a) x³ - 16x = 0

x(x² - 16) = 0

=> x = 0 hoặc x² - 16 = 0

x = 4

Vậy x = 0 hoặc x = 4

b) x⁴ -2x³ + 10x² - 20x = 0

x³ (x - 2) + 10x(x - 2) = 0

(x - 2)(x³ + 10x) = 0

=> x - 2 = 0 hoặc x³ + 10x = 0

x = 2 x(x² + 10) = 0

+ TH1: x = 0

+ TH2: x² + 10 = 0

x² = -10 (vô lí)

Vậy x = 2 hoặc x = 0

c) (2x - 3)² = (x + 5)²

(2x)² + 2 . 2x . 3 + 3² = x² + 2.x.5 + 5²

4x² + 12x + 9 = x² + 10x + 25

4x² + 12x - x² - 10x = 25 - 9

3x² + 2x = 16

x(3x + 2) = 16

Đến đây bạn làm nốt câu c nhé!

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

^^

\(o,x^2-9x+20=0\)

\(\Leftrightarrow x^2-4x-5x+20=0\)

\(\Leftrightarrow x\left(x-4\right)-5\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)

\(n,3x^3-3x^2-6x=0\)

\(\Leftrightarrow3x\left(x^2-x-2\right)=0\)

\(\Leftrightarrow3x\left(x^2+x-2x-2\right)=0\)

\(\Leftrightarrow3x\left[x\left(x+1\right)-2\left(x+1\right)\right]=0\)

\(\Leftrightarrow3x\left(x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}3x=0\\x+1=0\end{cases}}\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}x=0\\x=-1\end{cases}}\\x=2\end{cases}}\)

\(m,x^2-11x+28=0\)

\(\Leftrightarrow x^2-4x-7x+28=0\)

\(\Leftrightarrow x\left(x-4\right)-7\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-7=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=7\end{cases}}\)

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

\(\Leftrightarrow16x^2+24x+9=4x^2-8x+4\)

\(\Leftrightarrow16x^2+24x+9-4x^2+8x-4=0\)

\(\Leftrightarrow12x^2+32x+5=0\)

\(\Leftrightarrow\left(x+\frac{1}{6}\right)\left(x+\frac{5}{2}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{6}=0\\x+\frac{5}{2}=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{1}{6}\\x=-\frac{5}{2}\end{cases}}\)

A. \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x^2+3x+2x+6\right)-\left(x^2+5x-2x-10\right)=0\)
\(\Leftrightarrow x^2+3x+2x+6-x^2-5x+2x+10=0\)
\(\Leftrightarrow x^2+3x+2x-x^2-5x+2x=-6-10\)
\(\Leftrightarrow2x=-16\)
\(\Leftrightarrow x=-8\) .Vậy \(S=\left\{-8\right\}\)

B. \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x+5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x+5x-20\)
\(\Leftrightarrow2x^2-8x+3x+x^2-2x-5x-3x^2+12x-5x=12-10-20\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\) . Vậy \(S=\left\{\dfrac{18}{5}\right\}\)

C. \(\left(8-4x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow8x+16-4x^2-8x+4\left(x^2+x-2x-2\right)=0\)
\(\Leftrightarrow8x+16-4x^2-8x+4x^2+4x-8x-8=0\)
\(\Leftrightarrow8x-4x^2-8x+4x^2+4x-8x=-16+8\)
\(\Leftrightarrow-4x=-8\)
\(\Leftrightarrow x=2\) . Vậy \(S=\left\{2\right\}\)

D. \(\left(2x-3\right)\left(8x+2\right)=\left(4x+1\right)\left(4x-1\right)-3\)
\(\Leftrightarrow16x^2+4x-24x-6=16x^2+1^2-3\)
\(\Leftrightarrow16x^2+4x-24x-16x^2=6+1-3\)
\(\Leftrightarrow-20x=4\)
\(\Leftrightarrow x=-\dfrac{1}{5}\) . Vậy \(S=\left\{-\dfrac{1}{5}\right\}\)

a)(x+2)(x+3)-(x-2)(x+5)=0

\(\Leftrightarrow x^2+3x+2x+6-x^2-5x+2x+10=0\)

<=>2x=-16

<=>x=-8

b)(2x+3)(x-4)+(x-5)(x-2)=(3x-5)(x-4)

\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)

\(\Leftrightarrow3x^2-12x-2=3x^2-17x+20\)

\(\Leftrightarrow5x=22\Leftrightarrow x=\dfrac{22}{5}\)

c)(8-4x)(x+2)+4(x-2)(x+1)=0

\(\Leftrightarrow8x+16-4x^2-8x+4x^2+4x-8x-8=0\)

\(\Leftrightarrow-4x=-8\Leftrightarrow x=2\)

d)(2x-3)(8x+2)=(4x+1)(4x-1)-3

\(\Leftrightarrow16x^2+4x-24x-6=16x^2-4x+4x-1-3\)

\(\Leftrightarrow-20x=-2\Leftrightarrow x=\dfrac{-1}{10}\)

\(2x^2-7x+5=0\)

\(2x^2-2x-5x+5=0\)

\(2x\left(x-1\right)-5\left(x-1\right)=0\)

\(\left(x-1\right)\left(2x-5\right)=0\)

\(\left[\begin{array}{nghiempt}x-1=0\\2x-5=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=1\\2x=5\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=1\\x=\frac{5}{2}\end{array}\right.\)

