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a) \(1\frac{5}{8}:x-1\frac{1}{4}=2\)
=> \(\frac{13}{8}:x-\frac{5}{4}=2\)
=> \(\frac{13}{8}:x=\frac{13}{4}\)
=> \(x=\frac{13}{8}:\frac{13}{4}=\frac{13}{8}\cdot\frac{4}{13}=\frac{1}{2}\)
b) \(\left(x-\frac{1}{3}\right)^2-\frac{1}{2}=0\)
=> \(\left(x-\frac{1}{3}\right)^2=\frac{1}{2}\)
=> x không thỏa mãn
c) 2(x - 1) = 3x + 1
=> 2x - 2 = 3x + 1
=> 2x - 2 - 3x - 1 = 0
=> 2x - 3x - 2 - 1 = 0
=> -x = 3
=> x = -3
d) \(\frac{-2}{x}=\frac{x}{-8}\)=> x2 = 16 => x = \(\pm\)4
e) |7 - x| + 2x = 11
=> |7 - x| = 11 - 2x
=> 7 - x = 11 - 2x
=> 7 - x - 11 + 2x = 0
=> 7 - 11 - x + 2x = 0
=> -4 + x = 0
=> -4 = -x
=> x = 4
a)=> 13/8 : x-5/4 =2
<=> 13/8 : x= 13/4
<=> x=1/2
b)=>x2 - (1/3)2 = 1/2
=>x2-1/9=1/2
=>x2=11/18
=>x=0.78 (bằng xấp xỉ thôi nhé bạn :33)
c)=>2x-2=3x-1
=>2x-3x=2-1
=>-x=1
=>x=-1
còn 2 câu bạn làm nốt nhé :33
k đúng cho mk nhé!!!!

a) \(\frac{x-3}{3}-1=\frac{x}{-4}\)
\(\Leftrightarrow\frac{x-3}{3}-\frac{3}{3}=\frac{x}{-4}\)
\(\Leftrightarrow\frac{x-6}{3}=\frac{x}{-4}\)
\(\Leftrightarrow-4\left(x-6\right)=3x\)
\(\Leftrightarrow-4x+24=3x\)
\(\Leftrightarrow24=3x+4x\)
\(\Leftrightarrow7x=24\)
\(\Leftrightarrow x=\frac{24}{7}\)
b) \(\frac{5}{8}-\left(x-\frac{1}{2}\right)=\frac{-3}{4}\)
\(\Leftrightarrow\frac{5}{8}-x+\frac{1}{2}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{5}{8}+\frac{4}{8}-x=\frac{-3}{4}\)
\(\Leftrightarrow\frac{9}{8}-x=\frac{-3}{4}\)
\(\Leftrightarrow x=\frac{9}{8}+\frac{3}{4}\)
\(\Leftrightarrow x=\frac{15}{8}\)

a) 2012 x 8 - 8 x 2011
= 8 x (2012 - 2011)
= 8 x 1
= 8
b) \(\frac{7}{8}.2\frac{2}{3}-\frac{5}{8}.2\frac{2}{3}\)
\(=\frac{7}{8}.\frac{8}{3}-\frac{5}{8}.\frac{8}{3}\)
\(=\frac{8}{3}.\left(\frac{7}{8}-\frac{5}{8}\right)\)
\(=\frac{8}{3}.\frac{2}{8}\)
\(=\frac{16}{6}=\frac{8}{3}\)

\(8-12x+6x^2-x^3\)
\(=\left(2-x\right)^3\)
\(125x^3-75x^2+15x-1\)
\(=\left(5x-1\right)^3\)
\(x^2-xz-9y^2+3yz\)
\(=\left(x-3y\right)\left(x+3y\right)-z\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x+3y-z\right)\)
\(x^3-x^2-5x+125\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-5x+25-x\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
\(x^3+2x^2-6x-27\)
\(=x^3+5x^2+9x-3x^2-15x-27\)
\(=x\left(x^2+5x+9\right)-3\left(x^2+5x+9\right)\)
\(=\left(x-3\right)\left(x^2+5x+9\right)\)
\(12x^3+4x^2-27x-9\)
\(=4x^2\left(3x+1\right)-9\left(3x+1\right)\)
\(=\left(3x+1\right)\left(4x^2-9\right)\)
\(=\left(3x+1\right)\left(2x-3\right)\left(2x+3\right)\)
\(4x^4+4x^3-x^2-x\)
\(=4x^3\left(x+1\right)-x\left(x+1\right)\)
\(=x\left(x+1\right)\left(4x^2-1\right)\)
\(=x\left(x+1\right)\left(2x-1\right)\left(2x+1\right)\)

b; \(3\frac{1}{4}.x-1\frac{1}{6}.x-1\frac{2}{3}=\frac{5}{12}\)
\(\Rightarrow x.\left(3\frac{1}{4}-1\frac{1}{6}-1\frac{2}{3}\right)=\frac{5}{12}\)
\(\Rightarrow x.\left(\frac{13}{4}-\frac{7}{6}-\frac{5}{3}\right)=\frac{5}{12}\)
\(\Rightarrow x.\left(\frac{39}{12}-\frac{14}{12}-\frac{20}{12}\right)=\frac{5}{12}\)
\(\Rightarrow x.\frac{5}{12}=\frac{5}{12}\)
\(\Rightarrow x=\frac{5}{12}\div\frac{5}{12}=1\)
Vậy x=1

