a)\(=>2x=-10=>x=-5\)

b)\(=>-2x=-5=>x=\dfrac{-5}{-2}=\dfrac{5}{2}\)

c)\(4-x=0=>x=4-0=4\)

d)\(=>2x=-1=>x=-\dfrac{1}{2}\)

e)\(=>x^2=-2\)=> x ko tồn tại

f)\(=>x\left(2+1\right)=0=>3x=0=>x=0\)

2: \(3x\left(x-4\right)+2x-8=0\)

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

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

=>\(\left[{}\begin{matrix}x-4=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)

3: 4x(x-3)+x²-9=0

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

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

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

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

4: \(x\left(x-1\right)-x^2+3x=0\)

=>\(x^2-x-x^2+3x=0\)

=>2x=0

=>x=0

5: \(x\left(2x-1\right)-2x^2+5x=16\)

=>\(2x^2-x-2x^2+5x=16\)

=>4x=16

=>x=4

\(a,x^4-2x^3+5x^2-10x=0\\ \Leftrightarrow x^3\left(x-2\right)+5x\left(x-2\right)=0\\ \Leftrightarrow x\left(x^2+5\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x^2+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x\in\varnothing\left(x^2+5>0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(b,\left(3x+5\right)^2=\left(2x-2\right)^2\\ \Leftrightarrow\left(3x+5\right)^2-\left(2x-2\right)^2=0\\ \Leftrightarrow\left(3x+5+2x-2\right)\left(3x+5-2x+2\right)=0\\ \Leftrightarrow\left(5x+3\right)\left(x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=-7\end{matrix}\right.\)

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

\(d,x^2\left(x-1\right)-4x^2+8x-4=0\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a) \(x^4-2x^3+5x^2-10x=0\\ \Rightarrow\left(x^4-2x^3\right)+\left(5x^2-10x\right)=0\\ \Rightarrow x^3\left(x-2\right)+5x\left(x-2\right)=0\\ \Rightarrow\left(x^3+5x\right)\left(x-2\right)=0\\ \Rightarrow x\left(x^2+5\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^2+5=0\\x-2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\sqrt{5}\\x=2\end{matrix}\right.\)

Vậy \(x=\left\{-\sqrt{5};0;\sqrt{5};2\right\}\)

b) \(\left(3x+5\right)^2=\left(2x-2\right)^2\\ \Rightarrow\left[{}\begin{matrix}3x+5=2x-2\\3x+5=-2x+2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-7\\x=-\dfrac{3}{5}\end{matrix}\right.\)

c) \(x^3-2x^2+x=0\\ \Rightarrow x\left(x^2-2x+1\right)=0\\ \Rightarrow x\left(x-1\right)^2=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

vậy ...

d) \(x^2\left(x-1\right)-4x^2+8x-4=0\\ x^2\left(x-1\right)-\left(4x^2-8x+4\right)=0\\ x^2\left(x-1\right)-\left(2x-2\right)^2=0\\ \Rightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Rightarrow\left(x-1\right)\left[x^2-4\left(x-1\right)\right]=0\\ \Rightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Rightarrow\left(x-1\right)\left(x-2\right)^2=0\)

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

a: Ta có: \(x^4-2x^3+5x^2-10x=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

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

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

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

\(\Leftrightarrow\left(x+7\right)\left(5x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=-\dfrac{3}{5}\end{matrix}\right.\)

a: \(8x\left(x-2017\right)-2x+4034=0\)

\(\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)

Answer:

\(\left(2x-3\right).\left(x+1\right)-x.\left(2x+3\right)-9=0\)

\(\Rightarrow\left(2x^2+2x-3x-3\right)-2x^2-3x-9=0\)

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

\(\Rightarrow2x^2-x-3-2x^2-3x-9=0\)

\(\Rightarrow\left(2x^2-2x^2\right)-\left(x+3x\right)-\left(3+9\right)=0\)

\(\Rightarrow-4x-12=0\)

\(\Rightarrow x+3=0\)

\(\Rightarrow x=-3\)

\(2x.\left(x-3\right)-x+3=0\) (Sửa đề)

\(\Rightarrow2x.\left(x-3\right)-\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right).\left(2x-1\right)=0\)

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

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

\(\Rightarrow2x.\left(x^2-4\right)-6.\left(x^2-4\right)=0\)

\(\Rightarrow2.\left(x-3\right).\left(x+2\right).\left(x-2\right)=0\)

Trường hợp 1: \(x-3=0\Rightarrow x=3\)

Trường hợp 2: \(x+2=0\Rightarrow x=-2\)

Trường hợp 3: \(x-2=0\Rightarrow x=2\)

a) (2x + 1)(1 - 2x) + (1 - 2x)² = 18

= ( 1 - 2x) \(\left[\left(2x+1+1-2x\right)\right]\) = 18

= 2(1 - 2x)  - 18 = 0

= 2 - 4x - 18 = 0

= -16 - 4x = 0

= -4x = 16

= x = \(\dfrac{16}{-4}=-4\)

b) 2(x + 1)² -(x - 3)(x + 3) - (x - 4)² = 0

= 2 (x² + 2x + 1) - (x² - 9) - (x² - 8x + 16) = 0

= 2x² + 4x + 2 - x² + 9 - x² + 8x - 16 = 0

= 12x - 5 = 0

= 12x = 5

= x = \(\dfrac{5}{12}\)

c) (x - 5)² - x(x - 4) = 9

= x² - 10x + 25 - x² + 4x - 9 = 0

= -6x + 16 = 0

= -6x = -16

= x = \(\dfrac{-16}{-6}=\dfrac{8}{3}\)

d) (x - 5)² + (x - 4)(1 - x)

= x² - 10x + 25 + 5x - x² - 4 = 0

= -5x + 21 = 0

= -5x = -21

= x = \(\dfrac{-21}{-5}=\dfrac{21}{5}\) 

 Chúc bạn học tốt

a: \(x\in\left\{0;25\right\}\)

c: \(x\in\left\{0;5\right\}\)

Lời giải:

a. Đặt $x^2-2x=a$ thì pt trở thành:

$a^2+3a+2=0$

$\Leftrightarrow (a+1)(a+2)=0$

$\Leftrightarrow a+1=0$ hoặc $a+2=0$

$\Leftrightarrow x^2-2x+1=0$ hoặc $x^2-2x+2=0$

Nếu $x^2-2x+1=0\Leftrightarrow (x-1)^2=0\Leftrightarrow x=1$
Nếu $x^2-2x+2=0\Leftrightarrow (x-1)^2=-1<0$ (vô lý)

Vậy pt có nghiệm duy nhất $x=1$

b.

Đặt $x^2+x=a$ thì pt trở thành:

$a(a-4)+4=0$

$\Leftrightarrow a^2-4a+4=0$

$\Leftrightarrow (a-2)^2=0$

$\Leftrightarrow a-2=0$

$\Leftrightarrow x^2+x-2=0$

$\Leftrihgtarrow (x-1)(x+2)=0$

$\Rightarrow x=1$ hoặc $x=-2$