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Ta có: \(A=\dfrac{1}{101^2}+\dfrac{1}{102^2}+\dfrac{1}{103^2}+\dfrac{1}{104^2}+\dfrac{1}{105^2}\)
\(A>\dfrac{1}{100.101}+\dfrac{1}{101.102}+\dfrac{1}{102.103}+\dfrac{1}{103.104}+\dfrac{1}{104.105}\)\(A>\dfrac{1}{100}-\dfrac{1}{101}+\dfrac{1}{101}-\dfrac{1}{102}+\dfrac{1}{102}-\dfrac{1}{103}+\dfrac{1}{103}-\dfrac{1}{104}+\dfrac{1}{104}-\dfrac{1}{105}\)\(A>\dfrac{1}{100}-\dfrac{1}{105}\)
\(A>\dfrac{1}{2100}\)
Mà \(B=\dfrac{1}{2^2.3.5^2.7}\)=\(\dfrac{1}{2100}\)
=> \(A>B\)
Vậy \(A>B\)
\(A=\frac{1}{101^2}+\frac{1}{102^2}+\frac{1}{103^2}+\frac{1}{104^2}+\frac{1}{105^2}\)
\(A< \frac{1}{100\cdot101}+\frac{1}{101\cdot102}+\frac{1}{102\cdot103}+\frac{1}{103\cdot104}+\frac{1}{104\cdot105}\)
\(=\frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}+\frac{1}{102}-\frac{1}{103}+\frac{1}{103}-\frac{1}{104}+\frac{1}{104}-\frac{1}{105}\)
\(=\frac{1}{100}-\frac{1}{105}=\frac{1}{2100}=\frac{1}{2^2\cdot3\cdot5^2\cdot7}=B\)
Vậy \(A< B\)
2) -12:\(\left(-\dfrac{5}{6}\right)^2\)=\(-12:\dfrac{25}{36}=-12\cdot\dfrac{36}{25}=-\dfrac{432}{25}\)
s) \(-\dfrac{1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)=-\dfrac{1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)\)
= \(-\dfrac{1}{12}-\dfrac{55}{24}=-\dfrac{2}{24}-\dfrac{55}{24}=-\dfrac{57}{24}=-\dfrac{19}{8}\)
t) \(-1,75-\left(-\dfrac{1}{9}-2\dfrac{1}{18}\right)=-1,75-\left(-\dfrac{2}{18}-\dfrac{37}{18}\right)\)
= -1,75-(\(-\dfrac{13}{6}\)) = \(-\dfrac{7}{4}+\dfrac{13}{6}=\dfrac{5}{12}\)
c) \(\left(\sqrt{\dfrac{1}{9}}-0,5\right)^3+\dfrac{-1}{3}=\left(\dfrac{1}{3}-\dfrac{1}{2}\right)^3-\dfrac{1}{3}\)
= \(\left(-\dfrac{1}{6}\right)^3-\dfrac{1}{3}=\dfrac{-1}{216}-\dfrac{1}{3}=-\dfrac{73}{216}\)
d) \(\left(\dfrac{1}{2}-\sqrt{\dfrac{4}{25}}\right)^2-2\dfrac{1}{2}=\left(\dfrac{1}{2}-\dfrac{2}{5}\right)^2-\dfrac{5}{2}\)
= \(\left(\dfrac{1}{10}\right)^2-\dfrac{5}{2}=\dfrac{1}{100}-\dfrac{250}{100}=-\dfrac{249}{100}=-2,49\)
b) \(\dfrac{7}{15}-\dfrac{9}{19}\)\(-\dfrac{-8}{15}-\dfrac{10}{19}\)
=\(\left(\dfrac{7}{15}-\dfrac{8}{15}\right)\) \(-\left(\dfrac{9}{19}-\dfrac{10}{19}\right)\)
= \(-\dfrac{1}{15}\) - \(\left(-\dfrac{1}{19}\right)\)
\(=-\dfrac{1}{15}\) + \(\dfrac{1}{19}\)
= \(-\dfrac{4}{285}\)
c) \(1\dfrac{1}{3}\) \(\div\) \(\dfrac{4}{5}\) + 2\(\dfrac{2}{3}\) \(\div\)\(\dfrac{4}{5}\)
= \(\left(1\dfrac{1}{3}+2\dfrac{2}{3}\right)\) \(\div\dfrac{4}{5}\)
= \(\left[\left(1+2\right)+\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\right]\) \(\div\dfrac{4}{5}\)
= ( 3 + 1 ) \(\div\dfrac{4}{5}\)
= 4 \(\div\dfrac{4}{5}\)
= \(\dfrac{4.5}{4}\)
= 5
1,
\(\left(2x+1\right)^3=-0,001\\ \left(2x+1\right)^3=\left(-0.1\right)^3\\ \Leftrightarrow2x+1=-0.1\\ 2x=-1.1\\ x=-\dfrac{11}{10}:2\\ x=-\dfrac{11}{20}\\ Vậy...\)
2,
