1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)

=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)

b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)

c) TT

a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)

\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)

=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)

=> \(\left|50x-140\right|=\left|25x+24\right|\)

=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)

c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)

=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)

Bài 2 : a. |2x - 5| = x + 1

 TH1 : 2x - 5 = x + 1

    => 2x - 5 - x = 1

    => 2x - x - 5 = 1

    => 2x - x = 6

    => x = 6

TH2 : -2x + 5 = x + 1

   => -2x + 5 - x = 1

   => -2x - x + 5 = 1

   => -3x = -4

   => x = 4/3

Ba bài còn lại tương tự

cho mình

\(\left|\left(x+\frac{1}{2}\right).\left|2x-\frac{3}{4}\right|\right|=2x-\frac{3}{4}\)

\(\Rightarrow\left|x+\frac{1}{2}\right|.\left|2x-\frac{3}{4}\right|=2x-\frac{3}{4}\)

\(\Rightarrow2x-\frac{3}{4}\ge0\) (1)

Lúc này ta có: \(\left|x+\frac{1}{2}\right|.\left(2x-\frac{3}{4}\right)=2x-\frac{3}{4}\)

\(\Rightarrow\left|x+\frac{1}{2}\right|.\left(2x-\frac{3}{4}\right)-\left(2x-\frac{3}{4}\right)=0\)

\(\Rightarrow\left(2x-\frac{3}{4}\right).\left(\left|x+\frac{1}{2}\right|-1\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}2x-\frac{3}{4}=0\\\left|x+\frac{1}{2}\right|-1=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}2x=\frac{3}{4}\\\left|x+\frac{1}{2}\right|=1\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{3}{8}\\x+\frac{1}{2}=1\\x+\frac{1}{2}=-1\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{3}{8}\\x=\frac{1}{2}\\x=\frac{-3}{2}\end{array}\right.\)

Mà \(x\ge\frac{3}{8}\) do \(2x-\frac{3}{4}\ge0\)

Vậy \(x\in\left\{\frac{3}{8};\frac{1}{2}\right\}\)

1.

\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)

\(MC:12\)

Quy đồng :

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

\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)

\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)

\(\Leftrightarrow6x+9-3x=-4-9+16\)

\(\Leftrightarrow-7x=3\)

\(\Leftrightarrow x=\frac{-3}{7}\)

2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)

\(MC:20\)

Quy đồng :

\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)

\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)

\(\Leftrightarrow30x+15-20=15x-2\)

\(\Leftrightarrow15x=3\)

\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)

\(\frac{\left(x+1\right)^2-\frac{x}{2}}{4}=\frac{\left(2x-3\right)^2}{3}-\frac{\frac{x+1}{4}-\frac{x\left(3-2x\right)}{3}}{4}\)

\(\Rightarrow3\left[\left(x+1\right)^2-\frac{x}{2}\right]=4\left(2x-3\right)^2-3\left[\frac{x+1}{4}-\frac{x\left(3-2x\right)}{3}\right]\)

\(\Rightarrow3\left(x+1\right)^2-\frac{3x}{2}=4\left(2x-3\right)^2-\frac{3\left(x+1\right)}{4}+\frac{3x\left(3-2x\right)}{3}\)

\(\Rightarrow36\left(x+1\right)^2-18x=48\left(2x-3\right)^2-9\left(x+1\right)+12x\left(3-2x\right)\)

=> 36.(x² + 2x + 1) - 18x = 48.(4x² - 12x + 9) - 9(x + 1) + 12x(3 - 2x)

=> 36x² + 72x + 36 - 18x - 192x² + 576x - 432 + 9x + 9 - 36x + 24x² = 0

=> -132x² + 603x - 387 = 0

Có: \(\Delta=603^2-4.\left(-387\right)\left(-132\right)=159273\Rightarrow\sqrt{\Delta}=\sqrt{159273}\)

\(\Rightarrow x=\frac{-603+\sqrt{159273}}{-264}\)          hoặc          \(x=\frac{-603-\sqrt{159273}}{-264}\)

Vậy phương trình có 2 nghiệm : x = \(\left\{\frac{-603+\sqrt{159273}}{-264};\frac{-603-\sqrt{159273}}{-264}\right\}\)

Câu này không có nghiệm nguyên nha bạn.

Cảm ơn bn nhìu

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

\(2x^2-10x-3x-2x^2=26\)

-13x=26

x=-2