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\(x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-z^2x^2\left(z-x\right)\)
\(=x^2y^2\left(y-x\right)-y^2z^2\left[\left(y-x\right)-\left(z-x\right)\right]-z^2x^2\left(z-x\right)\)
\(=x^2y^2\left(y-x\right)-y^2z^2\left(y-x\right)+y^2z^2\left(z-x\right)-z^2x^2\left(z-x\right)\)
\(=y^2\left(y-x\right)\left(x-z\right)\left(x+z\right)-z^2\left(x-z\right)\left(y-x\right)\left(y+x\right)\)
\(=\left(y-x\right)\left(x-z\right)\left(xy^2+y^2z-z^2y-z^2x\right)\)
Xet \(xy^2+y^2z-z^2y-z^2x=x\left(y-z\right)\left(y+z\right)+yz\left(y-z\right)=\left(y-z\right)\left(xy+yz+zx\right)\)
Vay \(x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-z^2x^2\left(z-x\right)=\left(y-x\right)\left(x-z\right)\left(y-z\right)\left(xy+yz+zx\right)\)
\(x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-z^2x^2\left(z-x\right)\)
\(=x^2y^3-x^3y^2+y^2z^3-y^3z^2-z^3x^2+z^2x^3\)
\(=y^3\left(x^2-z^2\right)-y^2\left(x^3-z^3\right)+z^2x^2\left(x-z\right)\)
\(=y^3\left(x+z\right)\left(x-z\right)-y^2\left(x-z\right)\left(x^2+xz+z^2\right)+z^2x^2\left(x-z\right)\)
\(=\left(x-z\right)\left(xy^3+y^3z-y^2x^2-y^2xz-y^2z^2+z^2x^2\right)\)
.................
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) \(x^2-5x+4=x^2-x-4x+4=\left(x^2-x\right)-\left(4x-4\right)\)
\(=x\left(x-1\right)-4\left(x-1\right)=\left(x-1\right)\left(x-4\right)\)
\(x^3+2x^2y+xy^2-9x\)
\(=\)\(x\left[\left(x^2+2xy+y^2\right)-9\right]\)
\(=\)\(x\left[\left(x+y\right)^2-9\right]\)
\(=\)\(x\left(x+y-3\right)\left(x+y+3\right)\)
Chúc bạn học tốt ~
Phùng Minh Quân sai rồi nhé!Nhưng theo như lời hứa giữa hai ta,nên t sẽ không tích sai nhá!
\(x^3-2x^2y+xy^2-9x\)
\(=xy^2-x^2y-3xy-x^2y+x^3+3x^2+3xy-3x^2-9x\)
\(=xy\left(y-x-3\right)-x^2\left(y-x-3\right)+3x\left(y-x-3\right)\)
\(=\left(y-x-3\right)\left(xy-x^2+3x\right)\)
\(=x\left(y-x-3\right)\left(y-x+3\right)\)
\(\text{a)}x^3-6x^2+12x-8\)
\(=x^3-2x^2-4x^2+8x+4x-8\)
\(=\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(4x-8\right)\)
\(=x^2\left(x-2\right)+4x\left(x-2\right)+4\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+4x+4\right)\)
\(=\left(x-2\right)\left(x+2\right)^2\)
\(\text{b)}8x^2+12x^2y+6xy^2+y^3=\left(2x+y\right)^3\)
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:
\(\text{a) }69^2-31^2=\left(69-31\right)\left(69+31\right)=38.100=3800\)
\(\text{b) }1023^2-23^2=\left(1023-23\right)\left(1023+23\right)=1000.1046=1046000\)
Ta có x + y= 3 => x= 3 - y
=> (3 - y)^2 + y^2 \(\ge\)5
Giải bất phương trình trên, ta được: y \(\ge\)2
Chỉ biết giải đến đó, min P= 33 thì phải
cảm ơn bn , tôi nghĩ ra rồi
bn ra dc \(y\ge2\)thì thay vào \(x^2+y^2\ge5\) ra dc \(x\ge1\)
khi đó min P = 1+16+6.4.1=41 khi và chỉ khi x=1 và y=2
tks bn
a) \(4x^2-6x=2x\left(2x-3\right)\)
b) \(9x^4y^3+3x^2y^4=3x^2y^3\left(3x^2+y\right)\)
c) \(3\left(x-y\right)-5x\left(y-x\right)=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(5x+3\right)\left(x-y\right)\)
d) \(x^3-2x^2+5x=x\left(x^2-2x+5\right)\)
e) \(5\left(x+3y\right)-15x\left(x+3y\right)=\left(5-15x\right)\left(x+3y\right)\)
\(=5\left(1-3x\right)\left(x+3y\right)\)
f) \(2x^2\left(x+1\right)-4\left(x+1\right)=\left(2x^2-4\right)\left(x+1\right)\)
\(=\left(\sqrt{2}x-2\right)\left(\sqrt{2}x+2\right)\left(x+1\right)\)
- Toán lớp 8
Khóa học Toán lớp 8
\(B=\left(x^4+y^4+2x^2y^2\right)+z^4-2z^2\left(x^2+y^2\right)=\left(x^2+y^2\right)^2-2z^2\left(x^2+y^2\right)+z^4\)
\(=\left(x^2+y^2-z^2\right)^2\)