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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)
\(\frac{x^2-9}{x-3}=\frac{\left(x-3\right)\left(x+3\right)}{x-3}=x+3\)
\(\frac{x^2+5x+6}{x+2}=\frac{x^2+2x+3x+6}{x+2}=\frac{\left(x+2\right)\left(x+3\right)}{x+2}=x+3\)
Suy ra đpcm.