Bài 1: 

a) Ta có: \(2\left(3-4x\right)=10-\left(2x-5\right)\)

\(\Leftrightarrow6-8x-10+2x-5=0\)

\(\Leftrightarrow-6x+11=0\)

\(\Leftrightarrow-6x=-11\)

hay \(x=\dfrac{11}{6}\)

b) Ta có: \(3\left(2-4x\right)=11-\left(3x-1\right)\)

\(\Leftrightarrow6-12x-11+3x-1=0\)

\(\Leftrightarrow-9x-6=0\)

\(\Leftrightarrow-9x=6\)

hay \(x=-\dfrac{2}{3}\)

a) BPT \(\Leftrightarrow-2\left(x^2-\dfrac{3}{2}x+\dfrac{7}{2}\right)>0\)

            \(\Leftrightarrow-2\left(x^2-2x\cdot\dfrac{3}{4}+\dfrac{9}{16}+\dfrac{47}{16}\right)>0\)

            \(\Leftrightarrow-\dfrac{47}{8}-2\left(x-\dfrac{3}{4}\right)^2>0\)  (Vô lý)

b) Bạn xem lại đề !

x⁴ - 4x² + 12x - 9 = 0

<=> x⁴ - x³ + x³ - x² - 3x² + 3x + 9x - 9 = 0

<=> x³(x - 1) + x²(x - 1) - 3x(x - 1) + 9(x - 1) = 0

<=> (x - 1)(x³ + x² - 3x + 9) = 0

<=> (x - 1)(x³ + 3x² - 2x² - 6x + 3x + 9) = 0

<=> (x - 1)[ x²(x + 3) - 2x(x + 3) + 3(x + 3) ] = 0

<=> (x - 1)(x + 3)(x² - 2x + 3) = 0

<=> (x - 1)(x + 3)(x² - 2x + 1 + 2) = 0

<=> (x - 1)(x + 3)[ (x - 1)² + 2 ] = 0

<=> (x - 1)(x + 3) = 0 --> do (x - 1)² + 2 > 0 với mọi x

<=>

[ x - 1 = 0 =>[ x = 1

[ x + 3 = 0 =>[ x = -3

Bạn nên sửa >= là = vì giải bất phương trình mà

\(4x^2-4x-5\left|2x-1\right|-5=0\)

\(\Leftrightarrow-5\left|2x-1\right|=5-4x^2+4x\)

\(\Leftrightarrow\left|2x-1\right|=\frac{-4x^2+4x+5}{-5}\)

\(\Leftrightarrow\left|2x-1\right|=\frac{4x\left(x-1\right)}{5}-1\)

TH1 : \(2x-1=\frac{4x\left(x-1\right)}{5}-1\Leftrightarrow2x=\frac{4x\left(x-1\right)}{5}\)

\(\Leftrightarrow10x=4x^2-4x\Leftrightarrow14x-4x^2=0\)

\(\Leftrightarrow-2x\left(2x-7\right)=0\Leftrightarrow x=0;x=\frac{7}{2}\)

TH2 : \(2x-1=-\left(\frac{4x\left(x-1\right)}{5}-1\right)\Leftrightarrow2x-1=-\frac{4x\left(x-2\right)}{5}+1\)

\(\Leftrightarrow2x-2=-\frac{4x\left(x-2\right)}{5}\Leftrightarrow10x-10=-4x^2+8x\)

\(\Leftrightarrow2x-10+4x^2=0\Leftrightarrow2\left(2x^2+x-5\ne0\right)=0\)tự chứng minh 

Vậy tập nghiệm của phương trình là S = { 0 ; 7/2 }

\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)

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

\(\Leftrightarrow x^2-9-x^2+3x=0\)

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\left(n\right)\)

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

\(4x^2-4x+1>25\)

\(\Leftrightarrow\left(2x-1\right)^2-5^2>0\)

\(\Leftrightarrow\left(2x-1-5\right)\left(2x-1+5\right)>0\)

\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)>0\)

TH1 : \(\hept{\begin{cases}2x-6>0\\2x+4>0\end{cases}\Leftrightarrow\hept{\begin{cases}x>3\\x>-2\end{cases}\Leftrightarrow x>3}}\)

TH2 : \(\hept{\begin{cases}2x-6< 0\\2x+4< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 3\\x< -2\end{cases}\Leftrightarrow x< -2}}\)

Vậy....