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a)\(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)
\(=x^2y-x^2z-y^2x+z^2x+y^2z-z^2y\)
\(=x^2\left(y-z\right)-x\left(y^2-z^2\right)+yz\left(y-z\right)\)
\(=x^2\left(y-z\right)-x\left(y-z\right)\left(y+z\right)+yz\left(y-z\right)\)
\(=\left(y-z\right)\left(x^2-x\left(y+z\right)+yz\right)\)
\(=\left(y-z\right)\left(x^2-xy-xz+yz\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
đăng 1 lần thôi là người ta thấy rồi, ai biết ng` ta sẽ giúp ko biết thì thôi mắc gì đăng lắm thế đè hết câu người khác xuống
nhấn vào đây nhé có 2 cách làm: Chuyên đề Bồi dưỡng học sinh giỏi - Phân tích đa thức thành nhân tử - Giáo Án, Bài Giảng
t i c k mk!! 536546456545576768978045362546115346456575676868784675462552
Câu hỏi của Kim Lê Khánh Vy - Toán lớp 8 - Học toán với OnlineMath
Ta có :
\(\left(x+y\right)\left(x^2-y^2\right)+\left(y+z\right)\left(y^2-z^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)
\(=\left(x+y\right)^2.\left(x-y\right)+\left(y+z\right).\left(y^2-x^2+x^2-z^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)
\(=\left(x+y\right)\left(x^2-y^2\right)-\left(y+z\right)\left(x^2-y^2+z^2-x^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)
\(=\left(x+y\right)\left(x^2-y^2\right)-\left(y+z\right)\left(x^2-y^2\right)-\left(y+z\right)\left(z^2-x^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)
\(=\left(x^2-y^2\right)\left(x+y-y-z\right)-\left(z^2-x^2\right).\left(y+z-z-x\right)\)
\(=\left(x^2-y^2\right).\left(x-z\right)-\left(z^2-x^2\right).\left(y-x\right)\)
\(=\left(x-y\right)\left(x+y\right)\left(x-z\right)+\left(z-x\right)\left(z+x\right)\left(x-y\right)\)
\(=\left(x-y\right).\left[\left(x+y\right)\left(x-z\right)+\left(z-x\right).\left(x+z\right)\right]\)
\(=\left(x-y\right)\left(x^2-zx+xy-yz+zx+z^2-x^2-xz\right)\)
\(=\left(x-y\right)\left(z^2-zx+xy-yz\right)\)
\(=\left(x-y\right)\left[z.\left(z-x\right)-y.\left(z-x\right)\right]\)
\(=\left(x-y\right)\left(z-y\right)\left(z-x\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
Ta có :
\(\left(x+y\right)\left(x^2-y^2\right)+\left(y+z\right)\left(y^2-z^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)
\(=\left(x+y\right)^2.\left(x-y\right)+\left(y+z\right).\left(y^2-x^2+x^2-z^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)
\(=\left(x+y\right)\left(x^2-y^2\right)-\left(y+z\right)\left(x^2-y^2+z^2-x^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)
\(=\left(x+y\right)\left(x^2-y^2\right)-\left(y+z\right)\left(x^2-y^2\right)-\left(y+z\right)\left(z^2-x^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)
\(=\left(x^2-y^2\right)\left(x+y-y-z\right)-\left(z^2-x^2\right).\left(y+z-z-x\right)\)
\(=\left(x^2-y^2\right).\left(x-z\right)-\left(z^2-x^2\right).\left(y-x\right)\)
\(=\left(x-y\right)\left(x+y\right)\left(x-z\right)+\left(z-x\right)\left(z+x\right)\left(x-y\right)\)
