toán lớp 1 chưa học đâu bn

toán lớp 1 mà mình lớp 6 ko làm được

Ko hieu đề 

Ta có: a+b+c=1 <=>(a+b+c)2 = 1 <=> ab+bc+ca=0 (1)
Theo dãy tỉ số bằng nhau ta có:
xa=yb=zc=x+y+za+b+c=x+y+z1=x+y+zxa=yb=zc=x+y+za+b+c=x+y+z1=x+y+z
<=> x = a(x+y+z) ; y = b(x+y+z) ; z = c(x+y+z)
=> xy+yz+zx= ab(x+y+z)2+bc(x+y+z)2+ca(x + y + z)2
<=> xy+yz+zx =(ab+bc+ca)(x+y+z)2 (2)
từ (1) và (2) => xy + yz + zx = 0

Từ \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0=>\frac{ayz}{xyz}+\frac{bxz}{xyz}+\frac{cxy}{xyz}=0=>\frac{ayz+bxz+cxy}{xyz}=0=>ayz+bxz+cxy=0\)

Từ \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1=>\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1^2\)

\(=>\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1\)

\(=>\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1-2\left(\frac{xyc}{abc}+\frac{yza}{abc}+\frac{xzb}{abc}\right)=1-2.0=1\)

Vậy M=1

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

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