ĐK \(\hept{\begin{cases}5x-3\ne0\\4x-6\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne\frac{3}{5}\\x\ne\frac{3}{2}\end{cases}}}\)

Phương trình \(\Leftrightarrow\frac{\left(2x+3\right)\left(4x-6\right)-3\left(5x-3\right)}{\left(5x-3\right)\left(4x-6\right)}=\frac{2}{5}\)

\(\Leftrightarrow\frac{8x^2-12x+12x-18-15x+9}{\left(5x-3\right)\left(4x-6\right)}=\frac{2}{5}\)\(\Leftrightarrow\frac{8x^2-15x+9}{20x^2-42x+18}=\frac{2}{5}\)

\(\Leftrightarrow40x^2-84x+36=40x^2-75x-45\Leftrightarrow-9x=-81\Leftrightarrow x=9\left(tm\right)\)

Vậy x=9

7(2x - 5) - 5(7x - 2) + 2(5x - 7) = (x + 2) - (x + 4)

=> 14x - 35 - 35x + 10 + 10x -14 = x + 2 - x - 4

=> (14x - 35x + 10x) + (-35 + 10 - 14) = -2

=> -11x + (-39) = -2

=> -11x = -2 - (-39)

=> -11x = 37

=> x = \(\frac{-37}{11}\)

7(2x - 5) - 5(7x - 2) + 2(5x - 7)             = (x + 2) - (x + 4)

=> 14x - 35 - 35x + 10 + 10x -14         = x + 2 - x - 4

=> (14x - 35x + 10x) + (-35 + 10 - 14) = -6
=> -11x - 39                                         = -6
=> -11x                                                = -6+39
=> -11x                                                = 33
=>      x                                                = 33:(-11)
=>      x                                                = -3

\(x^4+2x^3+5x^2+4x-12=0\)

\(\Leftrightarrow\left(x^2+x^3-2x^2\right)+\left(x^3+x^2-2x\right)+\left(6x^2+6x-12\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+6=0\\x^2+x-2=0\end{matrix}\right.\)

* \(x^2+x+6=\left(x^2+x+\frac{1}{4}\right)+\frac{23}{4}=\left(x+\frac{1}{2}\right)^2+\frac{23}{4}>0\)

\(\Rightarrow x^2+x+6=0\) là vô lí

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

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

5x² - 7 = 38 => x² = 9 => x = \(\pm\)3

Từ đây thay x vào \(\dfrac{3x-2}{4}\) để tìm y,z

\(4\left(0,5-1,5\right)=\dfrac{\left(5x-6\right)}{3}\)

\(-4=\dfrac{\left(5x-6\right)}{3}\)

\(-12=5x-6\)

\(5x=-6\)

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