19

= 20

a: \(\left(x+3\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\)

\(=x^2-9-x^2-3x+10\)

\(=-3x+1\)

b: \(\dfrac{27x^3+1}{9x^2-3x+1}-\left(3x-19\right)\)

\(=3x+1-3x+19\)

=20

Với x = -3 thì (x -4).(x + 5) = (-3 -4 ).(-3 + 5) = (-7).2 = -14

Vậy chọn (d)

a: Ta có: \(27x^3-54x^2+36x=8\)

\(\Leftrightarrow27x^3-54x^2+36x-8=0\)

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

\(\Leftrightarrow3x-2=0\)

hay \(x=\dfrac{2}{3}\)

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

\(\Leftrightarrow\left(x+3\right)\cdot\left(x^2-4x+5\right)=0\)

\(\Leftrightarrow x=-3\)

Chia nhỏ ra cậu ơi :v

Cậu đặt câu hỏi free nên đặt nhỏ ra thì mới có người làm nha để như này dày cộp không ai dám làm đou =(((

a: \(=\dfrac{35x^3-14x^2+55x^2-22x+35x-14+9}{5x-2}\)

\(=7x^2-11x+7+\dfrac{9}{5x-2}\)

b: \(=\dfrac{\left(2x-3\right)\left(4x^2+6x+9\right)}{2x-3}=4x^2+6x+9\)