Ta có:

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

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

\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\)

\(\Leftrightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right)=0\)

\(\Leftrightarrow\left(x+y\right)\cdot\frac{xy+z\left(x+y+z\right)}{xyz\left(x+y+z\right)}=0\)

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

\(\Leftrightarrow x=-y\left(h\right)y=-z\left(h\right)z=-x\)

Xét \(x=-y\)

Ta có:

\(\frac{1}{x^{2017}}+\frac{1}{y^{2017}}+\frac{1}{z^{2017}}=\frac{1}{x^{2017}}+\frac{1}{-y^{2017}}+\frac{1}{y^{2017}}=\frac{1}{z^{2017}}\)

\(\frac{1}{x^{2017}+y^{2017}+z^{2017}}=\frac{1}{-x^{2017}+y^{2017}+z^{2017}}=\frac{1}{z^{2017}}\)

\(\Rightarrow\frac{1}{x^{2017}}+\frac{1}{y^{2017}}+\frac{1}{z^{2017}}=\frac{1}{x^{2017}+y^{2017}+z^{2017}}\left(dpcm\right)\)

Một cái chặt hơn nè:))

CMR nếu \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\) thì \(\frac{1}{x^n}+\frac{1}{y^n}+\frac{1}{z^n}=\frac{1}{x^n+y^n+z^n}\) với n lẻ.

a) (2600+6400) - 3x = 1200

=>    9000         - 3x = 1200

=>                       3x = 9000 - 1200

=>                       3x =       7800

=>                         x = 7800 : 3

=>                         x =   2600

                        Vậy x = 2600.

b)  [(6x - 72) : 2 - 84] . 28 = 5628

=> [(6x - 72) : 2 - 84]        = 5628 : 28

=>  (6x - 72) : 2 - 84         = 201

=>  (6x - 72) : 2                = 201 + 84

=>  (6x - 72) : 2                =    285

=>  (6x - 72)                     = 285 . 2

=>   6x - 72                      =    570

=>   6x                             = 570 + 72

=>   6x                             =        642

=>    x                              = 642 : 6

=>    x                              = 107

            Vậy x = 107.

còn on ko bn

vx on,dag doi bai

\(\left(x+1\right)^2+\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\)

\(=x^2+2x+1+\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\)

\(=x^2+2x+1+x^2-1+\left(x-1\right)^2\)

\(=x^2+2x+1+x^2-1+x^2-2x+1\)

\(=3x^2+1\) (1)

Thay x = 100 vào (1), ta có:

\(3x^2+1=\left(3.100\right)^2+1=90001\)

a) Ta có:  A = x² - 4x + 7

A = (x² - 4x + 4) + 3

A = (x - 2)² + 3 \(\ge\)3 \(\forall\)x

Dấu "=" xảy ra <=> x - 2  = 0 <=> x = 2

Vậy MinA = 3 <=> x = 2

b) Xem lại đề