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)

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

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

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

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

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

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

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

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

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

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

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

a) Ta có: 2x+33=-11

nên 2x=-44

hay x=-22

b) Ta có: \(\dfrac{x}{2}=\dfrac{-49}{14}\)

nên x=-7

c) Ta có: \(\dfrac{5}{6}x+\dfrac{10}{3}=\dfrac{7}{2}\)

nên \(\dfrac{5}{6}x=\dfrac{7}{2}-\dfrac{10}{3}=\dfrac{1}{6}\)

hay \(x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\)

Mk xin phép ko vt lại đề nx

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

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

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

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

\(\left(x+3\right)^2+\left(x-2\right)^2=2x^2\\ \Leftrightarrow x^2+6x+9+x^2-4x+4=2x^2\\ \Leftrightarrow2x=-13\Leftrightarrow x=-\dfrac{13}{2}\)

<=> x²+6x+9+x²+4x+4-2x² =0

<=>10x+13=0

<=> x= -13/10

Chúc bạn học tốt nha!

`(x+3)^2 +(x-2)^2=2x^2`

`<=>x^2+6x+9 +x^2-4x+4-2x^2=0`

`<=> 2x +13=0`

`<=>2x=-13`

`<=>x=(-13)/2`

vậy `x=(-13)/2`

\(\dfrac{3}{x-5}=\dfrac{-4}{x+2}\left(x\ne5;-2\right).\\ \Leftrightarrow\dfrac{3}{x-5}+\dfrac{4}{x+2}=0.\\ \Leftrightarrow\dfrac{3x+6+4x-20}{\left(x-5\right)\left(x+2\right)}=0.\\ \Rightarrow7x=14.\\ \Leftrightarrow x=2\left(TM\right).\)

a: \(=\dfrac{x+2-x+2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{2}=\dfrac{2}{x-2}\)

b: Để A=2 thì 2/x-2=2

=>x-2=1

=>x=3

\(a,50\%x-0,2+x=\dfrac{4}{5}\)

\(\Leftrightarrow\dfrac{1}{2}x-0,2+x=\dfrac{4}{5}\)

\(\Leftrightarrow\dfrac{1}{2}x+x=\dfrac{4}{5}+0,2\)

\(\Leftrightarrow\dfrac{3}{2}x=\dfrac{4}{5}+\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{3}{2}x=1\)

\(\Leftrightarrow x=\dfrac{2}{3}\)

\(b,\left(x-\dfrac{3}{4}\right):\dfrac{1}{2}+\dfrac{3}{2}=\dfrac{25}{2}\)

\(\Leftrightarrow\left(x-\dfrac{3}{4}\right).2=\dfrac{25}{2}-\dfrac{3}{2}\)

\(\Leftrightarrow\left(x-\dfrac{3}{4}\right).2=\dfrac{22}{2}\)

\(\Leftrightarrow x-\dfrac{3}{4}=11:2\)

\(\Leftrightarrow x=\dfrac{11}{2}+\dfrac{3}{4}\)

\(\Leftrightarrow x=\dfrac{25}{4}\)