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\(yz\left(y+z\right)+zx\left(z-x\right)-xy\left(x+y\right)\)
\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left[\left(y+z\right)-\left(z-x\right)\right]\)
\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left(y+z\right)+xy\left(z-x\right)\)
\(=y\left(y+z\right)\left(z-x\right)+x\left(z-x\right)\left(z-y\right)\)
\(=\left(z-x\right)\left(yz-xy+xz-xy\right)\)
a)\(4x^4+y^4=\left(4x^4+y^4+4x^2y^2\right)-4x^2y^2\)
\(=\left(2x^2+y^2\right)^2-\left(2xy\right)^2\)
\(=\left(2x^2+y^2-2xy\right)\left(2x^2+y^2+2xy\right)\)
b)\(\left(x^2-3x-1\right)^2-12\left(x^2-3x-1\right)+27\)
Đặt x^2 - 3x - 1 = A
\(\Rightarrow A^2-12A+27=\left(A^2-12A+36\right)-9\)
\(=\left(A-6\right)^2-9=\left(A-6-3\right)\left(A-6+3\right)\)
\(=\left(A-9\right)\left(A-3\right)\)
Hay \(=\left(x^2-3x-1-9\right)\left(x^2-3x-1-3\right)\)
\(=\left(x^2-3x-10\right)\left(x^2-3x-4\right)\)
\(=\left(x-5\right)\left(x+2\right)\left(x-4\right)\left(x+1\right)\)
c)\(x^3-x^2-5x+125\)
\(=\left(x^3+5^3\right)-\left(x^2+5x\right)\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-5x+25-x\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
d)\(xy\left(x+y\right)+yz\left(y+z\right)+zx\left(z+x\right)+2xyz\)
\(=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
Mình có việc bận nên chỉ đưa được kết quả ý d) thật lòng mong các bạn tự tham khảo và giải
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
a) xy(x + y) + yz(y + z) + xz(z + x) + 3xyz
= xy(X + y + z) + yz(x + y + z) + xz(X + y + z)
= (x + y +z)(xy + yz+ xz)
b) xy(x + y) - yz(y + z) - xz(z - x)
= x2y + xy2 - y2z - yz2 - xz2 + x2z
= x2(y + z) - yz(y + z) + x(y2 - z2)
= x2(y + z) - yz(y + z) + x(y + z)(y - z)
= (y + z)(x2 - yz + xy - xz)
= (y + z)[x(x + y) - z(x + y)]
= (y + z)(x + y)(x - z)
c) x(y2 - z2) + y(z2 - x2) + z(x2 - y2)
= x(y - z)(y + z) + yz2 - yx2 + x2z - y2z
= x(y - z)(y + z) - yz(y - z) - x2(y - z)
= (y - z)((xy + xz - yz - x2)
= (y - z)[x(y - x) - z(y - x)]
= (y - z)(x - z)(y -x)
Sửa đề chút :
\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+3\left(x+y\right)^2z+3\left(x+y\right)z^2+z^3-x^3-y^3-z^3\)
\(=x^3+3x^2y+3xy^2+y^3+3\left(x+y\right)^2z+3\left(x+y\right)z^2-x^3-y^3\)
\(=3x^2y+3xy^2+3\left(x+y\right)^2z+3\left(x+y\right)z^2\)
\(=3xy\left(x+y\right)+3\left(x+y\right)^2z+3\left(x+y\right)z^2\)
\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)
\(=3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)
\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
c) x3 + y3 + z3 - 3xyz
= x3 + 3x2y + 3xy2 + y3 + z3 - 3xyz - 3x2y - 3xy2
= (x+y)3 + z3 - 3xy.( z+x+y)
= (x+y+z).[(x+y)2 - (x+y).z + z2 ] - 3xy.(x+y+z)
= (x+y+z). ( x2 + 2xy + y2 - xz - yz + z2 - 3xy)
= (x+y+z) .(x2 + y2 + z2 - xy - xz -yz)
e) (a+b-c)2 - (a-c)2 - 2ab + 2bc
= (a+b-c - a+c).(a+b+c+a-c) - 2b.(a-c)
= b.(2a+b) - 2b.(a-c)
= b.(2a+b - a +c)
= b.( a+b+c)
xl bn nha! mk chỉ nghĩ đk 2 câu thoy, 1 câu bn kia làm r! 2 câu còn lại bn đợi người tiếp theo làm nhé
b \(x^8y^8+x^4y^4+1=x^8y^8+2x^4y^4+1-x^4y^4=\left(x^4y^4\right)^2+2x^4y^4+1-\left(x^2y^2\right)^2\)
\(=\left(x^4y^4+1\right)^2-\left(x^2y^2\right)^2=\left(x^4y^4-x^2y^2+1\right)\left(x^4y^4+x^2y^2+1\right)\)
c \(x^2y+xy^2+xz^2+x^2z+y^2z+yz^2+2xyz=\left(x^2y+x^2z+xyz+xy^2\right)+\left(xz^2+yz^2+xyz+y^2z\right)\)
\(=x\left(xy+xz+yz+y^2\right)+z\left(xz+yz+xy+y^2\right)=\left(x+z\right)\left(xy+xz+yz+y^2\right)\)
\(=\left(x+z\right)\left(x\left(y+z\right)+y\left(y+z\right)\right)=\left(x+z\right)\left(x+y\right)\left(y+z\right)\)
a \(3xyz+x\left(y^2+z^2\right)+y\left(x^2+z^2\right)+z\left(x^2+y^2\right)=3xyz+xy^2+xz^2+x^2y+yz^2+x^2z+y^2z\)
\(=\left(x^2y+x^2z+xyz\right)+\left(xy^2+xyz+y^2z\right)+\left(xyz+xz^2+yz^2\right)\)
\(=x\left(xy+xz+yz\right)+y\left(xy+xz+yz\right)+z\left(xy+xz+yz\right)=\left(x+y+z\right)\left(xy+xz+yz\right)\)
c) \(xy(x-y)+yz(y-z)+xz(z-x)\)
\(=xy(x-y)-yz[(x-y)+(z-x)]+zx(z-x)\)
\(=(xy-yz)(x-y)+(zx-yz)(z-x)\)
\(=y(x-z)(x-y)+z(x-y)(z-x)\)
\(=(x-y)(z-x)(z-y)\)
d) \(x^4+4a^4=(x^2)^2+(2a^2)^2\)
\(=(x^2)^2+(2a^2)^2+2x^2.2a^2-4x^2a^2\)
\(=(x^2+2a^2)^2-(2xa)^2\)
\(=(x^2+2a^2-2ax)(x^2+2a^2+2ax)\)
e)
\(x^5+x+1=x^5-x^2+x^2+x+1\)
\(=x^2(x^3-1)+x^2+x+1\)
\(=x^2(x-1)(x^2+x+1)+(x^2+x+1)\)
\(=(x^2+x+1)[x^2(x-1)+1]=(x^2+x+1)(x^3-x^2+1)\)
f)
\(x^4+2013x^2+2012x+2013\)
\(=x^4-x+2013x^2+2013x+2013\)
\(=x(x^3-1)+2013(x^2+x+1)\)
\(=x(x-1)(x^2+x+1)+2013(x^2+x+1)\)
\(=(x^2+x+1)[x(x-1)+2013]=(x^2+x+1)(x^2-x+2013)\)