d,

\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)

e,

\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)

\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)

\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)

Vậy không tồn tại $x$ thỏa mãn đề bài.

f, 

\(\frac{2x-1}{2}=\frac{5+3x}{3}\Leftrightarrow 3(2x-1)=2(5+3x)\)

\(\Leftrightarrow 6x-3=10+6x\)

\(\Leftrightarrow 13=0\) (vô lý)

Vậy không tồn tại $x$ thỏa mãn đề bài.

a,

$0-|x+1|=5$

$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)

Do đó không tồn tại $x$ thỏa mãn điều kiện đề.

b,

\(2-|\frac{3}{4}-x|=\frac{7}{12}\)

\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)

\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)

c, 

\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)

\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)

\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)

\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)

a: 0-|x+1|=5

=>|x+1|=0-5=-5<0(vô lý)

=>x∈∅

b: \(2-\left|\frac34-x\right|=\frac{7}{12}\)

=>\(\left|x-\frac34\right|=2-\frac{7}{12}=\frac{17}{12}\)

=>\(\left[\begin{array}{l}x-\frac34=\frac{17}{12}\\ x-\frac34=-\frac{17}{12}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{17}{12}+\frac{9}{12}=\frac{26}{12}=\frac{13}{6}\\ x=-\frac{17}{12}+\frac34=-\frac{17}{12}+\frac{9}{12}=-\frac{8}{12}=-\frac23\end{array}\right.\)

c: \(2\left|\frac12x-\frac13\right|-\frac32=\frac14\)

=>\(\left|2\left(\frac12x-\frac13\right)\right|=\frac32+\frac14=\frac74\)

=>\(\left|x-\frac23\right|=\frac74\)

=>\(\left[\begin{array}{l}x-\frac23=\frac74\\ x-\frac23=-\frac74\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac74+\frac23=\frac{21+8}{12}=\frac{29}{12}\\ x=-\frac74+\frac23=\frac{-21+8}{12}=-\frac{13}{12}\end{array}\right.\)

d: \(\left|x-\frac13\right|=\frac56\)

=>\(\left[\begin{array}{l}x-\frac13=\frac56\\ x-\frac13=-\frac56\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac56+\frac13=\frac76\\ x=-\frac56+\frac13=-\frac36=-\frac12\end{array}\right.\)

e: \(\frac34-2\left|2x-\frac23\right|=2\)

=>\(2\left|2x-\frac23\right|=\frac34-2=-\frac54<0\) (vô lý)

=>x∈∅

f: \(\frac{2x-1}{2}=\frac{5+3x}{3}\)

=>3(2x-1)=2(3x+5)

=>6x-3=6x+10

=>-3=10(vô lý)

=>x∈∅

a)

\(\begin{array}{l}x.\frac{{14}}{{27}} = \frac{{ - 7}}{9}\\x = \frac{{ - 7}}{9}:\frac{{14}}{{27}}\\x = \frac{{ - 7}}{9}.\frac{{27}}{{14}}\\x = \frac{{ - 3}}{2}\end{array}\)                

Vậy \(x = \frac{{ - 3}}{2}\).

b)

\(\begin{array}{l}\left( {\frac{{ - 5}}{9}} \right):x = \frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right):\frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right).\frac{3}{2}\\x = \frac{{ - 5}}{6}\end{array}\)

Vậy \(x = \frac{{ - 5}}{6}\).

c)

\(\begin{array}{l}\frac{2}{5}:x = \frac{1}{{16}}:0,125\\\frac{2}{5}:x = \frac{1}{{16}}:\frac{1}{8}\\\frac{2}{5}:x = \frac{1}{{16}}.8\\\frac{2}{5}:x = \frac{1}{2}\\x = \frac{2}{5}:\frac{1}{2}\\x = \frac{2}{5}.2\\x = \frac{4}{5}\end{array}\)      

Vậy \(x = \frac{4}{5}\)

d)

\(\begin{array}{l} - \frac{5}{{12}}x = \frac{2}{3} - \frac{1}{2}\\ - \frac{5}{{12}}x = \frac{4}{6} - \frac{3}{6}\\ - \frac{5}{{12}}x = \frac{1}{6}\\x = \frac{1}{6}:\left( { - \frac{5}{{12}}} \right)\\x = \frac{1}{6}.\frac{{ - 12}}{5}\\x = \frac{{ - 2}}{5}\end{array}\)

Vậy \(x = \frac{{ - 2}}{5}\).

Chú ý: Khi trình bày lời giải bài tìm x, sau khi tính xong, ta phải kết luận.

\(\frac{24}{-12}\) = \(\frac{x}{5}\) = \(\frac{-y}{3}\)

- 2 = \(\frac{x}{5}\) = \(\frac{-y}{3}\)

\(x=5.\left(-2\right)\) = -10

y = -2.3:(-1) = -6:(-1) = 6

Vậy (\(x;y\)) = (-10; 6)

d) \(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)

<=> \(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}-\frac{x+10}{2000}-\frac{x+11}{1999}-\frac{x+12}{1998}=0\)

<=> \(\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)-\left(\frac{x+10}{2000}+1\right)-\left(\frac{x+11}{1999}+1\right)-\left(\frac{x+12}{1998}+1=0\right)\)

<=> \(\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)

<=>\(\left(x+2010\right).\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)

<=> x+2010 = 0 vì \(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\ne0\)

<=> x = -2010

Làm câu khó nhất rồi, còn lại tự làm nha <(") /_\

1. \(A=\left(\frac{3}{2}-\frac{2}{5}+\frac{1}{10}\right):\left(\frac{3}{2}-\frac{2}{3}+\frac{1}{12}\right)=\frac{6}{5}:\frac{11}{12}=\frac{6}{5}.\frac{12}{11}=\frac{72}{55}\)

2. 2^x+2 . 3^x+1 . 5^x = 10800

=> 2^x . 2² . 3^x . 3 . 5^x = 10800

=> ( 2 . 3 . 5 )^x . 12 = 10800

=> 30^x = 900

=> 30^x = 30²

=> x = 2

$\textbf{a)}$

$\dfrac{x-1}{2}+\dfrac{x-2}{5}=\dfrac14+\dfrac{x-7}{10}$

$\Leftrightarrow\dfrac{5(x-1)+2(x-2)}{10}=\dfrac14+\dfrac{x-7}{10}$

$\Leftrightarrow\dfrac{7x-9}{10}=\dfrac14+\dfrac{x-7}{10}$

$\Leftrightarrow14x-18=5+2x-14$

$\Leftrightarrow12x=9$

$\Leftrightarrow x=\dfrac34.$

$\textbf{b)}$

$\dfrac{3-2}{2x-3}=\dfrac25+\dfrac1{2x-3}-\dfrac32$

$\Leftrightarrow1-\dfrac2{2x-3}=-\dfrac{11}{10}+\dfrac1{2x-3}$

$\Leftrightarrow\dfrac{21}{10}=\dfrac3{2x-3}$

$\Leftrightarrow21(2x-3)=30$

$\Leftrightarrow42x=93$

$\Leftrightarrow x=\dfrac{31}{14}.$

d: =>4x+6=15x-12

=>4x-15x=-12-6=-18

=>-11x=-18

hay x=18/11

e: =>\(45x+27=12+24x\)

=>21x=-15

hay x=-5/7

f: =>35x-5=96-6x

=>41x=101

hay x=101/41

g: =>3(x-3)=90-5(1-2x)

=>3x-9=90-5+10x

=>3x-9=10x+85

=>-7x=94

hay x=-94/7

làm rõ ra giúp với ạ, ghi v k hỉu j hết ;-;