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

\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}=\frac{-1}{z}\)

\(\Rightarrow\left(\frac{1}{x}+\frac{1}{y}\right)^3=\left(\frac{-1}{z}\right)^3\)

\(\Leftrightarrow\frac{1}{x^3}+3\frac{1}{x^2}\frac{1}{y}+3\frac{1}{x}\frac{1}{y^2}+\frac{1}{y^3}=\frac{-1}{z^3}\)

\(\Leftrightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=-3.\frac{1}{x}\frac{1}{y}\left(\frac{1}{x}+\frac{1}{y}\right)\)\(\Leftrightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=-3.\frac{1}{x}\frac{1}{y}\frac{-1}{z}\)

\(\Leftrightarrow\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)xyz=3.\frac{1}{x}\frac{1}{y}\frac{1}{z}.xyz\)

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

Ta có: \(\left(x+y\right)^5=x^5+y^5\)

\(\Leftrightarrow\left(x+y\right)^5-x^5-y^5=0\)

\(\Leftrightarrow x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5-x^5-y^5=0\)

\(\Leftrightarrow5x^4y+10x^3y^2+10x^2y^3+5xy^4=0\)

\(\Leftrightarrow\left(5x^4y+5xy^4\right)+\left(10x^3y^2+10x^2y^3\right)=0\)

\(\Leftrightarrow5xy\left(x^3+y^3\right)+10x^2y^2\left(x+y\right)=0\)

\(\Leftrightarrow5xy\left(x+y\right)\left(x^2-xy+y^2\right)+10x^2y^2\left(x+y\right)=0\)

\(\Leftrightarrow5xy\left(x+y\right)\left(x^2+xy+y^2\right)=0\)

\(\Rightarrow\)hoặc 5xy = 0 hoặc x + y = 0 hoặc \(x^2+xy+y^2=0\)

\(+)5xy=0\Rightarrow\orbr{\begin{cases}x=0\\y=0\end{cases}}\)

\(+)x+y=0\Rightarrow x=-y\)(hai số đối)

\(+)x^2+xy+y^2=0\)

\(\Leftrightarrow x^2+2.x.\frac{y}{2}+\frac{y^2}{4}+\frac{3y^2}{4}=0\)

\(\Leftrightarrow\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}=0\)

Mà \(\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}\ge0\)

(Dấu "="\(\Leftrightarrow x=y=0\))

Vậy x và y là hai số đối

A = 2x² + 6x - 1

A = 2( x² + 3x - 1 / 2 )

A = 2[ x² + 2 . 3 / 2 . x + ( 3 / 2 )² - ( 3 / 2 )² - 1 / 2 ]

A = 2[ ( x + 3 / 2 )² - 11 / 4 ]

A = ( x + 3 / 2 )² - 11 / 2 \(\ge\)- 11 / 2

Dấu " = " xảy ra\(\Leftrightarrow\)x + 3 / 2 = 0

                            \(\Rightarrow\)x              = 3 / 2

Min A = - 11 / 2 \(\Leftrightarrow\)x = 3 / 2

Cho đa thức \(f\left(x\right)\)bậc 3 với hệ số \(x^3\)là số nguyên dương thỏa mãn:

\(f\left(2019\right)=2020;f\left(2020\right)=2021\)

CMR \(f\left(2021\right)-f\left(2018\right)\)là hợp số

Cho xin cái đề ạ

Ta có:

\(x^3+ax+b=\left(x+1\right)\cdot P\left(x\right)+7\)

\(x^3+ax+b=\left(x-3\right)\cdot Q\left(x\right)+5\)

Theo Bezut ta có:

Với \(x=-1\Rightarrow b-a-1=7\)

Với \(x=3\Rightarrow3a+b+27=5\)

\(\Rightarrow4a+28=-2\Rightarrow4a=26\Rightarrow a=\frac{13}{2}\Rightarrow b=\frac{29}{2}\)

\(a.=a\left(a^2+a+1\right)\)

\(=a\left(a+\frac{1}{2}\right)^2+\frac{3}{4}\)

\(b.=\left(a+b\right)^2-9\)

\(=\left(a+b\right)^2-3^2\)

\(=\left(a+b-3\right)\left(a+b+3\right)\)

\(c.=a\left(b-1\right)+b\left(b-1\right)\)

\(=\left(a+b\right)\left(b-1\right)\)

\(d.=a^2-3a-4a+12\)

\(=a\left(a-3\right)-4\left(a-3\right)\)

\(=\left(a-4\right)\left(a-3\right)\)

\(=27a^6-18a^4b^3+12a^2b^6-18a^4b^3+12a^2b^6-8b^9\)

\(=27a^6-36a^4b^3+24a^2b^6-8a^9\)

\(A=\left(x+y\right)^2+\left(y-x\right)^2-2\left(x-y\right)\left(x+y\right)\)

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

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

\(=\left(2y\right)^2\)Thay \(y=\frac{1}{2}\)ta được:
\(\left(2.\frac{1}{2}\right)^2\)

\(=1\)

Vậy \(A=1\)tại \(x=2019\)và \(y=\frac{1}{2}\)

A = (x + y)^2 + (y - x)^2 - 2(x - y)(x + y)

A = x^2 + 2xy + y^2 + x^2 - 2xy + y^2 - 2x^2 + 2y^2

A = (x^2 + x^2 - 2x^2) + (2xy - 2xy) + (y^2 + y^2 + 2y^2)

A = 4y^2 (1)

Thay x = 2019 và y = 1/2 vào (1), ta có:

(4.1/2)^2 = 4