a) \(\left(\frac{2}{3}:x+\frac{1}{4}\right)^2=\frac{49}{64}\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}\left(\frac{2}{3}:x+\frac{1}{4}\right)^2=\left(\frac{7}{8}\right)^2\\\left(\frac{2}{3}:x+\frac{1}{4}\right)^2=-\left(\frac{7}{8}\right)^2\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}\frac{2}{3}:x+\frac{1}{4}=\frac{7}{8}\\\frac{2}{3}:x+\frac{1}{4}=-\frac{7}{8}\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}\frac{2}{3}:x=\frac{5}{8}\\\frac{2}{3}:x=-\frac{9}{8}\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{16}{15}\\x=-\frac{16}{27}\end{array}\right.\)

b) \(\frac{2}{5}:\left|x-2\right|-\frac{1}{3}=\frac{1}{2}\)

\(\Leftrightarrow\frac{2}{5}:\left|x-2\right|=\frac{5}{6}\)

\(\Leftrightarrow\left|x-2\right|=\frac{12}{25}\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=\frac{12}{25}\\x-2=-\frac{12}{25}\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{62}{25}\\x=\frac{38}{25}\end{array}\right.\)

c) \(\frac{1}{8}-\left(\frac{1}{12}-\frac{1}{3}x\right)^3=0\)

\(\Leftrightarrow\left(\frac{1}{12}-\frac{1}{3}x\right)^3=\frac{1}{8}\)

\(\Leftrightarrow\left(\frac{1}{12}-\frac{1}{3}x\right)^3=\left(\frac{1}{2}\right)^3\)

\(\Leftrightarrow\frac{1}{3}x=-\frac{5}{12}\)

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

ko biết

a. 60%x + 0,4x + x : 3 = 2

0.6x + 0,4x + x : 3 = 2

x(0,6 + 0,4 : 3 ) = 2

\(x.\frac{1}{3}=2=>x=2:\frac{1}{3}=\frac{1}{6}\)

câu B tự làm nha .

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

\(\frac{1}{3}x+\frac{2}{5}\left(x-1\right)=0\)

\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)

\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x=0+\frac{2}{5}\)

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

\(\Leftrightarrow x\left(\frac{5}{15}+\frac{6}{15}\right)=\frac{2}{5}\)

\(\Leftrightarrow\frac{11}{15}x=\frac{2}{5}\)

\(\Leftrightarrow x=\frac{2}{5}\div\frac{11}{15}\)

\(\Leftrightarrow x=\frac{6}{11}\)

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{49}{50}\)

\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{49}{50}\)

\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{49}{50}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{49}{50}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{49}{50}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{49}{50}\div2\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{49}{50}\times\frac{1}{2}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{49}{100}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{49}{100}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{50}{100}-\frac{49}{100}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{100}\)

\(\Leftrightarrow x+1=100\)

\(\Leftrightarrow x=100-1\)

\(\Leftrightarrow x=99\)

\(x-\left(\frac{11}{12}+x\right)=x-\frac{3}{4}\)

\(\Leftrightarrow x-\frac{11}{12}-x=x-\frac{3}{4}\)

\(\Leftrightarrow-\frac{11}{12}=x-\frac{3}{4}\)

\(\Leftrightarrow x=\frac{-11}{12}+\frac{3}{4}\)

\(\Leftrightarrow x=\frac{-11}{12}+\frac{9}{12}\)

\(\Leftrightarrow x=\frac{-2}{12}=\frac{-1}{6}\)

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

\(x=\frac{903}{391}\)

Bài này sử dụng MTCT đó bạn!

903/391 mình nghĩ vậy

x = 903/391

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, 

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

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

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

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