\(\left(2x-3\right)^4=\left(2x-3\right)^6\\ \Leftrightarrow\left(2x-3\right)^6-\left(2x-3\right)^4=0\\ \Leftrightarrow\left(2x-3\right)^4\cdot\left[\left(2x-3\right)^2-1\right]=0\\ \Rightarrow\left\{{}\begin{matrix}\left(2x-3\right)^4=0\\\left(2x-3\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x-3=0\\\left(2x-3\right)^2=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x=3\\2x-3=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\\ Vậyx\in\left\{\dfrac{3}{2};2\right\}\)
3, Làm tương tự câu 2
5,
\(9^x:3^x=3\\ \left(9:3\right)^x=3\\ 3^x=3\\ \Rightarrow x=1\\ Vậy...\)
6,
\(3^x+3^{x+3}=756\\ 3^x+3^x\cdot3^3\\ 3^x\cdot\left(1+27\right)=756\\ 3^x\cdot28=756\\ \Leftrightarrow3^x=27\\ 3^x=3^3\\ \Rightarrow x=3\\ vậy...\)
7,
\(5^{x+1}+6\cdot5^{x+1}=875\\ 5^{x+1}\cdot\left(1+6\right)=875\\ 5^{x+1}\cdot7=875\\ \Leftrightarrow5^{x+1}=125\\ \Leftrightarrow5^{x+1}=5^3\Leftrightarrow x+1=3\\ \Rightarrow x=2\\ Vậy...\)
9,
1.Tính
a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)
b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)
c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)
d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)
e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)
Bài 2
a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)
\(x=\dfrac{13}{49}\)
b.\(\left|x-1,5\right|=2\)
Xảy ra 2 trường hợp
TH1
\(x-1,5=2\)
\(x=3,5\)
TH2
\(x-1,5=-2\)
\(x=-0,5\)
Vậy \(x=3,5\) hoặc \(x=-0,5\) .
Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.
a, \(A=1+2+2^2+2^3+..........+2^{49}+2^{50}\)
\(\Leftrightarrow2A=2+2^2+..............+2^{50}+2^{51}\)
\(\Leftrightarrow2A-A=\left(2+2^2+.........+2^{51}\right)-\left(1+2+......+2^{50}\right)\)
\(\Leftrightarrow A=2^{51}-1\)
a) \(A=1+2+2^2+2^3+2^4+...+2^{49}+2^{50}\)
\(\Rightarrow2A=2\left(1+2+2^2+2^3+2^4+...+2^{49}+2^{50}\right)\)
\(2A=2+2^2+2^3+2^4+2^5+...+2^{50}+2^{51}\)
\(\Rightarrow2A-A=A=\left(2+2^2+2^3+2^4+2^5+...+2^{50}+2^{51}\right)-\left(1+2+2^2+2^3+2^4+...+2^{49}+2^{50}\right)\)
\(A=2^{51}-1\) vậy \(A=2^{51}-1\)
b) \(B=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+\left(\dfrac{1}{2}\right)^5+...+\left(\dfrac{1}{2}\right)^{99}+\left(\dfrac{1}{2}\right)^{100}\)
\(B=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+...+\dfrac{1}{2^{99}}+\dfrac{1}{2^{100}}\)
\(\Rightarrow2B=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+...+\dfrac{1}{2^{99}}+\dfrac{1}{2^{100}}\right)\)
\(2B=\dfrac{2}{2}+\dfrac{2}{2^2}+\dfrac{2}{2^3}+\dfrac{2}{2^4}+\dfrac{2}{2^5}+...+\dfrac{2}{2^{99}}+\dfrac{2}{2^{100}}\)
\(2B=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\)
\(\Rightarrow2B-B=B=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^{99}}+\dfrac{1}{2^{100}}\right)\)
\(B=1-\dfrac{1}{2^{100}}\) vậy \(B=1-\dfrac{1}{2^{100}}\)