a: =>(x+1)(x+5)>0 và (x+3)(x-2)<0

=>\(\left\{{}\begin{matrix}x\in\left(-\infty;-5\right)\cup\left(-1;+\infty\right)\\x\in\left(-3;2\right)\end{matrix}\right.\Leftrightarrow x\in\left(-1;2\right)\)

b: =>(x+2)(2x-3)>0và 3x^2-10x+3>=0

=>(x+2)(2x-3)>0 và (x-3)(3x-1)>=0

=>\(\left\{{}\begin{matrix}x\in\left(-\infty;-2\right)\cup\left(\dfrac{3}{2};+\infty\right)\\x\in(-\infty;\dfrac{1}{3}]\cup[3;+\infty)\end{matrix}\right.\Leftrightarrow x\in\left(-\infty;-2\right)\cup[3;+\infty)\)

a.\(\left\{{}\begin{matrix}4x+2y=14\\2x-2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x=18\\2x-2y=4\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=2\\4-2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\-2y=0\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)

vậy  hệ pt có ndn \(\left\{2;0\right\}\)

b.\(\left\{{}\begin{matrix}2x-4y=0\\3x+2y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-4y=0\\6x+4y=16\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}8x=16\\2x-4y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\4-4y=0\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=2\\-4y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

vậy hệ pt có ndn \(\left\{2;1\right\}\)

d.\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)

đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b\) ta có hệ pt:

\(\left\{{}\begin{matrix}a+b=\dfrac{1}{12}\\8a+15b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8a+8b=\dfrac{2}{3}\\8a+15b=1\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}7b=\dfrac{1}{3}\\8a+15b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{1}{21}\\8a+15\times\dfrac{1}{21}=1\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}b=\dfrac{1}{21}\\8a+\dfrac{5}{7}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{1}{21}\\8a=\dfrac{2}{7}\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}b=\dfrac{1}{21}\\a=\dfrac{1}{28}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{21}\\\dfrac{1}{x}=\dfrac{1}{28}\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}y=21\\x=28\end{matrix}\right.\)

vậy hệ pt có ndn\(\left\{28;21\right\}\)

\(\left(4x-5\right)\left(-x^2-3x+4\right)>=0\)

\(\Leftrightarrow\left(4x-5\right)\left(x^2+3x-4\right)< =0\)

=>(4x-5)(x+4)(x-1)<=0

BXD:

Theo BXD, ta được: x<=-4 hoặc 1<=x<=5/4

\(3x^2-7x+2>0\)

=>3x²-6x-x+2>0

=>(x-2)(3x-1)>0

=>x>2 hoặc x<1/3

=>x<=-4

các bn ơi giúp mình với

a)

\(\left\{{}\begin{matrix}x^2+x+5< 0\\x^2-6x+1>0\end{matrix}\right.\)

\(\)Ta có

\(x^2+x+5=\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{19}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}>0\)

=> Bất phương trình đàu tiên sai, hệ bất phương trình sai

b)

\(\left\{{}\begin{matrix}2x^2+x-6>0\\3x^2-10x+3\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-3\right)\left(x+2\right)>0\\\left(x-3\right)\left(3x-1\right)\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>2\\x< -3\end{matrix}\right.\\\left[{}\begin{matrix}x\le-\dfrac{1}{3}\\x\ge3\end{matrix}\right.\end{matrix}\right.\)

\(m< \dfrac{4}{3}\)

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

\(\Leftrightarrow x\left(x-1\right)< 0\Leftrightarrow0< x< 1\)

Xét \(3x^2-4x+m< 0\) trên \(\left(0;1\right)\)

\(\Leftrightarrow m< -3x^2+4x\) trên \(\left(0;1\right)\)

\(\Leftrightarrow m< \max\limits_{\left(0;1\right)}\left(-3x^2+4x\right)\)

Xét \(f\left(x\right)=-3x^2+4x\) trên \(\left(0;1\right)\)

\(a=-3< 0\); \(-\dfrac{b}{2a}=\dfrac{2}{3}\in\left(0;1\right)\)  \(\Rightarrow f\left(x\right)_{max}=f\left(\dfrac{2}{3}\right)=\dfrac{4}{3}\)

\(\Rightarrow m< \dfrac{4}{3}\)