đáp số:5/2

cậu làm bài nhanh đi chúc bạn học tốt nha

(2/3-1-2)x=5/12

1/6x=5/12

x=5/2

dap an la 5/12

`(-2 1/3 ) .( - (-6)/4 )`

`= - (2xx3+1)/3 . 6/4`

`= - 7/3 . 6/4`

`= -42/12`

`= -7/2`

\(\left(-2\dfrac{1}{3}\right).\left(-\dfrac{-6}{4}\right)\)

\(=\left(-\dfrac{7}{3}\right).\dfrac{3}{2}=-\dfrac{7}{2}\)

\(\left|3x+5\right|=x+1\)

TH1: \(3x+5=x+1\left(x\ge-\dfrac{5}{3}\right)\)

\(\Rightarrow3x-x=1-5\)

\(\Rightarrow2x=-4\)

\(\Rightarrow x=-2\left(ktm\right)\)

TH2: \(3x-5=-\left(x+1\right)\left(x< -\dfrac{5}{3}\right)\)

\(\Rightarrow3x-5=-x-1\)

\(\Rightarrow3x+x=-1+5\)

\(\Rightarrow4x=4\)

\(\Rightarrow x=1\)

Vậy không có x thõa mãn

_______

\(\left|2x-3\right|=2x-3\)

\(\Rightarrow2x-3=2x-3\left(x\ge\dfrac{3}{2}\right)\)

\(\Rightarrow0=0\) (luôn đúng)

Nên mọi x đề thỏa mãn khi \(x\ge\dfrac{3}{2}\)

Vậy: ... 

|3x + 5| = x + 1

TH1: x ≥log ) -5/3

(1) ⇒ 3x + 5 = x + 1

3x - x = 1 - 5

2x = -4

x = -2 (loại)

*) TH2: x < -5/3

(1) ⇒ 3x + 5 = -x - 1

3x + x = -1 - 5

4x = -6

x = -3/2 (loại)

Vậy không tìm được x thỏa mãn yêu cầu

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|2x - 3| = 2x - 3 (2)

*) TH1: x 3/2

(2) ⇒ 2x - 3 = 2x - 3

0x = 0 (luôn đúng với mọi x ≥ 3/2)

*) TH2: x < 3/2

(2) ⇒ 2x - 3 = 3 - 2x

2x + 2x = 3 + 3

4x = 6

x = 3/2 (loại)

Vậy x ≥  3/2

. 

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

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

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

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

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

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

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

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

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

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

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)

\(\frac{1}{x+2}+\frac{5}{x-2}=\frac{3x-12}{x^2-4}\)

ĐKXĐ \(x\ne\pm2\)

\(\frac{1}{x+2}+\frac{5}{x-2}=\frac{3x-12}{x^2-4}\)

\(\Leftrightarrow\frac{x-2+5\left(x+2\right)}{x^2-4}=\frac{3x-12}{x^2-4}\)

\(\Leftrightarrow x-2+5x+10=3x-12\)

\(\Leftrightarrow6x+8=3x-12\)

\(\Leftrightarrow3x=20\Leftrightarrow x=\frac{20}{3}\left(tm\right)\)