\(=\left(x-y\right).\left[\left(x+y\right)\left(x-z\right)+\left(z-x\right).\left(x+z\right)\right]\)
\(=\left(x-y\right)\left(x^2-zx+xy-yz+zx+z^2-x^2-xz\right)\)
\(=\left(x-y\right)\left(z^2-zx+xy-yz\right)\)
\(=\left(x-y\right)\left[z.\left(z-x\right)-y.\left(z-x\right)\right]\)
\(=\left(x-y\right)\left(z-y\right)\left(z-x\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
\(\left(x+y\right)\left(x^2-y^2\right)+\left(y+z\right)\left(y^2-z^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)
\(=-xy^2+yx^2-yz^2+zy^2-xz^2+zx^2\)
\(=xy^2\left(1-1\right)+yz^2\left(1-1\right)+zx^2\left(1-1\right)\)
\(=\left(xy^2+yz^2+zx^2\right).0\left(=0\right)\)
Ta có:
\(x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-z^2x^2\left(z-x\right)\)
\(=x^2y^2\left(y-x\right)+y^2z^2\left(z-x-y+x\right)-z^2x^2\left(z-x\right)\)
\(=x^2y^2\left(y-x\right)+y^2z^2\left(z-x\right)-y^2z^2\left(y-x\right)-z^2x^2\left(z-x\right)\)
\(=y^2\left(y-x\right)\left(x-z\right)\left(x+z\right)+z^2\left(z-x\right)\left(y-x\right)\left(y+x\right)\)
\(=\left(y-x\right)\left(x-z\right)\left(y^2x+y^2z-z^2y-z^2x\right)\)
\(=\left(y-x\right)\left(x-z\right)\left(y-z\right)\left(xy+yz+zx\right)\)
a) \([(x-y)3 + (y-z)3]+ (z-x)3\)=\(\left(x-y+y-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]-\left(x-z\right)^3\)
\(=\left(x-z\right)\left[\left(\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2-\left(x-z\right)^2\right)\right]\)
\(=\left(x-z\right)\left[\left(x-y\right)\left(x-y-y+z\right)+\left(y-z-x+z\right)\left(y-z+x-z\right)\right]=\left(x-z\right)\left[\left(x-2y+z\right)\left(x+z\right)-\left(x-y\right)\left(x+y-2z\right)\right]\)
\(=\left(x-z\right)\left(x-y\right)\left(x-2y+z-x-y+2z\right)=\left(x-z\right)\left(x-y\right)\left(z-y\right)3\)
b) \(=y^2\left(x^2y-x^3+z^3-z^2y\right)-z^2x^2\left(z-x\right)=y^2\left[-y\left(z^2-x^2\right)-\left(z^3-x^3\right)\right]-z^2x^2\left(z-x\right)\)
\(=y^2\left(z-x\right)\left(-yz-xy-z^2-zx-x^2\right)-z^2x^2\left(z-x\right)=\left(z-x\right)\left(-y^3z-xy^2-z^2y^2-xyz-x^2y^2-z^2x^2\right)\)
đến đây coi như là thành nhân tử rồi nha. em muốn gọn thì ráng ngồi nghĩ rồi tách nha. chỉ cần nhóm mấy cái có ngoặc giống nhau là đc. k khó đâu. chịu khó nghĩ để rèn luyện nha
c) \(x^8+2x^4+1-x^4=\left(x^4+1\right)^2-x^4=\left(x^4+1-x^2\right)\left(x^4+1+x^2\right)\)
\(\left(9a^3-6a^2\right)+\left(6a^2-4a\right)+\left(-9a+6\right)=3a^2\left(3a-2\right)+2a\left(3a-2\right)-3\left(3a-2\right)=\left(3a-2\right)\left(3a^2+2a-3\right)\)
d) em sửa đề đi. đề sai rồi. đồng nhất hệ số phải có dấu bằng nha.
có gì liên hệ chị. đúng nha ;)
y(x+y)+yz(y+z)+xz(x+z)+2xyz
= xy(x + y) + yz(y + z) + xyz + xz(x + z) + xyz
= xy(x + y) + yz(y + z + x) + xz(x + z + y)
= xy(x + y) + z(x + y + z)(y + x)
= (x + y)(xy + zx + zy + z²)
= (x + y)[x(y + z) + z(y + z)]
= (x + y)(y + z)(z + x)