Question

[Quânneverdie+『ʈєɑɱ❖๖ۣۜƝƘ☆』]

helpmee giúp 1+1=)

Answer 1

[Darlingg🥝]

Bài 1:

1) \(\left(3x^2y^3-2x^2y^2+6x^{^3}y^2\right):\left(-3x^2y^2\right)=-y+\frac{2}{3}-2x\)

2) a. \(3x\left(x-y\right)+2x-2y=3x\left(x-y\right)+2\left(x-y\right)=\left(3x+2\right)\left(x-y\right)\)

    b.\(x^2-2xy-25+y^2=\left(x^2-2xy+y^2\right)-5^2=\left(x-y\right)^2-5^2=\left(x-y-5\right)\left(x-y+5\right)\)

Bài 2:

1) a. \(\frac{6x^2+6xy}{2x^2-2y^2}=\frac{6x\left(x+y\right)}{2\left(x^2-y^2\right)}=\frac{6x\left(x+y\right)}{2\left(x-y\right)\left(x+y\right)}=\frac{3x}{x-y}\)

  b.\(\frac{x^2+7x+10}{x^2+4x+4}=\frac{x^2+2x+5x+10}{\left(x+2\right)^2}=\frac{x\left(x+2\right)+5\left(x+2\right)}{\left(x+2\right)^2}=\frac{\left(x+5\right)\left(x+2\right)}{\left(x+2\right)^2}\)

= x+5/x+2

2) CMR :

\(\frac{2x+2y}{x^2-y^2}=\frac{4x-4y}{2x^2-4xy+2y^2}\)

BĐ VT ta có: \(\frac{2x+2y}{x^2-y^2}=\frac{2\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}=\frac{2}{x-y}\) (1)

BĐ VP ta có:\(\frac{4x-4y}{2x^2-4xy+2y^2}=\frac{4\left(x-y\right)}{2\left(x^2-2xy+y^2\right)}=\frac{4\left(x-y\right)}{2\left(x-y\right)^2}=\frac{2}{x-y}\) (2)

Từ (1) và (2) => VT=VP = 2/x-y (đpcm)

Bài 3:

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

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

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

=> 2 TH

*2x-3=0=>2x=3=>x=3/2

*x+1=0=>x=-1

Vậy x=3/2 hoặc x=-1

b) x^2+x-6=0

=>x^2-2x+3x-6=0

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

=>(x+3).(x-2)=0

=> 2 TH:

*x+3=0=>x=-3

*x-2=0=>x=2

Vậy x=-3 hoặc x=2

Câu 2 bài 3;bài 4 làm riêng nhé

Bài 5: 

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

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

\(\Rightarrow A=2.\left(x^2+y^2+xy\right)-2\left(x-y+1\right)\)

\(\Rightarrow A=2x^2+2y^2+2xy-2x+2y+2\)

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

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

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

 \(\Rightarrow A=\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2\) > 0 với \(\forall\)x;y

Vậy A luôn o âm với mọi x,y (đpcm)

Answer 2

[Lê Minh Tuấn]

2 nnaggvc