Lũy thừa của một số hữu tỉ

Ta có: \(\left(x-\dfrac{1}{5}\right)^{2004}\ge0\forall x\)

\(\left(y+0.4\right)^{100}\ge0\forall y\)

\(\left(z-3\right)^{678}\ge0\forall z\)

Do đó: \(\left(x-\dfrac{1}{5}\right)^{2004}+\left(y+0.4\right)^{100}+\left(z-3\right)^{678}\ge0\forall x,y,z\)

Dấu '=' xảy ra khi

\(\left\{{}\begin{matrix}x-\dfrac{1}{5}=0\\y+0.4=0\\z-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-\dfrac{2}{5}\\z=3\end{matrix}\right.\)

Vậy: (x,y,z)=\(\left(\dfrac{1}{5};-\dfrac{2}{5};3\right)\)

\(x:3=y:5\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{y-x}{5-3}=\dfrac{24}{2}=12\)

=> \(\left\{{}\begin{matrix}x=36\\y=60\end{matrix}\right.\)

\(x:3=y:5 \Leftrightarrow \dfrac{x}{3}=\dfrac{y}{5}\)

Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{y-x}{5-3}=\dfrac{24}{2}=12 \\ \Rightarrow x=12.3=36 \\ y=12.5=60\)

Vậy...

Ta có: x:3=y:5

nên \(\dfrac{x}{3}=\dfrac{y}{5}\)

mà y-x=24

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{y-x}{5-3}=\dfrac{24}{2}=12\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{x}{3}=12\\\dfrac{y}{5}=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=36\\y=60\end{matrix}\right.\)

Vậy: (x,y)=(36;60)

Lời giải:

\(S=3^{x+1}+3^{x+3}+3^{x+5}+...+3^{x+101}\)

\(=3^{x+1}(1+3^2+3^4+..+3^{100})\)

\(9S=3^{x+1}(3^2+3^4+3^6+...+3^{102})\)

\(\Rightarrow 9S-S=3^{x+1}(3^{102}-1)=3^{x+1}(9^{51}-1)\)

hay $8S=3^{x+1}(9^{51}-1)$

Ta thấy $(10-1)^{51}-1=10k-1-1\not\vdots 10$

$\Rightarrow 8S=3^{x+1}(9^{51}-1)\not\vdots 10$

$\Rightarrow S\not\vdots 10$ nên $S$ cũng không thể chia hết cho $120$

Bạn xem lại đề.

Sửa đề: \(11\cdot5^{2n}+2^{3n+2}+2^{3n+1}\)

Ta có: \(11\cdot5^{2n}+2^{3n+2}+2^{3n+1}\)

\(=11\cdot25^n+8^n\cdot4+8^n\cdot2\)

\(=11\cdot25^n+6\cdot8^n\)

Vì \(25\equiv8\)(mod 17)

nên \(11\cdot25^n+6\cdot8^n\equiv11\cdot8^n+6\cdot8^n\equiv17\cdot8^n\equiv0\)(mod 17)

hay \(11\cdot5^{2n}+2^{3n+2}+2^{3n+1}⋮17\)(đpcm)

ta có :\(\dfrac{a}{b}\)=\(\dfrac{b}{c}\)=\(\dfrac{c}{a}\)=\(\dfrac{a+b+c}{b+c+a}\)=1

*\(\dfrac{a}{b}\)=1 =>a=b

*\(\dfrac{b}{c}\)=1 =>b=c

*\(\dfrac{c}{a}\)=1 =>c=a

=>a=b=c

=>\(a^{670}\)+\(b^{672}\)+\(c^{673}\)/\(a^{2015}\)=\(a^{2015}\)/\(a^{2015}\)=1

nhớ like nha 

Xem lại đề đi em!

Ta có: \(\left(-\dfrac{5}{8}\right)-x:3\dfrac{5}{6}+7\dfrac{3}{4}=-2\)

\(\Leftrightarrow\dfrac{-5}{8}-x:\dfrac{23}{6}+\dfrac{31}{4}+2=0\)

\(\Leftrightarrow x:\dfrac{-23}{6}+\dfrac{73}{8}=0\)

\(\Leftrightarrow x\cdot\dfrac{-6}{23}=-\dfrac{73}{8}\)

\(\Leftrightarrow x=-\dfrac{73}{8}:\dfrac{-6}{23}=\dfrac{-73}{8}\cdot\dfrac{23}{-6}\)

hay \(x=\dfrac{1679}{48}\)

Vậy: \(x=\dfrac{1679}{48}\)

kết quả bằng 1679/48 nhé bạn 

Ta có: \(2^{-1}+\left(5^2\right)^3\cdot5^{-6}+32-2\cdot\left(-3\right)^2\cdot\dfrac{1}{9}\)

\(=\dfrac{1}{2}+5^6\cdot5^{-6}+32-2\cdot9\cdot\dfrac{1}{9}\)

\(=\dfrac{65}{2}-2+5^0\)

\(=\dfrac{61}{2}+1=\dfrac{63}{2}\)

Ta có: \(\left|97\dfrac{2}{3}-123\dfrac{3}{5}+97\dfrac{2}{5}-125\dfrac{1}{3}\right|\)

\(=\left|97\left(\dfrac{2}{3}+\dfrac{2}{5}\right)-125\cdot\left(\dfrac{3}{5}+\dfrac{1}{3}\right)\right|\)

\(=\left|194\cdot\dfrac{8}{15}-125\cdot\dfrac{14}{15}\right|\)

\(=\left|\dfrac{-66}{5}\right|=\dfrac{66}{5}\)