Bài 1: 

a: \(\Leftrightarrow x-1\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{2;0;4;-2\right\}\)

\(|-2x+1,5|=\dfrac{1}{4}\Rightarrow-2x+1,5=\pm\dfrac{1}{4}\)

\(-2x+1,5=\dfrac{1}{4}\Rightarrow-2x=1,5-0,25\Rightarrow-2x=1,25\Rightarrow x=1,25:\left(-2\right)\Rightarrow x=...\)

\(-2x+1,5=-\dfrac{1}{4}\Rightarrow-2x=-0,25-1,5\Rightarrow-2x=1,75\Rightarrow x=1,75:\left(-2\right)\Rightarrow x=...\)

\(\dfrac{3}{2}-|1.\dfrac{1}{4}+3x|=\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{3}{2}-\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{5}{4}\)

\(\Rightarrow1.\dfrac{1}{4}+3x=\pm\dfrac{5}{4}\)

\(1.\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow3x=\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=1\Rightarrow x=3\)

\(1.\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow3x=-\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=-\dfrac{3}{2}x=...\)

a: ta có: \(\left|-2x+\dfrac{3}{2}\right|=\dfrac{1}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x+\dfrac{3}{2}=\dfrac{1}{4}\\-2x+\dfrac{3}{2}=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-2x=-\dfrac{5}{4}\\-2x=-\dfrac{7}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{7}{8}\end{matrix}\right.\)

b: Ta có: \(\dfrac{3}{2}-\left|\dfrac{5}{4}+3x\right|=\dfrac{1}{4}\)

\(\Leftrightarrow\left|3x+\dfrac{5}{4}\right|=\dfrac{5}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{5}{4}=\dfrac{5}{4}\\3x+\dfrac{5}{4}=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=0\\3x=-\dfrac{5}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{6}\end{matrix}\right.\)

Mk xin phép ko vt lại đề nx

\(\Rightarrow A=\left[\left(3x-2\right)\left(x+1\right)-\left(2x+5\right)\left(x^2-1\right)\right]\div x+1\)

\(\Rightarrow A=3x-2-\left(2x-5\right)\left(x-1\right)\)

\(\Rightarrow x=\dfrac{1}{2}\)

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

\(a\\ -5x^2+3x.\left(x+2\right)=-5x^2+3x^2+6x=-2x^2+6x\\ b\\ -2x.\left(1-x^2\right)-2x^3=-2x+2x^3-2x^3=-2x\\ c\\ 4x.\left(x-1\right)-4.\left(x^2+2x-1\right)\\ =4x^2-4x-4x^2-8x+4=-12x+4\)

\(d\\ 6x^3-2x^2.\left(-x^2-3x\right)=6x^3+2x^4+6x^3=2x^4+12x^3\\ e\\ 3x.\left(x-1\right)-\left(1+2x\right).5x\\ =3x^2-3x-5x-10x^2=-7x^2-8x\\ f\\ -5x^2-\left(x-6\right).\left(-2x^2\right)=-5x^2+2x^3-12x^2=2x^3-17x^2\)

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)

1) \(\Rightarrow16x^2+24x+9+9x^2-24x+16+4-25x^2=x\)

\(\Rightarrow x=29\)

2)

a) \(=x^2-9-x^2+6x-9=6x-18\)

b) \(=\left(3x-1+2x+1\right)^2=\left(5x\right)^2=25x^2\)

Bài 5: 

a. 1 - 2y + y²

= (1 - y)²

b. (x + 1)² - 25

= (x + 1)² - 5²

= (x + 1 - 5)(x + 1 + 5)

= (x - 4)(x + 6)

c. 1 - 4x²

= 1² - (2x)²

= (1 - 2x)(1 + 2x)

d. 8 - 27x³

= 2³ - (3x)³

= (2 - 3x)(4 + 6x + 9x²)

e. (đề hơi khó hiểu ''x3'' !?)

g. x³ + 8y³

= (x + 2y)(x² - 2xy + y²)