a: Ta có: \(\left(3x+5\right)^2-4x^2=0\)

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

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

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

a)5(x+1)(x-x-2)=0

=>5(x+1).-2=0

=>5(x+1)=0

=>x+1=0

=>x=-1

a)5x.(x+1)-5.(x+1).(x-2)=0

⇒5x(x+1)-(5x-10)(x+1)=0

⇒(x+1)(5x-5x+10)=0

⇒10(x+1)=0

⇒x+1=0⇒x=-1

e: ta có: \(4x^2+4x-6=2\)

\(\Leftrightarrow4x^2+4x-8=0\)

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

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

f: Ta có: \(2x^2+7x+3=0\)

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

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

a) 3x(4x-3)-2x(5-6x)=0

\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)

\(\Leftrightarrow24x^2-19x=0\)

\(\Leftrightarrow x\left(24x-19\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{19}{24}\)

b) 5(2x-3)+4x(x-2)+2x(3-2x)=0

\(\Leftrightarrow\)10x-15+4x²-8x+6x-4x²=0

\(\Leftrightarrow8x-15=0\)

\(\Leftrightarrow8x=15\)

\(\Leftrightarrow x=\dfrac{15}{8}\)

vậy x=\(\dfrac{15}{8}\)

Bài 1:

a. $3x^3-12x^2+12x=3x(x^2-4x+4)=3x(x-2)^2$

b. $x^2-25+4xy+4y^2=(x^2+4xy+4y^2)-25=(x+2y)^2-5^2=(x+2y-5)(x+2y+5)$

c. $4x^3-x=x(4x^2-1)=x[(2x)^2-1^2]=x(2x-1)(2x+1)$

d. $x^2-x+2y-4y^2=(x^2-4y^2)-(x-2y)=(x-2y)(x+2y)-(x-2y)=(x-2y)(x+2y+1)$

Bài 2: 

a. $3x(x-1)+x-1=0$

$\Leftrightarrow (x-1)(3x+1)=0$

$\Leftrightarrow x-1=0$ hoặc $3x+1=0$

$\Leftrightarrow x=1$ hoặc $x=\frac{-1}{3}$

b. $x(2x+1)-4x^2+1=0$

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

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

$\Leftrightarrow (2x+1)[x-(2x-1)]=0$

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

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

$\Leftrightarrow x=\frac{-1}{2}$ hoặc $x=1$

a)   \(\left(2x-1\right)^2-25=0\)

⇔ \(\left(2x-1\right)^2-5^2=0\)

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

⇒  \(2x-1-5=0\) hoặc \(2x-1+5=0\)

⇔      \(x=3\)           hoặc  \(x=-2\)

Bài 1: Tìm x

a) (2x-1) ² - 25 = 0

<=> (2x-1)² =  25

<=>  2x-1 = 5  hay 2x-1 =-5

<=>  2x= 6      hay  2x=-4

<=>   x=3     hay    x= -2

Vậy S={3; -2}
b) 3x (x-1) + x - 1 = 0

<=> (x-1)(3x+1)=0

<=> x-1=0  hay  3x+1=0

<=> x=1 hay 3x=-1

<=> x=1 hay x=\(\dfrac{-1}{3}\)

Vậy S={1;\(\dfrac{-1}{3}\)}

c) 2(x+3) - x ² - 3x = 0

<=> 2(x+3)- x(x+3)=0

<=> (x+3)(2-x)=0

<=> x+3=0 hay 2-x=0

<=> x=-3  hay  x=2

Vậy S={-3;2}
d) x(x - 2) + 3x - 6 = 0

<=> x(x-2)+3(x-2)=0

<=> (x-2)(x+3)=0

<=> x-2=0 hay x+3=0

<=> x=2 hay x=-3

Vậy S={2;-3}
e) 4x ² - 4x +1 = 0

<=> (2x-1)²=0

<=> 2x-1=0

<=> 2x=1

<=> x=\(\dfrac{1}{2}\)

Vậy S={\(\dfrac{1}{2}\)}
f) x +5x²  = 0

<=> x(1+5x)=0

<=>x=0 hay 1+5x=0

<=> x=0 hay 5x=-1

<=> x=0 hay x= \(\dfrac{-1}{5}\)

Vậy S={0;\(\dfrac{-1}{5}\)}
g) x ²+ 2x -3 = 0

<=> x²-x+3x-3=0

<=> x(x-1)+3(x-1)=0

<=>  (x-1)(x+3)=0

<=> x-1=0 hay x+3=0

<=> x=1  hay x=-3

Vậy S={1;-3}

a, \(2x\left(x-3\right)-15+5x=0\\ \Rightarrow2x\left(x-3\right)-\left(15-5x\right)=0\\ \Rightarrow2x\left(x-3\right)-5\left(3-x\right)=0\\ \Rightarrow\left(2x+5\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=3\end{matrix}\right.\)

b, \(x^3-7x=0\\ \Rightarrow x\left(x^2-7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm7\end{matrix}\right.\)

c, \(\left(2x-3\right)^2-\left(x+5\right)^2=0\\ \Rightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\\ \Rightarrow\left(x-8\right)\left(3x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)

Xem lại đề câu d 

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

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

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

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

a) \(\left|4x-1\right|-\left|3x-\dfrac{1}{2}\right|=0\\ \Leftrightarrow\left|4x-1\right|=\left|3x-\dfrac{1}{2}\right|\\ \Leftrightarrow\left[{}\begin{matrix}4x-1=3x-\dfrac{1}{2}\\4x-1=\dfrac{1}{2}-3x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}4x-3x=1-\dfrac{1}{2}\\4x+3x=\dfrac{1}{2}+1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\7x=\dfrac{3}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{14}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{1}{2};\dfrac{3}{14}\right\}\) là nghiệm của pt.

b) \(\left|x-1\right|-2x=\dfrac{1}{2}\\ \Leftrightarrow\left|x-1\right|=2x+\dfrac{1}{2}\left(ĐK:x\ge\dfrac{-1}{4}\right)\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2x+\dfrac{1}{2}\\x-1=-2x-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-2x=1+\dfrac{1}{2}\\x+2x=1-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-x=\dfrac{3}{2}\\3x=\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\left(ktmđk\right)\\x=\dfrac{1}{6}\left(tmđk\right)\end{matrix}\right.\)

Vậy \(x=\dfrac{1}{6}\) là nghiệm của pt.

Lời giải:

a.

$|4x-1|-|3x-\frac{1}{2}|=0$

$\Leftrightarrow |4x-1|=|3x-\frac{1}{2}$

\(\Leftrightarrow \left[\begin{matrix} 4x-1=3x-\frac{1}{2}\\ 4x-1=\frac{1}{2}-3x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{1}{2}\\ x=\frac{3}{14}\end{matrix}\right.\)

b. Nếu $x\geq 1$ thì:

$|x-1|-2x=\frac{1}{2}$

$\Leftrightarrow x-1-2x=\frac{1}{2}$
$\Leftrightarrow -x-1=\frac{1}{2}$

$\Leftrightarrow x=\frac{-3}{2}$ (vô lý vì $x\geq 1$)

Nếu $x< 1$ thì:

$1-x-2x=\frac{1}{2}$

$\Leftrightarrow x=\frac{1}{6}$ (tm)