\(x\left(2x-5\right)-4x+10=0\)

\(x\left(2x-5\right)-2\left(2x-5\right)=0\)

\(\left(2x-5\right)\left(x-2\right)=0\)

\(\left[\begin{array}{nghiempt}x-2=0\\2x-5=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=2\\2x=5\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=2\\x=\frac{5}{2}\end{array}\right.\)

\(\left(x-5\right)\left(x+5\right)-x\left(x-2\right)=15\)

\(x^2-25-x^2+2x=15\)

\(2x=15+25\)

\(2x=40\)

\(x=\frac{40}{2}\)

\(x=20\)

\(x^2\left(2x-3\right)-12+8x=0\)

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

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

\(2x-3=0\) (vì \(x^2\ge0\Rightarrow x^2+4\ge4>0\))

\(2x=3\)

\(x=\frac{3}{2}\)

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

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

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

\(\left[\begin{array}{nghiempt}x-1=0\\x+5=0\end{array}\right.\)

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

\(\left(2x-3\right)^2-4x\left(x-1\right)=5\)

\(4x^2-12x+9-4x^2+4x=5\)

\(-8x=5-9\)

\(-8x=-4\)

\(x=\frac{4}{8}\)

\(x=\frac{1}{2}\)

\(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(5x-2x^2+2x^2-2x=13\)

\(3x=13\)

\(x=\frac{13}{3}\)

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

\(\left(2x+10\right)\left(2x-5\right)-\left(x-1\right)\left(2x-5\right)=0\)

\(\left(2x-5\right)\left(2x+10-x+1\right)=0\)

\(\left(2x-5\right)\left(x+11\right)=0\)

\(\left[\begin{array}{nghiempt}2x-5=0\\x+11=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}2x=5\\x=-11\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-11\end{array}\right.\)

\(a,2x^2-7x+5=0\Leftrightarrow2x^2-2x-5x+5=0\Leftrightarrow2x\left(x-1\right)-5\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(2x-5\right)=0\Rightarrow\left[{}\begin{matrix}x-1=0\\2x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\2x=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2,5\end{matrix}\right.\)\(b,x\left(2x-5\right)-4x+10=0\Rightarrow x\left(2x-5\right)-2\left(2x-5\right)=0\Leftrightarrow\left(x-2\right)\left(2x-5\right)=0\Rightarrow\left[{}\begin{matrix}x-2=0\\2x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\2x=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=2,5\end{matrix}\right.\)\(c,\left(x-5\right)\left(x+5\right)-x\left(x-2\right)=15\Leftrightarrow x^2-25-x^2+2x-15=0\Leftrightarrow2x-40=0\Rightarrow2x=40\Rightarrow x=20\)\(d,x^2\left(2x-3\right)-12+8x=0\Rightarrow2x^3-3x^2-12+8x=0\Leftrightarrow2x^3+8x-3x^2-12=0\Leftrightarrow2x\left(x^2+4\right)-2\left(x^2+4\right)=0\Leftrightarrow\left(2x-2\right)\left(x^2+4\right)=0\Rightarrow\left[{}\begin{matrix}2x-2=0\\x^2+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=2\\x^2=-4\end{matrix}\right.\Rightarrow x=1\)

a. \(\left(2x-1\right)\left(3x+2\right)\left(5-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+2=0\\5-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{-2}{3}\\x=5\end{matrix}\right.\)

\(\Rightarrow S=\left\{\dfrac{1}{2};\dfrac{-2}{3};5\right\}\)

b. \(\left(2x+5\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)

\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)\)

\(\Leftrightarrow3x\left(x-4\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

\(\Rightarrow S=\left\{0;4\right\}\)

c. \(16x^2-8x+1=4\left(x+3\right)\left(4x-1\right)\)

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

\(\Leftrightarrow\left(4x-1\right)\left(4x-1-4x-3\right)=0\)

\(\Leftrightarrow-4\left(4x-1\right)=0\Leftrightarrow4x-1=0\Leftrightarrow x=\dfrac{1}{4}\)

d. \(27x^2\left(x+3\right)-12\left(x^2+3x\right)=0\)

\(\Leftrightarrow27x^2\left(x+3\right)-12x\left(x+3\right)=0\)

\(\Leftrightarrow x\left(27x-12\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\27x-12=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\\x=-3\end{matrix}\right.\)

\(\Rightarrow S=\left\{0;\dfrac{4}{9};-3\right\}\)

e. \(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)

\(\Leftrightarrow2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(6x+1-x+2\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(7x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\7x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=\dfrac{-3}{7}\end{matrix}\right.\)

\(\Rightarrow S=\left\{\dfrac{-1}{3};\dfrac{-3}{7}\right\}\)

g. \(\left(2x-1\right)^2=49\)

\(\Leftrightarrow2x-1=7\Leftrightarrow x=4\)

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

\(\Rightarrow\left(2x-3\right)^2-\left(x+5\right)^2=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)

* \(x^3-16x=0\)

\(\Rightarrow x\left(x^2-16\right)=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

*\(x^4-2x^3+10x^2-20x=0\)

\(\Rightarrow\left(x^4+10x^2\right)-\left(2x^3+20x\right)=0\)

\(\Rightarrow x^2\left(x^2+10\right)-2x\left(x^2+10\right)=0\)

\(\Rightarrow\left(x^2+10\right)\left(x^2-2x\right)=0\)

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