\(M=\frac{42-x}{x-15}=\frac{-\left(x-15\right)+27}{x-15}=-1+\frac{27}{x-15}\)

Để M∈Z⇔x−15∈Ư(27)={±1;±3;±9}

Mà để M min ⇔27x−15⇔27x−15 min ⇔x−15⇔x−15 max ⇔x−15=9⇔x=24

Vậy MinM=−1+279=2⇔x=24

giup tui ik

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