1,
\(S=-\dfrac{7}{20}-\dfrac{7}{200}-\dfrac{7}{2000}-\dfrac{7}{20000}\\ =-\dfrac{7}{20}\left(1+\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{1000}\right)\\ =-\dfrac{7}{20}\left(\dfrac{1000+100+10+1}{1000}\right)\\ =-\dfrac{7}{20}\cdot\dfrac{1111}{1000}\\ =\dfrac{7777}{20000}\)
2,
a, \(Tacó:\\ 9^{2000}=\left(3^2\right)^{2000}=3^{4000}\\ \Rightarrow9^{2000}=3^{4000}\)
b,
\(2^{225}=\left(2^{15}\right)^{15}=32768^{15}\\ 3^{150}=\left(3^{10}\right)^{15}=59049^{15}\\ Vì32768< 59049nên32768^{15}< 59049^{15}\\ \Rightarrow2^{225}< 3^{150}\)
3,
\(\left|x-7\right|=x-7\\ Vì\left|x-7\right|\ge0\forall x\\ \Rightarrow x-7\ge0\forall x\\ \Leftrightarrow x-7\ge0\\ \Leftrightarrow x\ge7\\ Vậyx\ge7\)
Bài 3:
\(\left|x-7\right|=x-7\)
Khi giá trị tuyệt đối của \(x-7\) bằng chính nó, thì \(x-7\) phải \(\ge0\)
Suy ra: \(x-7\ge0\Rightarrow x\ge7\)
Vậy \(x\ge 7\)
