\(\left(2x-5\right)\left(x-3\right)+\left(2x-5\right)^2=0\)

\(\Rightarrow\left(2x-5\right)\left(x-3+2x-5\right)=0\)

\(\Rightarrow\left(2x-5\right)\left(3x-8\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-5=0\\3x-8=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{8}{3}\end{cases}}\)

\(\frac{3x-5}{4}+\frac{2x-3}{6}=\frac{x}{3}-1\)

\(\Leftrightarrow\frac{18x-30+8x-12}{24}=\frac{x-3}{3}\)

\(\Leftrightarrow\frac{26x-42}{24}=\frac{x-3}{3}\)

\(\Leftrightarrow78x-126=24x-72\)

Chuyển vế các kiểu

\(\frac{2x-9}{2x-5}+\frac{3x}{3x-2}=2\)

\(\Leftrightarrow\left(2x-6\right)\left(3x-2\right)+3x\left(2x-5\right)=2\left(2x-5\right)\left(3x-2\right)\)

\(\Leftrightarrow6x^2-4x-18x+12+6x^2-15x=2\left(6x^2-4x-15x+10\right)\)

\(\Leftrightarrow12x^2-27x+12=12x^2-38x+20\)

Đến đây EZ rồi

câu c) mang tính mua vui hay gì hả bn

mếu thật thì x=0,x=số nào cx đc(câu trả lời này mang tính mua vui thôi nhé)

1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)

=-27x^3-18x^2+4x+10

2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27

=7x^3+37x^2+46x+33

5:

\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)

\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)

=7x^3-48x^2+8x-35

\(\dfrac{x-2}{x+1}-\dfrac{3}{x+2}>0.\left(x\ne-1;-2\right).\\ \Leftrightarrow\dfrac{x^2-4-3x-3}{\left(x+1\right)\left(x+2\right)}>0.\\ \Leftrightarrow\dfrac{x^2-3x-7}{\left(x+1\right)\left(x+2\right)}>0.\)    

Đặt \(f\left(x\right)=\dfrac{x^2-3x-7}{\left(x+1\right)\left(x+2\right)}>0.\)

Ta có: \(x^2-3x-7=0.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt{37}}{2}.\\x=\dfrac{3-\sqrt{37}}{2}.\end{matrix}\right.\)

          \(x+1=0.\Leftrightarrow x=-1.\\ x+2=0.\Leftrightarrow x=-2.\)

Bảng xét dấu:

\(\Rightarrow f\left(x\right)>0\Leftrightarrow x\in\left(-\infty-2\right)\cup\left(\dfrac{3-\sqrt{37}}{2};-1\right)\cup\left(\dfrac{3+\sqrt{37}}{2};+\infty\right).\)

\(\sqrt{x^2-3x+2}\ge3.\\ \Leftrightarrow x^2-3x+2\ge9.\\ \Leftrightarrow x^2-3x-7\ge0.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3-\sqrt{37}}{2}.\\x=\dfrac{3+\sqrt{37}}{2}.\end{matrix}\right.\)

Đặt \(f\left(x\right)=x^2-3x-7.\)

\(f\left(x\right)=x^2-3x-7.\)

\(\Rightarrow f\left(x\right)\ge0\Leftrightarrow x\in(-\infty;\dfrac{3-\sqrt{37}}{2}]\cup[\dfrac{3+\sqrt{37}}{2};+\infty).\)

\(\Rightarrow\sqrt{x^2-3x+2}\ge3\Leftrightarrow x\in(-\infty;\dfrac{3-\sqrt{37}}{2}]\cup[\dfrac{3+\sqrt{37}}{2};+\infty).\)

g: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)

\(\Leftrightarrow14x=0\)

hay x=0

w

         Bài 1:

- \(\dfrac{11}{2}x\) + 1 = \(\dfrac{1}{3}x-\dfrac{1}{4}\)

- \(\dfrac{11}{2}\)\(x\) - \(\dfrac{1}{3}\)\(x\) = - \(\dfrac{1}{4}\) - 1

-(\(\dfrac{33}{6}\) + \(\dfrac{2}{6}\))\(x\) = - \(\dfrac{5}{4}\)

- \(\dfrac{35}{6}\)\(x\) = - \(\dfrac{5}{4}\)

  \(x=-\dfrac{5}{4}\) : (- \(\dfrac{35}{6}\))

 \(x\) = \(\dfrac{3}{14}\)

Vậy \(x=\dfrac{3}{14}\)

Bài 2: 2\(x\) - \(\dfrac{2}{3}\) - 7\(x\) = \(\dfrac{3}{2}\) - 1

         2\(x\) - 7\(x\) = \(\dfrac{3}{2}\) - 1 + \(\dfrac{2}{3}\)

         - 5\(x\)    = \(\dfrac{9}{6}\) - \(\dfrac{6}{6}\) + \(\dfrac{4}{6}\) 

        - 5\(x\)    = \(\dfrac{7}{6}\)

           \(x\)    = \(\dfrac{7}{6}\) : (- 5) 

          \(x\)    = - \(\dfrac{7}{30}\)

Vậy \(x=-\dfrac{7}{30}\)

Bài 3: \(\dfrac{3}{2}\)\(x\) - \(\dfrac{2}{5}\) = \(\dfrac{1}{3}x\) - \(\dfrac{1}{4}\)

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

           (\(\dfrac{9}{6}\) - \(\dfrac{2}{6}\))\(x\) = \(\dfrac{-5}{20}\) + \(\dfrac{8}{20}\)

            \(\dfrac{7}{6}x\)       = \(\dfrac{3}{20}\)

            \(x\)       = \(\dfrac{3}{20}\) : \(\dfrac{7}{6}\)

            \(x\)      = \(\dfrac{9}{70}\)

Vậy  \(x=\dfrac{9}{70}\)