\(a,PT\Leftrightarrow3x^2+3x-2x^2-4x=-1-x\Leftrightarrow x^2=-1\left(\text{vô nghiệm}\right)\)

Vậy: ...

\(b,PT\Leftrightarrow4x\left(x-2019\right)-\left(x-2019\right)=0\Leftrightarrow\left(x-2019\right)\left(4x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2019\\x=\dfrac{1}{4}\end{matrix}\right.\)

Vậy: ...

\(c,PT\Leftrightarrow\left(x-4-6\right)\left(x-4+6\right)=0\Leftrightarrow\left(x-10\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)

Vậy: ...

\(d,PT\Leftrightarrow\left(x+4\right)^2=0\Leftrightarrow x=-4\)

Vậy: ...

\(e,PT\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)

Vậy: ...

\(f,PT\Leftrightarrow\left(5x-3\right)\left(5x+3\right)=0\Leftrightarrow x=\pm\dfrac{3}{5}\)

Vậy: ...

câu c sao tính ra vậy đc vậy k hiểu giải thích hộ e đi 36 đâu mất òi

a) \(0x+9=0\)

\(\Leftrightarrow0x=-9\)

So \(x\in\left\{\varnothing\right\}\)

b) \(\frac{1}{2}x=0\)

\(\Leftrightarrow x=0\)

So x = 0

c) \(\frac{2}{x}+7=0\)

\(\Leftrightarrow\frac{2}{x}=-7\)

\(\Leftrightarrow x=\frac{-2}{7}\)

So \(x=\frac{-2}{7}\)

a, x \(\in\) N

b, x \(\in\) N*

a: Ta có: \(2x^2+y^2-2xy-10x+6y+13=0\)

=>\(y^2-2xy+x^2+6y-6x+x^2-4x+13=0\)

=>\(\left(y-x\right)^2+6\left(y-x\right)+9+x^2-4x+4=0\)

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

=>x-2=0 và y-x+3=0

=>x=2 và y=x-3=2-3=-1

b: \(x^2+7y^2-4xy-2x-2y+4=0\)

=>\(x^2-4xy+4y^2-2x+4y+3y^2-6y+4=0\)

=>\(\left(x-2y\right)^2-2\left(x-2y\right)+1+3y^2-6y+3=0\)

=>\(\left(x-2y-1\right)^2+3\left(y-1\right)^2=0\)

=>x-2y-1=0 và y-1=0

=>y=1 và x=2y+1=2*1+1=3

c: \(11x^2+y^2-6xy-14x+2y+9=0\)

=>\(y^2-6xy+9y^2+2y-6x+2x^2-8x+9=0\)

=>\(\left(y-3x\right)^2+2\left(y-3x\right)+1+2x^2-8x+8=0\)

=>\(\left(y-3x+1\right)^2+2\left(x-2\right)^2=0\)

=>x-2=0 và y-3x+1=0

=>x=2 và y=3x-1=3*2-1=5

0x+111=0

\(\Leftrightarrow\)0x=-111

\(\Rightarrow\)Không có x

****nha                              

SAI ĐỀ RỒI BẠN ƠI

\(x^6+2x^3+1=0\)

\(\Leftrightarrow\left(x^3\right)^2+2x^3+1=0\)

\(\Leftrightarrow\left(x^3+1\right)^2=0\)

\(\Leftrightarrow x^3=\left(-1\right)^3\)

\(\Leftrightarrow x=-1\)

___________

\(x\left(x-5\right)=4x-20\)

\(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

_____________

\(x^4-2x^2=8-4x^2\)

\(\Leftrightarrow x^2\left(x^2-2\right)+\left(4x^2-8\right)=0\)

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

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

\(\Leftrightarrow x^2=2\)

\(\Leftrightarrow x=\pm\sqrt{2}\)

_______________

\(\left(x^3-x^2\right)-4x^2+8x-4\)

\(\Leftrightarrow x^2\left(x-1\right)-4\left(x^2-2x+1\right)=0\)

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

\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)

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

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

a: \(x^2-8x+17\)

\(=x^2-8x+16+1=\left(x-4\right)^2+1\ge1>0\forall x\)

b: \(x^2+4x+5\)

\(=x^2+4x+4+1=\left(x+2\right)^2+1\ge1>0\forall x\)

c: \(x^2-x+1=x^2-x+\frac14+\frac34=\left(x-\frac12\right)^2+\frac34\ge\frac34>0\forall x\)

d: \(-x^2-4x-5=-\left(x^2+4x+5\right)\)

\(=-\left(x^2+4x+4+1\right)=-\left(x+2\right)^2-1\le-1<0\forall x\)

e: \(-x^2-3x-4=-\left(x^2+3x+4\right)\)

\(=-\left(x^2+3x+\frac94+\frac74\right)=-\left(x+\frac32\right)^2-\frac74\le-\frac74<0\forall x\)

f: \(-x^2+10x-27\)

\(=-\left(x^2-10x+27\right)\)

\(=-\left(x^2-10x+25+2\right)=-\left(x-5\right)^2-2\le-2<0\forall x\)

Đáp án D là đáp án đúng

Thế tọa độ O lần lượt vào các đáp án thì A: \(2\le0\) (sai), B: \(2\le0\) (sai), C:\(-2\ge0\) (sai)

D: \(2\ge0\) (đúng)