\(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\Rightarrow\frac{xy}{ab}=\frac{yz}{bc}=\frac{xz}{ac}=\frac{xy+yz+xz}{ab+bc+ac}.\)(1)

Ta có

\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)

\(\Leftrightarrow1=1+2\left(ab+bc+ac\right)\Rightarrow ab+bc+ac=0\) => (1) vô nghĩa bạn xem lại đề bài

ta có: \(x+y+z=a\Rightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)=a^2\)

\(\Rightarrow b+2\left(xy+yz+xz\right)=a^2\Rightarrow xy+yz+xz=\frac{a^2-b}{2}\)

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{c}\Rightarrow\frac{xy+yz+xz}{xyz}=\frac{1}{c}\Rightarrow c\left(xy+yz+xz\right)=xyz\)

Ta có:\(x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)+3xyz\)

\(=a\left(b-\frac{a^2-b}{2}\right)+\frac{3c\left(a^2-b\right)}{2}\)

\(a,=\left(xy-1-x-y\right)\left(xy-1+x+y\right)\\ b,Sửa:a^3+2a^2+2a+1\\ =a^3+a^2+a^2+a+a+1=\left(a+1\right)\left(a^2+a+1\right)\\ c,=1-4a^2-a\left(a^2-4\right)=1-4a^2-a^3+4a\\ =\left(1-a\right)\left(1+a+a^2\right)+4a\left(1-a\right)\\ =\left(1-a\right)\left(1+5a+a^2\right)\\ d,=\left(a^2-a^2b^2\right)+\left(b^2-b\right)+\left(ab-a\right)\\ =a^2\left(1-b\right)\left(1+b\right)+b\left(b-1\right)+a\left(b-1\right)\\ =\left(b-1\right)\left(-a^2-ab+b+a\right)\\ =\left(b-1\right)\left(b-1\right)\left(a+b\right)\left(1-a\right)\)

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

\(f,=xyz-xy-yz-xz+x+y+z-1\\ =xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(x-1\right)\\ =\left(z-1\right)\left(xy-y-x+1\right)=\left(z-1\right)\left(x-1\right)\left(y-1\right)\)

Đặt  \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\)  thì  \(x=ak;\)  \(y=bk;\)  \(z=ck\)

Khi đó,  \(xy+yz+xz=abk^2+ack^2+bck^2=k^2\left(ab+bc+ac\right)\)  \(\left(1\right)\)

Vì  \(a+b+c=1\)  nên  suy ra  \(\left(a+b+c\right)^2=1\)

Hay  \(a^2+b^2+c^2+2\left(ab+bc+ac\right)=1\)  \(\left(2\right)\)

Do   \(a^2+b^2+c^2=1\)  (theo giả thiết) nên  \(\left(2\right)\)   \(\Rightarrow\)   \(ab+bc+ac=0\)

Thay vào   \(\left(1\right)\), ta được  \(xy+yz+xz=0\)

1, mk nhớ k lầm thì mk  đã từng làm cho bn rồi ,kq=1/2

2,Dễ CM \(x^2+y^2+z^2\ge xy+yz+xz\) ,dấu "=" xảy ra <=>x=y=z

\(=>\left(x+y+z\right)^2\ge\left(xy+yz+xz\right)+2\left(xy+yz+xz\right)=3\left(xy+yz+xz\right)\)

\(=>9\ge3\left(xy+yz+xz\right)=>xy+yz+xz\le\frac{9}{3}=3\)

=>GTLN của xy+yz+xz=3

3)x³+y³+z³=3xyz

<=>x³+y³+z³-3xyz=0

<=>(x+y+z)(x²+y²+z²-xy-yz-xz)=0

<=>x+y+z=0 hoặc x²+y²+z²-xy-yz-xz=0

(+)x+y+z=0 thì x+y=-z;y+z=-x;x+z=-y

thế vô P =-1

(+)x²+y²+z²-xy-yz-xz=0

TH này thì x=y=z

thế vô P=2