a, \(\dfrac{\left(2x-5\right)\left(x+2\right)}{4x-3}< 0\)

⇔ \(\left[{}\begin{matrix}\left\{{}\begin{matrix}\left(2x-5\right)\left(x+2\right)< 0\\4x-3>0\end{matrix}\right.\\\left\{{}\begin{matrix}\left(2x-5\right)\left(x+2\right)>0\\4x-3< 0\end{matrix}\right.\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}\left\{{}\begin{matrix}-2< x< \dfrac{5}{2}\\x>\dfrac{3}{4}\end{matrix}\right.\\\left\{{}\begin{matrix}\left[{}\begin{matrix}x< -2\\x>\dfrac{5}{2}\end{matrix}\right.\\x< \dfrac{3}{4}\end{matrix}\right.\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}\dfrac{3}{4}< x< \dfrac{5}{2}\\x< -2\end{matrix}\right.\)

Vậy tập nghiệm của bất phương trình là

S = \(\left(\dfrac{3}{4};\dfrac{5}{2}\right)\cup\left(-\infty;-2\right)\)

b, Pt

⇔ \(\left\{{}\begin{matrix}x^2-5x+6=x^2+6x+5\\x\in R\backslash\left\{-1;2\right\}\end{matrix}\right.\)

⇔ x = \(\dfrac{1}{11}\)

Vậy S = \(\left\{\dfrac{1}{11}\right\}\)

Bài 1: Giải các phương trình sau:

a) 3(2,2-0,3x)=2,6 + (0,1x-4)

<=> 6.6 - 0.9x = 2,6 + 0,1x - 4

<=> - 0.9x - 0,1x = -6.6 -1,4

<=> -x = -8

<=> x = 8

Vậy x = 8

b) 3,6 -0,5 (2x+1) = x - 0,25(22-4x)

<=> 3,6 - x - 0,5 = x - 5,5 + x

<=> - x - 3,1 = -5,5

<=> - x = -2.4

<=> x = 2.4

Vậy  x = 2.4

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

=>13x-4=90

=>13x=94

hay x=94/13

b: \(\Leftrightarrow2x-4-x-1=3x-11\)

=>3x-11=x-5

=>2x=6

hay x=3(nhận)

a: Ta có: \(4x-2\left(1-x\right)=5\left(x-4\right)\)

\(\Leftrightarrow4x-2+2x=5x-20\)

\(\Leftrightarrow x=-18\)

b: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)

\(\Leftrightarrow6x+4\left(1-3x\right)=3\left(-x+1\right)\)

\(\Leftrightarrow6x+4-12x=-3x+3\)

\(\Leftrightarrow-3x=-1\)

hay \(x=\dfrac{1}{3}\)

c: Ta có: \(\left(x+2\right)^2-3\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)

d: Ta có: \(\dfrac{x-5}{x}+\dfrac{x-3}{x+5}=\dfrac{x-25}{x\left(x+5\right)}\)

\(\Leftrightarrow x^2-25+x^2-3x=x-25\)

\(\Leftrightarrow2x^2-4x=0\)

\(\Leftrightarrow2x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

\(1,\) thiếu đề

\(2,\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)

\(\Leftrightarrow\dfrac{5\left(5x+2\right)}{30}-\dfrac{10\left(8x-1\right)}{30}=\dfrac{6\left(4x+2\right)}{30}-\dfrac{150}{30}\)

\(\Leftrightarrow5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-150\)

\(\Leftrightarrow25x+10-80x+10=24x+12-150\)

\(\Leftrightarrow-55x+20=24x-138\)

\(\Leftrightarrow24x-138+55x-20=0\)

\(\Leftrightarrow79x-158=0\)

\(\Leftrightarrow x=2\)

\(3,ĐKXĐ:\left\{{}\begin{matrix}x\ne1\\x\ne-1\\x\ne3\end{matrix}\right.\\ \dfrac{x}{2x-6}+\dfrac{x}{2x-2}=\dfrac{-2x}{\left(x+1\right)\left(3-x\right)}\)

\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x-1\right)}+\dfrac{2x}{\left(x+1\right)\left(3-x\right)}=0\)

\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x-1\right)}-\dfrac{2x}{\left(x+1\right)\left(x-3\right)}=0\)

\(\Leftrightarrow x\left(\dfrac{1}{2\left(x-3\right)}+\dfrac{1}{2\left(x-1\right)}-\dfrac{2}{\left(x+1\right)\left(x-3\right)}\right)=0\)

\(\Leftrightarrow x\left(\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}+\dfrac{\left(x-3\right)\left(x+1\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}-\dfrac{4\left(x-1\right)}{2\left(x+1\right)\left(x-3\right)\left(x-1\right)}\right)=0\)

\(\Leftrightarrow x\left(\dfrac{x^2-1}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}+\dfrac{x^2-2x-3}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}-\dfrac{4x-4}{2\left(x+1\right)\left(x-3\right)\left(x-1\right)}\right)=0\)

\(\Leftrightarrow x.\dfrac{x^2-1+x^2-2x-3-4x+4}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)

\(\Leftrightarrow x.\dfrac{2x^2-6x}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)

\(\Leftrightarrow x.\dfrac{2x\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)

\(\Leftrightarrow x.\dfrac{x}{\left(x-1\right)\left(x+1\right)}=0\)

\(\Leftrightarrow\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}=0\)

\(\Leftrightarrow x=0\)

1a.

