Question

 

giúp mik vs

Answer 1

Bài 11:

\(A=\dfrac{1}{7}+\dfrac{1}{7^2}+\dfrac{1}{7^3}+...+\dfrac{1}{7^{100}}\)

\(7A=7\cdot\left(\dfrac{1}{7}+\dfrac{1}{7^2}+...+\dfrac{1}{7^{100}}\right)\)

\(7A=1+\dfrac{1}{7}+\dfrac{1}{7^2}+...+\dfrac{1}{7^{99}}\)

\(7A-A=1+\dfrac{1}{7}+\dfrac{1}{7^2}+...+\dfrac{1}{7^{99}}-\dfrac{1}{7}-...-\dfrac{1}{7^{100}}\)

\(6A=1-\dfrac{1}{7^{100}}\)

\(A=\dfrac{1-\dfrac{1}{7^{100}}}{6}\)

\(A=\dfrac{1}{6}-\dfrac{1}{6\cdot7^{100}}\)

Answer 2

Bài 8:

a: ĐKXĐ: x>0

Để A là số nguyên thì \(7⋮\sqrt{x}\)

=>\(\sqrt{x}\in\left\{1;7\right\}\)

=>\(x\in\left\{1;49\right\}\)

b: ĐKXĐ: x>=0 và x<>1

Để B là số nguyên thì \(1+\dfrac{3}{\sqrt{x}-1}\in Z\)

=>\(3⋮\sqrt{x}-1\)

=>\(\sqrt{x}-1\in\left\{1;-1;3;-3\right\}\)

=>\(\sqrt{x}\in\left\{2;0;4;-2\right\}\)

=>\(\sqrt{x}\in\left\{2;0;4\right\}\)

=>\(x\in\left\{4;0;16\right\}\)

c: ĐKXĐ: x>=0 và x<>9

Để C là số nguyên thì \(2+\dfrac{\sqrt{x}-1}{\sqrt{x}-3}\in Z\)

=>\(\dfrac{\sqrt{x}-1}{\sqrt{x}-3}\in Z\)

=>\(\sqrt{x}-1⋮\sqrt{x}-3\)

=>\(\sqrt{x}-3+2⋮\sqrt{x}-3\)

=>\(\sqrt{x}-3\inƯ\left(2\right)\)

=>\(\sqrt{x}-3\in\left\{1;-1;2;-2\right\}\)

=>\(\sqrt{x}\in\left\{4;2;5;1\right\}\)

=>\(x\in\left\{16;4;25;1\right\}\)

Bài 9:

\(\left|x+\dfrac{3}{5}\right|-\left|x-\dfrac{7}{3}\right|=0\)

=>\(\left|x+\dfrac{3}{5}\right|=\left|x-\dfrac{7}{3}\right|\)

=>\(\left[{}\begin{matrix}x+\dfrac{3}{5}=x-\dfrac{7}{3}\\x+\dfrac{3}{5}=-x+\dfrac{7}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{5}=-\dfrac{7}{3}\left(vôlý\right)\\x+\dfrac{3}{5}=-x+\dfrac{7}{3}\end{matrix}\right.\)

=>\(x+\dfrac{3}{5}=-x+\dfrac{7}{3}\)

=>\(2x=\dfrac{7}{3}-\dfrac{3}{5}=\dfrac{35-9}{15}=\dfrac{24}{15}=\dfrac{8}{5}\)

=>x=4/5