a. \(\frac{x+1}{2}=\frac{8}{x+1}\)
\(\Leftrightarrow\left(x+1\right).\left(x+1\right)=8.2\)
\(\Leftrightarrow\left(x+1\right)^2=16\)
\(\Leftrightarrow\left(x+1\right)^2=2^4\)
\(\Leftrightarrow\left(x+1\right)=2^2\)
\(\Leftrightarrow\left(x+1\right)=4\)
\(\Leftrightarrow x=4-1=3\)
b. \(x:\left(9\frac{1}{2}-\frac{3}{2}\right)=\frac{0,4+\frac{2}{9}-\frac{2}{11}}{1,6+\frac{8}{9}-\frac{8}{11}}\)
\(\Leftrightarrow x:\left(\frac{10}{2}-\frac{3}{2}\right)=\frac{0,4+0,2-0,18}{1,6+0,8-0,72}\)
\(\Leftrightarrow x:\frac{7}{2}=\frac{\frac{21}{50}}{\frac{42}{25}}\)
\(\Leftrightarrow x=\frac{\frac{21}{50}}{\frac{42}{25}}.\frac{7}{2}\Leftrightarrow x=\frac{1}{4}.\frac{7}{2}=\frac{7}{8}\)
a ) \(\frac{x+1}{2}=\frac{8}{x+1}\)
\(\Rightarrow\left(x+1\right).\left(x+1\right)=2.8\)
\(\Rightarrow\left(x+1\right)^2=16\)
\(\Rightarrow\orbr{\begin{cases}x+1=4\\x+1=-4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=4-1\\x=-4-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)
Dấu " \(\orbr{\begin{cases}\\\end{cases}}\)là hoặc nha !!!

\(\frac{x+1}{2}=\frac{x+2}{5}\)
\(\Leftrightarrow\)\(5.\left(x+1\right)=2\left(x+2\right)\)
\(\Leftrightarrow\)\(5x+5=2x+4\)
\(\Leftrightarrow\)\(5x-2x=4-5\)
\(\Leftrightarrow\)\(3x=-1\)
\(\Leftrightarrow\)\(x=-\frac{1}{3}\)
Vậy....
\(\frac{x+1}{2}=\frac{x+2}{5}\)
\(\Leftrightarrow5.\left(x+1\right)=2\left(x+2\right)\)
\(\Leftrightarrow5x+5=2x+4\)
\(\Leftrightarrow5x-2x=4-5\)
\(\Leftrightarrow3x=-1\)
\(\Leftrightarrow x=\frac{-1}{3}\)
Vậy ........

c) pt <=> \(x-\frac{21}{5}=\frac{23}{7}< =>x=\frac{23}{7}+\frac{21}{5}=\frac{262}{35}\)
vậy x = \(\frac{262}{35}\)
d) \(x-\frac{3}{4}=\frac{51}{8}< =>x=\frac{51}{8}+\frac{3}{4}=\frac{57}{8}\)
vậy x = \(\frac{57}{8}\)
e) pt <=> \(\frac{7}{8}:x=\frac{7}{2}< =>\frac{7}{8}.\frac{1}{x}=\frac{7}{2}< =>\frac{7}{8x}=\frac{7}{2}< =>56x=14< =>x=\frac{14}{56}=\frac{1}{4}\)
vậy x = \(\frac{1}{4}\)
a) pt <=> \(x+\frac{11}{4}=\frac{17}{3}< =>x=\frac{17}{3}-\frac{11}{4}=\frac{35}{12}\)
vậy x = \(\frac{35}{12}\)
b) pt <=> \(\frac{x.7}{2}=\frac{19}{4}< =>x=\frac{19.2}{4.7}=\frac{38}{28}=\frac{19}{14}\)
vậy x = \(\frac{19}{14}\)

3x-|2x+1|=2
<=>|2x+1|=3x-2
=>2x+1=3x-2 <=> 3x-2x=2+1 <=>x=3
=>2x+1=-3x+2 <=> 5x=2-1 <=> x=1/5
\(\frac{x-1}{2}\) = \(\frac{8}{x-1}\)
(\(x-1\)).(\(x-1\)) = 8.2
(\(x\) - 1)\(^2\) = 16
(\(x-1\))\(^2\) = 4\(^2\)
\(\left[\begin{array}{l}x-1=-4\\ x-1=4\end{array}\right.\)
\(\left[\begin{array}{l}x=-4+1\\ x=4+1\end{array}\right.\)
\(\left[\begin{array}{l}x=-3\\ x=5\end{array}\right.\)
Vậy \(x\in\) {-3; 5}
\(\dfrac{x-1}{2}=\dfrac{8}{x-1}\)
=>\(\left(x-1\right)\left(x-1\right)=2\cdot8=16\)
=>\(\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)