Question

cho x,y,z>0 và xy+yz+xz=1
tính \(A=x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}+y\sqrt{\frac{\left(1+z^2\right)\left(1+x^2\right)}{1+y^2}}+z\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}\)

Answer 1

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 :)

Answer 2

Bài khó thế, mình chịu.

Answer 3

Dệ thôi bạn, bạn chứ thay 1 dưới mẫu= xy+yz+xz rồi PT thành NT xong

Answer 4

nói chung bạn chỉ cần thay 1=xy+yz+zx rồi phân tích thành nhân tử là dc
 

Answer 5

\(xy+yz+zx=1\left(x;y;z>0\right)\Rightarrow x\left(y+z\right)=1-yz\Rightarrow\left(1-yz\right)^2=x^2\left(y+z\right)^2.\)

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: \(A=x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}+y\sqrt{\frac{\left(1+z^2\right)\left(1+x^2\right)}{1+y^2}}+z\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}=2\left(xy+yz+zx\right)=2\)

Answer 6

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

Answer 7

đề này nhìn vất vả :D

Answer 8

tui cx bị thế 

Answer 9

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:

\(S=x\sqrt{\frac{\left(y+z\right)\left(x+y\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\)\(+y\sqrt{\frac{\left(x+y\right)\left(x+z\right)\left(x+z\right)\left(y+z\right)}{\left(y+z\right)\left(x+y\right)}}\)\(+z\sqrt{\frac{\left(x+y\right)\left(x+z\right)\left(y+z\right)\left(x+y\right)}{\left(x+z\right)\left(y+z\right)}}\)

\(=x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}\)\(x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\)

\(=xy+xz+xy+yz+xz+zy\)

\(=2\left(xy+yz+xz\right)\)

\(=2\)

Answer 10

máy mk lag quá viết bị thiếu chờ tí mk làm l

Answer 11

hay thay1=xy+yz+zx roi phan h thanh tu la duoc

Answer 12

kho the ban

Answer 13

hhhhhhhhhhh