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b)\(x^2-2xy+y^2-z^2\)
\(=\left(x-y\right)^2-z^2\)
\(=\left(x-y-z\right)\left(x-y+z\right)\)
a) \(x^2-x-y^2-y\)
\(=\left(x^2-y^2\right)-\left(x+y\right)\)
\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)
a) 3x2 - 5x - 3y2 + 5y
= 3(x2- y2) -5(x-y)
=3(x+y)(x-y) - 5(x-y)
=(x-y)(3x+3y-5)
b) 49 - x2+2xy-y2
= 72 - (x-y)2
=(7-x+y)(7+x-y)
a) \(3x^2-5x-3y^2+5y\)
\(=\left(3x^2-3y^2\right)-\left(5x-5y\right)\)
\(=3\left(x^2-y^2\right)-5\left(x-y\right)\)
\(=3\left[\left(x-y\right).\left(x+y\right)\right]-5\left(x-y\right)\)
\(3\left(x-y\right).\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right).\left[3.\left(x+y\right)-5\right]\)
\(=\left(x-y\right)\left(3x+3y-5\right)\)
b) \(49-x^2+2xy-y^2\)
\(=7^2-x^2+2xy-y^2\)
\(=7^2-\left(x^2-2xy+y^2\right)\)
\(=7-\left(x-y\right)^2\)
\(=\sqrt{7}^2-\left(x-y\right)^2\)
\(=\left[7-\left(x-y\right).-7+\left(x-y\right)\right]\)
\(=\left(7-x+y\right).\left(-7+x-y\right)\)
a) \(x^2+4x+3\)
\(=x^2+3x+x+3\)
\(=x\left(x+3\right)+\left(x+3\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
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) \(x^2-5x+4=x^2-x-4x+4=\left(x^2-x\right)-\left(4x-4\right)\)
\(=x\left(x-1\right)-4\left(x-1\right)=\left(x-1\right)\left(x-4\right)\)
\(\text{a)}x^3-6x^2+12x-8\)
\(=x^3-2x^2-4x^2+8x+4x-8\)
\(=\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(4x-8\right)\)
\(=x^2\left(x-2\right)+4x\left(x-2\right)+4\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+4x+4\right)\)
\(=\left(x-2\right)\left(x+2\right)^2\)
\(\text{b)}8x^2+12x^2y+6xy^2+y^3=\left(2x+y\right)^3\)
Bài 2:
\(\text{a) }x^7+1=\left(x^{\frac{7}{3}}\right)^3+1^3=\left(x^{\frac{7}{3}}+1\right)\left[\left(x^{\frac{7}{3}}\right)^2-x^{\frac{7}{3}}+1\right]=\left(x^{\frac{7}{3}}+1\right)\left(x^{\frac{14}{3}}-x^{\frac{7}{3}}+1\right)\)
\(\text{b) }x^{10}-1=\left(x^5\right)^2-1^2=\left(x^5-1\right)\left(x^5+1\right)\)
Bài 3:
\(\text{a) }69^2-31^2=\left(69-31\right)\left(69+31\right)=38.100=3800\)
\(\text{b) }1023^2-23^2=\left(1023-23\right)\left(1023+23\right)=1000.1046=1046000\)
Thời gian có hạn copy cái này hộ mình vào google xem nha :
https://lazi.vn/quiz/d/16491/nhac-edm-la-loai-nhac-the-loai-gi
Vào xem xong các bạn nhận được 1 thẻ cào mệnh giá 100k nhận thưởng bằng cách nhắn tin vs mình và 1 phần thưởng bí mật là chiếc áo đá bóng,....
Có 300 giải nhanh nha đã có 241 người nhận rồi
OKthanks
\(x^4+5x^3-7x^2-41x-30\)
\(=x^4+x^3+4x^3+4x^2-11x^2-11x-30x-30\)
\(=x^3\left(x+1\right)+4x^2\left(x+1\right)-11x\left(x+1\right)-3x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3+4x^2-11x-30\right)\)
\(=\left(x+1\right)\left(x^3-3x^2+7x^2-21x+10x-30\right)\)
\(=\left(x+1\right)\left[x^2\left(x-3\right)+7x\left(x-3\right)+10x\left(x-3\right)\right]\)
\(=\left(x+1\right)\left(x-3\right)\left(x^2+2x+5x+10\right)\)
\(=\left(x+1\right)\left(x-3\right)\left(x+2\right)\left(x+5\right)\)
x^2-2xy-4z^2+y^2
=(x^2-2xy+y^2)-(2z)^2
=(x-y)^2-2z^2
(x-y+2z)(x-y-2z)