\(x\left(y^2+z^2\right)+y\left(z^2+x^2\right)+z\left(x^2+y^2\right)+2xzy\) pt đt thành nt
\(\left(x+y\right)\left(x^2-y^2\right)+\left(y+z\right)\left(y^2-z^2\right)+\left(x+z\right)\left(z^2-x^2\right)\) pt đt thành nt
\(\left(x+y\right)\left(x^2-y^2\right)+\left(y+z\right)\left(y^2-z^2\right)+\left(x+z\right)\left(z^2-x^2\right)\)
\(=\left(x+y\right)\left(x^2-y^2\right)-\left(y+z\right)\left[\left(x^2-y^2\right)+\left(z^2-x^2\right)\right]+\left(x+z\right)\left(z^2-x^2\right)\)
\(=\left(x+y\right)\left(x^2-y^2\right)-\left(y+z\right)\left(x^2-y^2\right)-\left(y+z\right)\left(z^2-x^2\right)+\left(x+z\right)\left(z^2-x^2\right)\)
\(=\left(x^2-y^2\right)\left(x+y-y-z\right)-\left(z^2-x^2\right)\left(y+z-x-z\right)\)
\(=\left(x^2-y^2\right)\left(x-z\right)-\left(z^2-x^2\right)\left(y-x\right)\)
\(=\left(x-y\right)\left(x+y\right)\left(x-z\right)-\left(z-x\right)\left(z+x\right)\left(y-x\right)\)
\(=-\left(y-x\right)\left(x+y\right)\left(x-z\right)+\left(x-z\right)\left(z+x\right)\left(y-x\right)\)
\(=\left(y-x\right)\left(x-z\right)\left[-\left(x+y\right)+\left(z+x\right)\right]\)
\(=\left(y-x\right)\left(x-z\right)\left(-x+y+z+x\right)\)
\(=\left(y-x\right)\left(x-z\right)\left(y+z\right)\)
pt đt thành nt
a)\(x^3\left(z-y^2\right)+y^3\left(x-z^2\right)+z^3\left(y-z^2\right)+xyz\left(xyz-1\right)\)
b)\(a\left(b-c\right)^3+b\left(c-a\right)^3+c\left(a-b\right)^2\)
Phân tích đt sau thành nt
a) \(x\left(y^2-2^2\right)+b\left(z^2+x^2\right)+c\left(X^2+Y^2\right)\)
b) \(x^2\left(1-x^2\right)-4-4x^2\)
c) \(\left(1+2x\right)\left(1-2x\right)-x\left(x+2\right)\left(x-2\right)\)
phân tích đa thức thành nhân tử;
a)\(x\left(y^2+z^2\right)+y\left(z^2+x^2\right)+z\left(x^2+y^2\right)+2abc\)
b)\(\left(x+y\right)\left(x^2-y^2\right)+\left(y+z\right)\left(y^2-z^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)
phân tích đa thức thành nhân tử:\(2\left(x^2+y^4+z^4\right)-\left(x^2+y^2+z^2\right)^2-2\left(x^2+y^2+z^2\right)\left(x+y+z\right)^2+\left(x+y+z\right)^4\)
nâng cao phát triển toán 8 tập 1 mình ngại viết nên bạn vào đó xem nhé
ta có : \(x^2+1=x^2+xy+yz+zx=x\left(x+y\right)+z\left(x+y\right)=\left(x+y\right)\left(x+z\right)\)
Tương tự ta đc \(y^2+1=\left(y+x\right)\left(y+z\right)\)
\(z^2+1=\left(z+x\right)\left(z+y\right)\)
ĐẶt \(A=x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{\left(1+x^2\right)}}+y\sqrt{\frac{\left(1+z^2\right)\left(1+x^2\right)}{\left(1+y^2\right)}}+z\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{\left(1+z^2\right)}}\)
\(\Rightarrow A=x\sqrt{\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}+y\sqrt{\frac{\left(z+x\right)\left(z+y\right)\left(x+y\right)\left(x+z\right)}{\left(x+y\right)\left(y+z\right)}}+z\sqrt{\frac{\left(x+y\right)\left(x+z\right)\left(y+z\right)\left(y+x\right)}{\left(z+x\right)\left(z+y\right)}}\)
\(\Rightarrow A=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)=2\left(xy+yz+zx\right)=2\)
phân tích đa thức thành nhân tử:
\(\left(x+y\right)\left(x^2-y^2\right)+\left(y+z\right)\left(y^2-z^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)
\(\left(x+y\right)\left(x^2-y^2\right)+\left(y+z\right)\left(y^2-z^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)
\(=-y^3-xy^2+x^2y+x^3-z^3-yz^2+y^2z+y^3-x^3-zx^2+z^2x+z^3\)
\(=-xy^2+x^2y-yz^2+y^2z-zx^2+z^2x\)
\(=\left(x-y\right)\left(z-x\right)\left(z-y\right)\)
Phân tích đa thức thành nhân tử:
\(M=2\left(x^4+y^4+z^4\right)-\left(x^2+y^2+z^2\right)^2-2\left(x^2+y^2+z^2\right)\left(x+y+z\right)^2+\left(x+y+z\right)^4\)
Ây za,mik ko bt có đúng ko nhưng mik thử làm nhé.
Đặt \(x^4+y^4+z^4=a;x^2+y^2+z^2=b;x+y+z=c\)
\(\Rightarrow M=2a-b^2-2bc^2+c^4\)
\(M=2a-2b^2+b^2-2bc^2+c^4\)
\(M=2\left(a-b^2\right)+\left(b-c^2\right)^2\)
Mà:
\(a-b^2=-2\left(x^2y^2+y^2z^2+z^2x^2\right)\)
\(b-c^2=-2\left(xy+yz+zx\right)\)
Khi đó:
\(M=-4\left(x^2y^2+y^2z^2+z^2x^2\right)+4\left(xy+yz+zx\right)^2\)
\(M=-4x^2y^2-4y^2z^2-4z^2x^2+4x^2y^2++4y^2z^2+4z^2x^2+4z^2x^2+8x^2yz+8xy^2z+8xyz^2\)
\(M=8xyz\left(x+y+z\right)\)
Tính:
\(\dfrac{x^2-yz}{\left(x+y\right)\left(x+z\right)}+\dfrac{y^2-xz}{\left(y+z\right)\left(y+x\right)}+\dfrac{z^2-xy}{\left(z+x\right)\left(z+y\right)}\)
\(\dfrac{x^2}{\left(x-y\right)\left(x-z\right)}+\dfrac{y^2}{\left(y-x\right)\left(y-z\right)}+\dfrac{z^2}{\left(z-x\right)\left(z-y\right)}\)