Ta có:
1+x2=xy+yz+xz+x2=(x+y)(x+z)
1+y2=xy+yz+xz+y2=(y+z)(x+y)
1+z2=xy+yz+zx+z2=(x+z)(y+z)
Thay vào A ta được:
\(A=x\sqrt{\frac{\left(y+z\right)\left(x+y\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\)\(+y\sqrt{\frac{\left(x+y\right)\left(x+z\right)\left(x+z\right)\left(y+z\right)}{\left(y+z\right)\left(x+y\right)}}\)\(+z\sqrt{\frac{\left(x+y\right)\left(x+z\right)\left(y+z\right)\left(x+y\right)}{\left(x+z\right)\left(y+z\right)}}\)
\(=x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\left(x+y\right)^2\)
\(=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\)
\(=xy+xz+xy+yz+xz+zy\)
\(=2\left(xy+yz+xz\right)\)
\(=2\)
Đây ms là chuẩn :)
Dệ thôi bạn, bạn chứ thay 1 dưới mẫu= xy+yz+xz rồi PT thành NT xong
nói chung bạn chỉ cần thay 1=xy+yz+zx rồi phân tích thành nhân tử là dc
\(xy+yz+zx=1\left(x;y;z>0\right)\Rightarrow x\left(y+z\right)=1-yz\Rightarrow\left(1-yz\right)^2=x^2\left(y+z\right)^2.\)
Mặt khác: \(\left(1+y^2\right)\left(1+z^2\right)=1+y^2+z^2+y^2z^2=1-2yz+y^2z^2+y^2+z^2+2yz=\)
\(=\left(1-yz\right)^2+\left(y+z\right)^2=x^2\left(y+z\right)^2+\left(y+z\right)^2=\left(y+z\right)^2\left(1+x^2\right)\)
\(\Rightarrow x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}=x\sqrt{\frac{\left(1+x^2\right)\left(y+z\right)^2}{1+x^2}}=x\left(y+z\right)\)(vì y+z>0)
Tương tự: \(y\sqrt{\frac{\left(1+z^2\right)\left(1+x^2\right)}{1+y^2}}=y\left(x+z\right)\)
Và: \(z\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}=z\left(x+y\right)\)
Do đó: \(A=x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}+y\sqrt{\frac{\left(1+z^2\right)\left(1+x^2\right)}{1+y^2}}+z\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}=2\left(xy+yz+zx\right)=2\)
1 + x2 = xy + yz + zx + x2 = y(z+x) + x(z+x) = (x+y)(z+x) (1)
cm tương tự: 1 + y2 =(x+y)(y+z) (2)
1 + z2 =(z+x)(y+z) (3)
Từ (1),(2) và (3) =>A=x(y+z) + y(z+x) + z(x+y) = 2
Ta có:
\(1+x^2=xy+yz+xz+x^2=\left(x+y\right)\left(x+z\right)\)
\(1+y^2=xy+yz+xz+y^2=\left(y+z\right)\left(x+y\right)\)
\(1+z^2=xy+yz+zx+z^2=\left(x+z\right)\left(y+z\right)\)
Thay vào S ta được:
\(S=x\sqrt{\frac{\left(y+z\right)\left(x+y\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\)\(+y\sqrt{\frac{\left(x+y\right)\left(x+z\right)\left(x+z\right)\left(y+z\right)}{\left(y+z\right)\left(x+y\right)}}\)\(+z\sqrt{\frac{\left(x+y\right)\left(x+z\right)\left(y+z\right)\left(x+y\right)}{\left(x+z\right)\left(y+z\right)}}\)
\(=x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}\)\(x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\)
\(=xy+xz+xy+yz+xz+zy\)
\(=2\left(xy+yz+xz\right)\)
\(=2\)
máy mk lag quá viết bị thiếu chờ tí mk làm l
hay thay1=xy+yz+zx roi phan h thanh tu la duoc
