RỒI J NỮA ??
KO CÓ cÂU Hỏi à

Answer:

`S=(-1/7)^0+(-1/7)^1+(-1/7)^2+...+(-1/7)^2007`

`=>S=1-1/7+(1/7)^2-...-(1/7)^2007`

`=>7S=7-1+1/7-...-(1/7)^2006`

`=>S+7S=(1-1/7+(1/7)^2-...-(1/7)^2007)+(7-1+1/7-...-(1/7)^2006)`

`=>8S=7-(1/7)^2017`

`=>8S=7-\frac{1}{7^2007}`

`=>8S=\frac{7^2008-1}{7^2007}`

`=>S=\frac{7^2008-1}{8.7^2007}`

Vì:

\(\left(x-\frac{1}{5}\right)^{2004}\ge0∀x\)

\(\left(y+0,4\right)^{100}\ge0∀y\)

\(\left(z-3\right)^{678}\ge0∀z\)

Do đó

\(\left(x-\frac{1}{5}\right)^{2004}+\left(y+0,4\right)^{100}+\left(z-3\right)^{678}\ge0\)\(\text{ ∀ }x,y,z\)

Mà theo đề bài ta có: \(\left(x-\frac{1}{5}\right)^{2004}+\left(y+0,4\right)^{100}+\left(z-3\right)^{678}=0\)

Nên: \(\hept{\begin{cases}\left(x-\frac{1}{5}\right)^{2004}=0\\\left(y+0,4\right)^{100}=0\\\left(z-3\right)^{678}=0\end{cases}}\)

\(=>\hept{\begin{cases}x=\frac{1}{5}\\y=-0,4\\z=3\end{cases}}\)

Vậy \(\left(x,y,z\right)\in\left\{\frac{1}{5};-0,4;3\right\}\)

Nếu sai mong bạn thông cảm

HT bạn nhé

câu hỏi tương tự cs 

\(P=x^4+x^3-x^2-x^4+3x^2-5x-1.\)

\(\Leftrightarrow P=\left(x^4-x^4\right)+x^3+\left(3x^2-x^2\right)-5x-1\)

\(\Leftrightarrow P=x^3+2x^2-5x-1\)

Vậy chọn  D

hhhhhhhhhh

x⁴ - x³ - 3x - x⁴ = 0

=> -x³ - 3x = 0

=> - (x³ + 3x) = 0

=> x³ + 3x = 0

=> x (x² + 3)=0

=> \(\orbr{\begin{cases}x=0\\x^2=-3\rightarrow VL\end{cases}}\)

Vậy x = 0

Answer:

`x^4-x^3-3x-x^4=0`

`=>(x^4-x^4)-x^3-3x=0`

`=>-x^3-3x=0`

`=>-(x^3+3x)=0`

`=>-x(x^2+3)=0`

Trường hợp 1: `-x=0=>x=0`

Trường hợp 2: `x^2+3=0=>x^2=0-3=>x^2=-3` (Vô lý)

Answer:

Đặt \(k=\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)

\(\Rightarrow\hept{\begin{cases}x=3k\\y=4k\\z=5k\end{cases}}\)

\(-2x^2+y^2-3z^2=-77\)

\(\Rightarrow-2\left(3k\right)^2+\left(4k\right)^2-3\left(5k\right)^2=-77\)

\(\Rightarrow-18k^2+16k^2-75k^2=-77\)

\(\Rightarrow-77k^2=-77\)

\(\Rightarrow k=\pm1\)

Với \(k=1\)

\(\Rightarrow\hept{\begin{cases}x=3\\y=4\\z=5\end{cases}}\)

Với \(k=-1\)

\(\Rightarrow\hept{\begin{cases}x=-3\\y=-4\\z=-5\end{cases}}\)