(Nó có hơi dài dòng)

Cho 3 số x,y,z thỏa mãn: x/2020=y/2021=z/2022.Chứng minh rằng: (x-z)^3 =

(x-z)^3= (2020 - 2022)^3 = -8

8(x-y)^2.(y-z)= 8(2020 - 2021)^2 . (2021 - 2022) = -8.

Vì (x-z)^3 = -8

 8(x-y)^2.(y-z) = -8

==> (x-z)^3 = 8(x-y)^2.(y-z)

-Ủa lòi đâu ra a,b,c vậy bạn?

PT (1) \(\Leftrightarrow2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

Nhận thấy VT\(\ge\)0 với mọi x,y,z

Dấu = xảy ra <=> x=y=z

Thay x=y=z vào pt (2) ta được:

\(3x^{2021}=3^{2022}\) \(\Leftrightarrow x^{2021}=3^{2021}\) \(\Leftrightarrow x=3\)

\(\Rightarrow x=y=z=3\)

Vậy (x;y;z)=(3;3;3)

Lấy điểm O trong ΔABC sao cho \(\hat{AOB}=\hat{BOC}=\hat{AOC}=120^0\)

Đặt OA=x; OB=y; OC=z

Xét ΔOAB có \(cosAOB=\frac{OA^2+OB^2-AB^2}{2\cdot OA\cdot OB}\)

=>\(x^2+y^2-AB^2=2\cdot x\cdot y\cdot\frac{-1}{2}=-xy\)

=>\(AB^2=x^2+y^2+xy=1\)

=>AB=1

Xét ΔOAC có \(cosAOC=\frac{OA^2+OC^2-AC^2}{2\cdot OA\cdot OC}\)

=>\(x^2+z^2-AC^2=2\cdot x\cdot z\cdot cos120=-xz\)

=>\(AC^2=x^2+z^2+xz=\frac34\)

=>\(AC=\frac{\sqrt3}{2}\)

Xét ΔOBC có \(cosBOC=\frac{OB^2+OC^2-BC^2}{2\cdot OB\cdot OC}\)

=>\(\frac{y^2+z^2-BC^2}{2\cdot y\cdot z}=cos120=-\frac12\)

=>\(y^2+z^2-BC^2=-yz\)

=>\(BC^2=y^2+z^2+yz=\frac14\)

=> BC=1/2

Vì \(CA^2+CB^2=AB^2\)

nên ΔCAB vuông tại C

=>\(S_{CAB}=\frac12\cdot CA\cdot CB=\frac12\cdot\frac12\cdot\frac{\sqrt3}{2}=\frac{\sqrt3}{8}\)

\(S_{ABC} = \frac{1}{2}xy\sin(120^\circ) + \frac{1}{2}yz\sin(120^\circ) + \frac{1}{2}xz\sin(120^\circ)\)

=>\(S_{ABC} = \frac{\sqrt{3}}{4}(xy + yz + xz)\)

=>\(\frac{\sqrt{3}}{4}(xy+yz+xz)=\frac{\sqrt{3}}{8}\)

=>\(xy+yz+xz=\frac{1}{2}\)

\((x^2 + xy + y^2) + (y^2 + yz + z^2) + (x^2 + xz + z^2) = 1 + \frac{1}{4} + \frac{3}{4}\)

=>\(2(x^2 + y^2 + z^2) + (xy + yz + xz) = 2\)

=>\(2(x^2+y^2+z^2)+\frac{1}{2}=2\)

=>\(x^2+y^2+z^2=\frac{3}{4}\)

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

\(=\frac34+2\cdot\frac12=\frac74\)

=>\(x+y+z=\frac{\sqrt7}{2}\)

=>\(B=\frac{\sqrt7}{2}\)

ta có bđt phụ ,,,,,,,,  x²+y²+z² >= xy+yz+zx

thay vào thôi,,,cái bđt dễ cm mà,,,nhân 2 2 vế rồi dùng tương đương

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

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