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Bài 2:
a) \(x^2-y^2+3x-3y=\left(x^2-y^2\right)+\left(3x-3y\right)\)
\(=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\)
b) \(5x-5y+x^2-2xy+y^2=\left(5x-5y\right)+\left(x^2-2xy+y^2\right)\)
\(=5\left(x-y\right)+\left(x-y\right)^2=\left(x-y\right)\left(x-y+5\right)\)
c) \(x^2-5x+4=x^2-x-4x+4=\left(x^2-x\right)-\left(4x-4\right)\)
\(=x\left(x-1\right)-4\left(x-1\right)=\left(x-1\right)\left(x-4\right)\)
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Bài 2
a. (x-2y)2 =2x-4y
b. (2x^2 +3)2 =4x^2+6
c. (x-2) (x^2+2x+4) = x^3-8 (hằng đẳng thức)
d. (2x-1)3 = 6x-3
Xin lỗi mik chỉ lm ổn bài 2 thôi!
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a, \(x^3-2x^2+3x-6=x\left(x^2+3\right)-2\left(x^2+3\right)=\left(x-2\right)\left(x^2+3\right)\)
b, \(x^2+2x+1-4y^2=\left(x+1\right)^2-\left(2y\right)^2=\left(x+1-2y\right)\left(x+1+2y\right)\)
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Ta có:\(y=100...05=100...00+5=x\cdot9+1+5=9x+6\)
\(\Rightarrow xy+1=\cdot\left(9x+6\right)x+1=9x^2+6x+1=\left(3x+1\right)^2\)Là số chính phương
a) \(\left(x^2-2x+1\right)\left(x-1\right)=\left(x-1\right)^2\left(x-1\right)=\left(x-1\right)^3=x^3-3x^2+27x-1\)
\(a,\)\(\left(x^2-2x+1\right)\left(x-1\right)\)
\(=\left(x-1\right)^2\left(x-1\right)\)
\(=\left(x-1\right)^3\)
\(=x^3-3x^2+3x-1\)
\(b,\)\(\left(x^3-2x^2+x-1\right)\left(5-x\right)\)
\(=5x^3-x^4-10x^2+2x^3+5x-x^2-5+x\)
\(-x^4+7x^3-11x^2+6x-5\)