a) (x + 3)² - (x - 2)² = 2x

=> (x + 3 - x + 2)(x + 3 + x - 2) = 2x

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

=> 10x + 5 = 2x

=> 10x - 2x = -5

=> 8x = -5

=> x = -5/8

b) 7x(x - 2) = x - 2

=> 7x(x - 2) - (x - 2) = 0

=> (7x - 1)(x - 2) = 0

=> \(\orbr{\begin{cases}7x-1=0\\x-2=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{1}{7}\\x=2\end{cases}}\)

c) 8x³ - 12x² + 6x - 1 = 0

=> (2x - 1)³ = 0

=> 2x - 1 = 0

=> 2x = 1

=> x = 1/2

b) \(7x\left(x-2\right)-\left(x-2\right)=0\) 

<=>  \(\left(7x-1\right)\left(x-2\right)=0\)

=> x=1/7  hoặc x=2

c) <=>  (2x-1)³   =0 

=> x=1/2

d)<=>  \(\left(2x-3\right)\left(2x+3\right)-x\left(2x-3\right)=0\)

<=>  \(\left(2x-3\right)\left(x+3\right)=0\)

=> x=3/2  hoặc x=-3

e) <=>\(x^2\left(x+5\right)+9\left(x+5\right)=0\)

<=> \(\left(x+5\right)\left(x^2+9\right)=0\)

=> x=-5

f) \(x^3-6x^2-x+30=0\)

<=>\(x^3+2x^2-8x^2-16x+15x+30=0\)

<=>\(x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)=0\)

<=>\(\left(x+2\right)\left(x^2-5x-3x+15\right)=0\)

<=> \(\left(x+2\right)\left(x-5\right)\left(x-3\right)=0\)

=> x=-2 hoặc x=5 hoặc x=3

Ik mk nha, hôm nay ngày mai, ngày kia mk ik 3 lần lại cho bạn (thành 9 lần)

Nhớ kb với mìn lun nha!! Mk rất vui đc làm quen vs bạn, cảm ơn mn nhìu lắm

a)\(7x\left(x-2\right)=\left(x-2\right)\)

\(\Leftrightarrow7x\left(x-2\right)-\left(x-2\right)=0\)

\(\Leftrightarrow\left(7x-1\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}7x-1=0\\x-2=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}\\x=2\end{matrix}\right.\)

b)\(4x^2-9-x\left(2x-3\right)=0\)

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

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

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

\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\x+3=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-3\end{matrix}\right.\)

c)\(x^3+5x^2+9x=-45\)

\(\Leftrightarrow x^3+9x+5x^2+45=0\)

\(\Leftrightarrow x\left(x^2+9\right)+5\left(x^2+9\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x^2+9\right)=0\)

Dễ thấy: \(x^2+9\ge 9 >0\forall x\)

\(\Rightarrow x+5=0\Rightarrow x=-5\)

d,e tương tự

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

^^

e: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow2x=-7\)

hay \(x=-\dfrac{7}{2}\)

f: Ta có: \(x^3-6x^2+12x-19=0\)

\(\Leftrightarrow x^3-6x^2+12x-8-11=0\)

\(\Leftrightarrow\left(x-2\right)^3=11\)

hay \(x=\sqrt[3]{11}+2\)

c: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)