ĐKXĐ: \(x\ne\left\{1;3\right\}\)

\(\Leftrightarrow\dfrac{6}{x-1}=\dfrac{4}{x-3}+\dfrac{4}{x-3}\)

\(\Leftrightarrow\dfrac{3}{x-1}=\dfrac{4}{x-3}\Leftrightarrow3\left(x-3\right)=4\left(x-1\right)\)

\(\Leftrightarrow3x-9=4x-4\Rightarrow x=-5\)

b.

ĐKXĐ: \(x\ne\left\{-1;2\right\}\)

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

\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{4}{2-x}\Leftrightarrow5\left(2-x\right)=4\left(x+1\right)\)

\(\Leftrightarrow10-2x=4x+4\Leftrightarrow6x=6\Rightarrow x=1\)

1c.

ĐKXĐ: \(x\ne\left\{2;5\right\}\)

\(\Leftrightarrow\dfrac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}=\dfrac{-3x}{\left(x-2\right)\left(x-5\right)}\)

\(\Leftrightarrow3x\left(x-5\right)-x\left(x-2\right)=-3x\)

\(\Leftrightarrow2x^2-10x=0\Leftrightarrow2x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\left(loại\right)\end{matrix}\right.\)

2a.

\(\Leftrightarrow-4x^2-5x+6=x^2+4x+4\)

\(\Leftrightarrow5x^2+9x-2=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{5}\end{matrix}\right.\)

2b.

\(2x^2-6x+1=0\Rightarrow x=\dfrac{3\pm\sqrt{7}}{2}\)

\(a.ĐK:x\ne3;1\)

\(\Rightarrow\dfrac{1}{2\left(x-3\right)}+\dfrac{3x-10}{\left(x-1\right)\left(x-3\right)}=\dfrac{7}{2}\)

\(\Leftrightarrow\dfrac{\left(x-1\right)+2\left(3x-10\right)}{2\left(x-1\right)\left(x-3\right)}=\dfrac{7\left(x-1\right)\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}\)

\(\Leftrightarrow x-1+2\left(3x-10\right)=7\left(x-1\right)\left(x-3\right)\)

\(\Leftrightarrow x-1+6x-20=7\left(x^2-4x+3\right)\)

\(\Leftrightarrow7x-21=7x^2-28x+21\)

\(\Leftrightarrow7x^2-35x+42=0\)

\(\Leftrightarrow7\left(x^2-5x+6\right)=0\)

\(\Leftrightarrow x^2-5x+6=0\)

\(\Leftrightarrow x^2-2x-3x+6=0\)

\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=3\left(ktm\right)\end{matrix}\right.\)

b.\(ĐK:x\ne2;4\)

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

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

\(\Leftrightarrow\left(x-1\right)\left(4-x\right)-\left(x+3\right)\left(x-2\right)=2\)

\(\Leftrightarrow4x-x^2-4+x-x^2+2x-3x+6-2=0\)

\(\Leftrightarrow-2x^2+4x=0\)

\(\Leftrightarrow-2x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=2\left(ktm\right)\end{matrix}\right.\)

a: \(\Leftrightarrow\dfrac{1}{2\left(x-3\right)}+\dfrac{3x-10}{\left(x-1\right)\left(x-3\right)}=\dfrac{7}{2}\)

\(\Leftrightarrow x-1+2\left(3x-10\right)=7\left(x-1\right)\left(x-3\right)\)

\(\Leftrightarrow7\left(x^2-4x+3\right)=x-1+6x-20=7x-21\)

\(\Leftrightarrow\left(x-3\right)\left(7x-7\right)-7\left(x-3\right)=0\)

=>(x-3)(7x-14)=0

=>x=3(loại) hoặc x=2(nhận)

b: \(\Leftrightarrow\left(x-1\right)\left(x-4\right)+\left(x+3\right)\left(x-2\right)=-2\)

\(\Leftrightarrow x^2-5x+4+x^2+x-6=-2\)

\(\Leftrightarrow2x^2-4x=0\)

=>2x(x-2)=0

=>x=0(nhận) hoặc x=2(loại)

bn đăng bên toán nhé