Hummmm

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+2x-3\ge0\\2x^2-3x+1\ge0\\x^2+2x-3\le2x^2-3x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge1\\x\le-3\end{matrix}\right.\\\left[{}\begin{matrix}x\ge1\\x\le\dfrac{1}{2}\end{matrix}\right.\\x^2-5x+4\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge1\\x\le-3\end{matrix}\right.\\\left[{}\begin{matrix}x\ge4\\x\le1\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x\le-3\\x\ge4\end{matrix}\right.\)

e: \(\begin{cases}x\left(x+5\right)<4x+2\\ \left(2x-1\right)\left(x+3\right)\ge4x\end{cases}\Rightarrow\begin{cases}x^2+5x-4x-2<0\\ 2x^2+6x-x-3-4x\ge0\end{cases}\)

=>\(\begin{cases}x^2+x-2<0\\ 2x^2+x-3\ge0\end{cases}\Rightarrow\begin{cases}\left(x+2\right)\left(x-1\right)<0\\ 2x^2+3x-2x-3\ge0\end{cases}\)

=>\(\begin{cases}-2

=>-2<x<=-1

f: ĐKXĐ: x∉{1;4;2;5}

Ta có: \(\frac{1}{x^2-5x+4}\le\frac{1}{x^2-7x+10}\)

=>\(\frac{1}{x^2-5x+4}-\frac{1}{x^2-7x+10}\le0\)

=>\(\frac{x^2-7x+10-x^2+5x-4}{\left(x-1\right)\left(x-4\right)\left(x-2\right)\left(x-5\right)}\le0\)

=>\(\frac{-2x+6}{\left(x-1\right)\left(x-4\right)\left(x-2\right)\left(x-5\right)}\le0\)

=>\(\frac{x-3}{\left(x-1\right)\left(x-2\right)\left(x-4\right)\left(x-5\right)}\ge0\)

Đặt \(A=\frac{x-3}{\left(x-1\right)\left(x-2\right)\left(x-4\right)\left(x-5\right)}\)

Đặt x-3=0

=>x=3

Đặt x-1=0

=>x=1

Đặt x-2=0

=>x=2

Đặt x-4=0

=>x=4

Đặt x-5=0

=>x=5

Bảng xét dấu:

Theo bãng xét dấu, ta có: A>=0 khi 1<x<2; 3<=x<4; x>5

Đặt \(\hept{\begin{cases}a=\sqrt{4x+1}\\b=\sqrt{3x-2}\end{cases}\ge}0\) thì có:

\(\Rightarrow a^2-b^2=x+3\)\(\Rightarrow a-b=\frac{a^2-b^2}{5}\)

\(\Rightarrow a-b-\frac{\left(a-b\right)\left(a+b\right)}{5}=0\)

\(\Rightarrow\left(a-b\right)\left(1-\frac{a+b}{5}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}a=b\\a+b=5\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\sqrt{4x+1}=\sqrt{3x-2}\\\sqrt{4x+1}+\sqrt{3x-2}=5\end{cases}}\)\(\Rightarrow x=2\)

@Thắng Nguyễn