http://olm.vn/hoi-dap/question/595391.html

Bài giải đây bạn nhé! Mà bạn xem lại đề bài , sao lại từ a,b,c lại chuyển qua x,y,z vậy?

cảm ơn bạn nhìu nhé

2: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

=>\(\frac{xy+yz+xz}{xyz}=0\)

=>xy+yz+xz=0

=>xy=-xz-yz; yz=-xy-xz; xz=-xy-yz

\(x^2+2yz=x^2+yz+yz\)

\(=x^2+yz-xy-xz=x\left(x-y\right)-z\left(x-y\right)=\left(x-y\right)\left(x-z\right)\)

\(y^2+2xz=y^2+xz+xz\)

\(=y^2+xz-xy-yz=y^2-xy+xz-yz\)

=y(y-x)+z(x-y)

=z(x-y)-y(x-y)=(x-y)(z-y)

\(z^2+2xy\)

\(=z^2+xy+xy\)

\(=z^2+xy-yz-xz\)

\(=z^2-xz+xy-yz=z\left(z-x\right)+y\left(x-z\right)=\left(x-z\right)\left(y-z\right)\)

\(A=\frac{yz}{x^2+2yz}+\frac{xz}{y^2+2xz}+\frac{xy}{z^2+2xy}\)

\(=\frac{yz}{\left(x-y\right)\left(x-z\right)}+\frac{xz}{\left(x-y\right)\left(z-y\right)}+\frac{xy}{\left(x-z\right)\left(y-z\right)}\)

\(=\frac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\frac{y^2z-yz^2-x^2z+xz^2+x^2y-xy^2}{\left(x-y\right)\cdot\left(x-z\right)\left(y-z\right)}\)

\(=\frac{z\left(y^2-x^2\right)+z^2\left(x-y\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\frac{\left(x-y\right)\left\lbrack-z\left(x+y\right)+z^2+xy\right\rbrack}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)

\(=\frac{-xz-yz+z^2+xy}{\left(x-z\right)\left(y-z\right)}=\frac{z^2-yz-xz+xy}{\left(x-z\right)\left(y-z\right)}=\frac{z\left(z-y\right)-x\left(z-y\right)}{\left(x-z\right)\left(y-z\right)}=\frac{\left(z-x\right)\left(z-y\right)}{\left(z-x\right)\left(z-y\right)}\)

=1

Đặt:⎧⎩⎨⎪⎪⎪⎪⎪⎪a=13xb=45yc=32z{a=13xb=45yc=32z (x,y,z>0)(x,y,z>0)
Khi đó điều kiện đã cho trở thành:3x+5y+7z≤15xyz3x+5y+7z≤15xyz
Áp dụng AM−GMAM−GM ta có:
3x+5y+7z≥15x3y5z7−−−−−−√153x+5y+7z≥15x3y5z715
=>15xyz≥15x3y5z7−−−−−−√15=>x6y5z4≥1.=>15xyz≥15x3y5z715=>x6y5z4≥1.
Ta có:
P=3x+2.54y+3.23z=12(6x+5y+4z)≥12.15x6y5z4−−−−−−√15≥152P=3x+2.54y+3.23z=12(6x+5y+4z)≥12.15x6y5z415≥152   (AM−GM)   (AM−GM)
Dấu ′=′′=′ xảy ra <=><=> x=y=z=1x=y=z=1 hay a=13;b=45;c=32