Bài 1: 

a: \(x^2+5x=x\left(x+5\right)\)

Để biểu thức này âm thì \(x\left(x+5\right)< 0\)

hay -5<x<0

b: \(3\left(2x+3\right)\left(3x-5\right)< 0\)

\(\Leftrightarrow-\dfrac{3}{2}< x< \dfrac{5}{3}\)

Bài 2: 

a: \(2y^2-4y>0\)

\(\Leftrightarrow y\left(y-2\right)>0\)

\(\Leftrightarrow\left[{}\begin{matrix}y>2\\y< 0\end{matrix}\right.\)

b: \(5\left(3y+1\right)\left(4y-3\right)>0\)

\(\Leftrightarrow\left[{}\begin{matrix}y>\dfrac{3}{4}\\y< -\dfrac{1}{3}\end{matrix}\right.\)

x^2-8x+20=(x^2-8x+16)+4

                 =(x-4)^2+4>0(vì (x-4)^2>=0)

4x^2-12x+11=4x^2-12x+9+2

                     =(2x-3)^2+2>0

x^2-x+1=x^2-x+1/4+3/4

             =(x-1/2)^2+3/4>0

x^2-2x+y^2+4y+6

=x^2-2x+1+y^2+4y+4+1

=(x-1)^2+(y+2)^2+1>0

a: \(x^2-8x+20\)

\(=x^2-8x+16+4\)

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

b: Ta có: \(4x^2-12x+11\)

\(=4x^2-12x+9+2\)

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

c: Ta có: \(x^2-x+1\)

\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\)

d: Ta có: \(x^2-2x+y^2+4y+6\)

\(=x^2-2x+1+y^2+4y+4+1\)

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

\(A=2x^2-20x+7=2\left(x^2-10x+25\right)-43=2\left(x-5\right)^2-43\ge-43\left(\forall x\right)\)

=> Chưa thể khẳng định A dương

\(B=9x^2-6xy+2y^2+1\)

\(B=\left(9x^2-6xy+y^2\right)+y^2+1\)

\(B=\left(3x-y\right)^2+y^2+1\ge1>0\left(\forall x\right)\)

=> đpcm

\(C=x^2-2x+y^2+4y+6\)

\(C=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+1\)

\(C=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1>0\left(\forall x\right)\)

=> đpcm

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

=> đpcm

\(a.\)

\(A=9x^2-6xy+2y^2+1\)

\(A=\left(3x\right)^2-2\cdot3x\cdot y+y^2+y^2+1\)

\(A=\left(3x-y\right)^2+\left(y^2+1\right)\ge0\)

\(b.\)

\(B=x^2-2x+y^2+4y+6\)

\(B=x^2-2x+1+y^2+4y+4+1\)

\(B=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\)

\(c.\)

\(C=x^2-2x+2\)

\(C=x^2-2x+1+1\)

\(C=\left(x-1\right)^2+1\ge1\)

a) A=9x²-6xy+2y²+1

    A=(3x)²-2.3x.y+y²+y²+1

    A=(3x-y)²+(y²+1)≥0

Câu b, c tương tự câu a

a) Ta có: \(A=9x^2-6xy+2y^2+1\)

\(=\left(3x-1\right)^2+2y^2\ge0\forall x,y\)

b) Ta có: \(B=x^2-2x+y^2+4y+6\)

\(=x^2-2x+1+y^2+4y+4+1\)

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

c) Ta có: \(C=x^2-2x+2\)

\(=x^2-2x+1+1\)

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

a)\(A=x^2+x+1=\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)

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

\(\left(x^2-2x+1\right)+\left(y^2-4y+4\right)+1\) 

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

Xét nữa là xong

a) \(A=x^2-x+1=\left(x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)

b) \(B=\left(x-2\right)\left(x-4\right)+3=x^2-6x+8+3=\left(x-3\right)^2+2\ge2>0\)

c) \(C=2x^2-4xy+4y^2+2x+5=\left(x-2y\right)^2+\left(x+1\right)^2+4\ge4>0\)

a: =x^2-x+1/4+3/4

=(x-1/2)^2+3/4>=3/4>0 với mọi x

b: B=x^2-6x+8+3

=x^2-6x+11

=x^2-6x+9+2

=(x-3)^2+2>=2>0 với mọi x

c: =x^2-4xy+4y^2+x^2+2x+1+4

=(x-2y)^2+(x+1)^2+4>=4>0 với mọi x,y