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\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\)
\(\Leftrightarrow\frac{x^2-yz}{x-xyz}=\frac{y^2-xz}{y-xyz}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{x^2-yz}{x-xyz}=\frac{y^2-xz}{y-xyz}=\frac{x^2-y^2+xz-yz}{x-xyz-y+xyz}=\frac{\left(x-y\right)\left(x+y\right)+z\left(x-y\right)}{x-y}=\frac{\left(x-y\right)\left(x+y+z\right)}{x-y}=x+y+z\)
\(\Rightarrow\frac{x^2-yz}{x-xyz}=x+y+z\)
\(\Rightarrow x^2-yz=\left(x-xyz\right)\left(x+y+z\right)\)
\(\Rightarrow x^2-yz=x\left(x-xyz\right)+y\left(x-xyz\right)+z\left(x-xyz\right)\)
\(\Rightarrow x^2-yz=x^2-x^2yz+xy-xy^2z+xz-xyz^2\)
\(\Rightarrow-yz-xy-xz=-x^2yz-xy^2z-xyz^2\)
\(\Rightarrow-\left(yz+xy+xz\right)=-\left(x^2yz+xy^2z+xyz^2\right)\)
\(\Rightarrow yz+xy+xz=x^2yz+xy^2z+xyz^2\)
\(\Rightarrow yz+xy+xz=xyz\left(x+y+z\right)\)
Vậy nếu \(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\) thì \(yz+xy+xz=xyz\left(x+y+z\right)\)
Áp dụng holder ta có:
\(\left(1+1+1\right)\left(x^2y+y^2z+z^2x\right)\left(xy^2+yz^2+zx^2\right)\)
\(\ge\left(\sqrt[3]{x^4yz}+\sqrt{y^4zx}+\sqrt{z^4xy}\right)^3=xyz\left(x+y+z\right)^3\)
Dạo này bận lắm nên cũng lười luôn nên thông cảm.
Bài này làm được theo 1 cách khác nhưng phải áp dụng 2 lần bđt
lần 1 dùng bđt Schur
lần 2 dùng AM-GM
Em(mình) thử nhé, ko chắc đâu
3/ Ta có \(\left(a+b\right)\left(b+c\right)\left(c+a\right)=ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc\)
\(=\left[ab\left(a+b\right)+abc\right]+\left[bc\left(b+c\right)+abc\right]+\left[ca\left(c+a\right)+ca\right]-abc\)
\(=\left(a+b+c\right)ab+\left(a+b+c\right)bc+\left(a+b+c\right)ca-abc\)
\(=\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)= -abc
Suy ra \(P=\frac{-abc}{abc}=-1\)
Vậy..
Ta có :
\(VT=\left(x+y\right)\left(y+z\right)\left(z+x\right)+xyz\)
\(=\left(xy+y^2+xz+yz\right)\left(z+x\right)+xyz\)
\(=xyz+y^2z+xz^2+yz^2+x^2y+y^2x+x^2z+xyz+xyz\)
\(=\left(x^2y+xyz+x^2z\right)+\left(y^2x+y^2z+xyz\right)+\left(xyz+z^2y+z^2x\right)\)\(=x\left(xy+yz+zx\right)+y\left(xy+yz+zx\right)+z\left(xy+yz+zx\right)\)
\(=\left(x+y+z\right)\left(xy+yz+zx\right)=VP\)
\(\left(đpcm\right)\)
:D
a) \(\left(x-y\right)^2+\left(y-z\right)^3+\left(z-x\right)^3\)
\(=\left(x-y\right)^2+\left(y-z+z-x\right)\left[\left(y-z\right)^2-\left(y-z\right)\left(z-x\right)+\left(z-x\right)^2\right]\)
\(=\left(x-y\right)^2+\left(y-x\right)\left(x^2+y^2+3z^2-3yz+xy-3xz\right)\)
\(=\left(x-y\right)\left(x-y-x^2-y^2-3z^2+3yz-xy+3xz\right)\)
Cô nghĩ phân tích đa thức này sẽ đẹp hơn:
\(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)
\(=\left(x-y+y-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]+\left(z-x\right)^3\)
\(=\left(x-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]+\left(z-x\right)^3\)
\(=\left(x-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2-\left(z-x\right)^2\right]\)
\(=\left(x-z\right)\left(3y^2-3xy+3zx-3xyz\right)\)
\(=3\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
b) \(\left(x+y+z\right)\left(xy+yz+zx\right)-xyz\)
\(=\left(xy+yz+zx\right)\left(x+y+z\right)-xyz\)
\(=xy\left(x+y+z\right)+\left(yz+zx\right)\left(x+y+z\right)-xyz\)
\(=xy\left(x+y+z-z\right)+\left(yz+zx\right)\left(x+y+z\right)\)
\(=xy\left(x+y\right)+z\left(y+x\right)\left(x+y+z\right)\)
\(=\left(x+y\right)\left[xy+z\left(x+y+z\right)\right]\)
\(=\left(x+y\right)\left(xy+zx+zy+z^2\right)\)
\(=\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)
\(=\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
a) \left(x-y\right)^2+\left(y-z\right)^3+\left(z-x\right)^3(x−y)2+(y−z)3+(z−x)3
=\left(x-y\right)^2+\left(y-z+z-x\right)\left[\left(y-z\right)^2-\left(y-z\right)\left(z-x\right)+\left(z-x\right)^2\right]=(x−y)2+(y−z+z−x)[(y−z)2−(y−z)(z−x)+(z−x)2]
=\left(x-y\right)^2+\left(y-x\right)\left(x^2+y^2+3z^2-3yz+xy-3xz\right)=(x−y)2+(y−x)(x2+y2+3z2−3yz+xy−3xz)
=\left(x-y\right)\left(x-y-x^2-y^2-3z^2+3yz-xy+3xz\right)=(x−y)(x−y−x2−y2−3z2+3yz−xy+3xz
\left(x-y\right)^3+\left(y-z\right)^3+\left
=\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(z-x\right)^3
=\left(x-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]+\l
=\left(x-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2-\left(z-x\
=\left(x-z\right)\left(
=3\left(x-y\right)\lefb) \left(x+y+z\right)\left(xy+yz+zx\right)-xyzb)(x+y+z)(xy+yz+zx)−xyz
=\left(xy+yz+zx\right)\left(x+y+z\right)-xyz=(xy+yz+zx)(x+y+z)−xyz
=xy\left(x+y+z\right)+\left(yz+zx\right)\left(x+y+z\right)-xyz=xy(x+y+z)+(yz+zx)(x+y+z)−xyz
=xy\left(x+y+z-z\right)+\left(yz+zx\right)\left(x+y+z\right)=xy(x+y+z−z)+(yz+zx)(x+y+z)
=xy\left(x+y\right)+z\left(y+x\right)\left(x+y+z\right)=xy(x+y)+z(y+x)(x+y+z)
=\left(x+y\right)\left[xy+z\left(x+y+z\right)\right]=(x+y)[xy+z(x+y+z)]
=\left(x+y\right)\left(xy+zx+zy+z^2\right)=(x+y)(xy+zx+zy+z2)
=\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]=(x+y)[x(y+z)+z(y+z)]
=\left(x+y\right)\left(y+z\right)\left(z+x\right)=(x+y)(y+z)(z+x)