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\(x\left(y+z\right)^2+y\left(x+z\right)^2+z\left(x+y\right)^2-4xyz\)
\(=x\left(y^2+2yz+z^2\right)+y\left(x^2+2xz+z^2\right)+z\left(x+y\right)^2-4xyz\)
\(=xy^2+2xyz+xz^2+x^2y+2xyz+yz^2+z\left(x+y\right)\left(x+y\right)-4xyz\)
\(=\left(xy^2+x^2y\right)+\left(xz^2+yz^2\right)+z\left(x+y\right)^2\)
\(=xy\left(x+y\right)+z^2\left(x+y\right)+\left(xz+yz\right)\left(x+y\right)\)
\(=\left(x+y\right)\left(z^2+xz+yz+xy\right)\)
\(=\left(x+y\right)\left[z\left(x+z\right)+y\left(x+z\right)\right]\)
\(=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
a: \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b+c\right)^3-a^3\right]-\left(b^3+c^3\right)\)
\(=\left(a+b+c-a\right)\left[\left(a+b+c\right)^2+a\left(a+b+c\right)+a^2\right]-\left(b+c\right)\left(b^2-bc+c^2\right)\)
\(=\left(b+c\right)\left[a^2+b^2+c^2+a^2+a^2+2ab+2bc+2ac+ab+ac-b^2+bc-c^2\right]\)
\(=\left(b+c\right)\left(3a^2+3ab+3bc+3ac\right)\)
\(=3\left(b+c\right)\left(a+b\right)\left(a+c\right)\)
b: \(=\left(2x+2y+2z\right)^3-\left(x+y\right)^3-\left[\left(y+z\right)^3+\left(x+z\right)^3\right]\)
\(=\left(x+y+2z\right)\left[\left(2x+2y+2z\right)^2+2\left(x+y+z\right)\left(x+y\right)+\left(x+y\right)^2\right]-\left(x+y+2z\right)\left[\left(y+z\right)^2-\left(y+z\right)\left(x+z\right)+\left(x+z\right)^2\right]\)
\(=3\left(x+y+2z\right)\left(x+z+2y\right)\left(y+z+2x\right)\)
Hướng dẫn: đặt \(A=\dfrac{y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\dfrac{z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\dfrac{x^4}{\left(z^2+x^2\right)\left(z+x\right)}\)
Khi đó \(F-A=x-y+y-z+z-x=0\Rightarrow F=A\)
\(\Rightarrow2F=F+A=\sum\dfrac{x^4+y^4}{\left(x^2+y^2\right)\left(x+y\right)}\ge\sum\dfrac{\left(x^2+y^2\right)^2}{2\left(x^2+y^2\right)\left(x+y\right)}\ge\sum\dfrac{\left(x+y\right)^2\left(x^2+y^2\right)}{4\left(x^2+y^2\right)\left(x+y\right)}\)
\(\Rightarrow2F\ge\dfrac{x+y+z}{2}\Rightarrow F\ge\dfrac{x+y+z}{4}\)
\(\left(x+y+z\right)^7-x^7=\left(x+y+z-x\right)\times\int\left(x,y,z\right)=\left(y+z\right)\times\int\left(x,y,z\right)\)
\(y^7+z^7=\left(y+z\right)\left(y^6-y^5z+...\right)\)
Từ đây đưa nhân tử chung là y+z ra ngoài là xong
Mình chỉ làm được đến đó thôi
Các công thức trên thì search wikipedia là xong
Có \(x\left(y+z\right)^2+y\left(x+z\right)^2+z\left(x+y\right)^2-4xyz\)
\(=xy^2+2xyz+xz^2+yx^2+2xyz+yz^2+zx^2+2xyz+zy^2-4xyz\)
\(=xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+xyz+xyz\)
\(=xy\left(x+y+z\right)+yz\left(x+y+z\right)+xz\left(x+z\right)\)
\(=\left(xy+yz\right)\left(x+y+z\right)+xz\left(x+z\right)\)
\(=y\left(x+y+z\right)\left(x+z\right)+xz\left(x+z\right)\)
\(=\left(x+z\right)\left(xy+y^2+yz+xz\right)\)
\(=\left(x+z\right)\left(x\left(x+y\right)+z\left(x+y\right)\right)\)
\(=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
Chúc bạn học tốt!