c)

\(\left\{\begin{matrix} -x^2+4x-7< 0\\ x^2-2x-1\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x^2-4x+7>0\\ x^2-2x+1\geq 2\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} (x-2)^2+3>0\\ (x-1)^2-2\geq 0\end{matrix}\right.\Leftrightarrow (x-1)^2-2\geq 0\Leftrightarrow \left[\begin{matrix} x-1\geq \sqrt{2}\\ x-1\leq -\sqrt{2}\end{matrix}\right.\)

\(\Leftrightarrow \left[\begin{matrix} x\geq \sqrt{2}+1\\ x\leq 1-\sqrt{2}\end{matrix}\right.\)

d)

\(\left\{\begin{matrix} -2x^2-5x+4< 0\\ -x^2-3x+10>0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 2x^2+5x-4>0\\ (2-x)(x+5)>0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 2(x+\frac{5}{4})^2-\frac{57}{8}>0\\ (2-x)(x+5)>0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} (x+\frac{5}{4}-\frac{\sqrt{57}}{4})(x+\frac{5}{4}+\frac{\sqrt{57}}{4})>0\\ (2-x)(x+5)>0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} \left[\begin{matrix} x>\frac{-5+\sqrt{57}}{4}\\ x< \frac{-5-\sqrt{57}}{4}\end{matrix}\right.\\ -5< x< 2\end{matrix}\right.\) \(\Rightarrow \left[\begin{matrix} -5< x< \frac{-5-\sqrt{57}}{4}\\ \frac{\sqrt{57}-5}{4}< x< 2\end{matrix}\right.\)

a)

\(\left\{\begin{matrix} 2x^2+9x+7>0\\ x^2+x-6< 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} (x+1)(2x+7)>0\\ (x-2)(x+3)< 0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} \left[\begin{matrix} x>-1\\ x< \frac{-7}{2}\end{matrix}\right.\\ -3< x< 2\end{matrix}\right.\Rightarrow -1< x< 2\)

b) \(\left\{\begin{matrix} 2x^2+x-6>0\\ 3x^2-10x+3\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} (2x-3)(x+2)>0\\ (x-3)(3x-1)\geq 0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} \left[\begin{matrix} x>\frac{3}{2}\\ x< -2\end{matrix}\right.\\ \left[\begin{matrix} x\geq 3\\ x\leq \frac{1}{3}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow \left[\begin{matrix} x\geq 3\\ x< -2\end{matrix}\right.\)

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)

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

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

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

bạn ơi giải giúp mình câu c, e, f giùm mình với ạ .

Vì $3x^2-x+1>0,x^2+1>0$

$\to \begin{cases}x^2 \geq 4\x<-1\\\end{cases}$

$\to \begin{cases}\left[ \begin{array}{l}x \geq 2\\x \leq -2\end{array} \right.\\x<-1\\\end{cases}$

$\to x \leq -2$

Vậy tập xác định của phương trình là `(-oo,-2]`

Ghi nhầm ;-;

\(x^2-4xy+3y^2-x+3y=0\)

=>(x-3y)(x-y)-(x-3y)=0

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

TH1: x-3y=0

=>x=3y

4xy+3x+2y=-2

=>4y*3y+3*3y+2y=-2

=>12y^2+9y+2y+2=0

=>12y^2+11y+2=0

=>12y^2+8y+3y+2=0

=>(3y+2)(4y+3)=0

=>y=-2/3 hoặc y=-3/4

Nếu \(y=-\frac23\) thì \(x=3y=3\cdot\frac{-2}{3}=-2\)

Nếu \(y=-\frac34\) thì \(x=3y=3\cdot\frac{-3}{4}=-\frac94\)
TH2: x-y-1=0

=>y=x-1

4xy+3x+2y=-2

=>4x(x-1)+3x+2(x-1)=-2

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

=>\(4x^2+2x=0\)

=>2x(2x+1)=0

=>x=0 hoặc x=-1/2

Khi x=0 thì y=0-1=-1

Khi x=-1/2 thì y=-1/2-1=-3/2

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

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

=>(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

\(\left\{{}\begin{matrix}x^3y^2+x^2y^3+x^3y+2x^2y^2+xy^3-30=0\\x^2y+xy^2+xy+x+y-11=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2y^2\left(x+y\right)+xy\left(x+y\right)^2-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)\left[xy+x+y\right]-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}xy\left(x+y\right)=u\\xy+x+y=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}uv-30=0\\u+v-11=0\end{matrix}\right.\)  \(\Rightarrow\left(u;v\right)=\left(6;5\right);\left(5;6\right)\)

TH1: \(\left\{{}\begin{matrix}xy\left(x+y\right)=6\\xy+x+y=5\end{matrix}\right.\)

Theo Viet đảo \(\Rightarrow\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(1;2\right);\left(2;1\right)\)hoặc \(\left\{{}\begin{matrix}x+y=2\\xy=3\end{matrix}\right.\)(vô nghiệm)

TH2: \(\left\{{}\begin{matrix}xy\left(x+y\right)=5\\xy+x+y=6\end{matrix}\right.\) 

\(\Rightarrow\left\{{}\begin{matrix}x+y=5\\xy=1\end{matrix}\right.\) \(\Rightarrow...\) hoặc \(\left\{{}\begin{matrix}x+y=1\\xy=5\end{matrix}\right.\) (vô nghiệm)

2 câu dưới hình như em hỏi rồi?

\(3x^2-13xy-10y^2=0\)

=>\(3x^2-15xy+2xy-10y^2=0\)

=>3x(x-5y)+2y(x-5y)=0

=>(x-5y)(3x+2y)=0

TH1: x-5y=0

=>x=5y

\(2x^2-y^2+x=-22\)

=>\(2\left(5y\right)^2-y^2+5y+22=0\)

=>\(50y^2-y^2+5y+22=0\)

=>\(49y^2+5y+22=0\)

\(\Delta=5^2-4\cdot49\cdot22=-4287<0\)

=>Phương trình vô nghiệm

TH2: 3x+2y=0

=>2y=-3x

=>y=-1,5x

\(2x^2-y^2+x=-22\)

=>\(2x^2-\left(-1,5x\right)^2+x+22=0\)

=>\(-0,25x^2+x+22=0\)

=>\(x^2-4x-88=0\)

=>\(x^2-4x+4-92=0\)

=>\(\left(x-2\right)^2=92\)

=>\(\left[\begin{array}{l}x-2=2\sqrt{23}\\ x-2=-2\sqrt{23}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\sqrt{23}+2\\ x=2-2\sqrt{23}\end{array}\right.\)

Khi \(x=2+2\sqrt{23}\) thì \(y=-1,5x=-1,5\left(2+2\sqrt{23}\right)=-3-3\sqrt{23}\)

Khi \(x=2-2\sqrt{23}\) thì y=-1,5x=\(-1,5\left(2-2\sqrt{23}\right)=-3+3\sqrt{23}\)