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c)
Ta có :\(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)
\(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{\dfrac{8}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{3}{8}}\) \(=2+\dfrac{1}{\dfrac{11}{8}}\) \(=2+\dfrac{8}{11}\) \(=\dfrac{30}{11}\)
d) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\dfrac{1}{4}:2\)
\(=3-1+\dfrac{1}{8}\)
\(=\dfrac{17}{8}\)
1:
a: =7/5(40+1/4-25-1/4)-1/2021
=21-1/2021=42440/2021
b: =5/9*9-1*16/25=5-16/25=109/25
1: \(\dfrac{11}{24}-\dfrac{5}{41}+\dfrac{13}{24}+0,5-\dfrac{36}{41}\)
\(=\left(\dfrac{11}{24}+\dfrac{13}{24}\right)-\left(\dfrac{5}{41}+\dfrac{36}{41}\right)+\dfrac{1}{2}\)
\(=1-1+\dfrac{1}{2}=\dfrac{1}{2}\)
2: \(12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)
\(=12:\left(\dfrac{9}{12}-\dfrac{10}{12}\right)^2\)
\(=12:\left(-\dfrac{1}{12}\right)^2=12:\dfrac{1}{144}=12\cdot144=1368\)
3: \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\cdot\left(0,8-\dfrac{3}{4}\right)^2\)
\(=\dfrac{12+8-3}{12}\cdot\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)
\(=\dfrac{17}{12}\cdot\left(\dfrac{16-15}{20}\right)^2\)
\(=\dfrac{17}{12}\cdot\dfrac{1}{400}=\dfrac{17}{4800}\)
4: \(16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)+28\dfrac{2}{7}:\dfrac{3}{5}\)
\(=\dfrac{5}{3}\cdot\left(-16-\dfrac{2}{7}\right)+\dfrac{5}{3}\cdot\left(28+\dfrac{2}{7}\right)\)
\(=\dfrac{5}{3}\left(-16-\dfrac{2}{7}+28+\dfrac{2}{7}\right)\)
\(=12\cdot\dfrac{5}{3}=20\)
5: \(\left(2^2:\dfrac{4}{3}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)
\(=\left(4\cdot\dfrac{3}{4}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)
\(=\dfrac{5}{2}\cdot\dfrac{6}{5}-17=3-17=-14\)
6: \(\left(\dfrac{1}{3}\right)^{50}\cdot\left(-9\right)^{25}-\dfrac{2}{3}:4\)
\(=\left(\dfrac{1}{3}\right)^{50}\cdot\left(-1\right)\cdot3^{50}-\dfrac{2}{3\cdot4}\)
\(=-1-\dfrac{2}{12}=-1-\dfrac{1}{6}=-\dfrac{7}{6}\)
3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
\(a,0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{5}+\dfrac{5}{7}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\dfrac{5}{6}+\dfrac{39}{35}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\left(\dfrac{5}{6}-\dfrac{1}{6}\right)+\left(\dfrac{39}{35}-\dfrac{4}{35}\right)\\ =\dfrac{2}{3}+1\\ =\dfrac{4}{3}.\)
\(b,\left(3-\dfrac{1}{4}+\dfrac{2}{3}\right)-\left(5+\dfrac{1}{3}-\dfrac{6}{5}\right)-\left(-6-\dfrac{7}{4}+\dfrac{3}{2}\right)\\ =3-\dfrac{1}{4}+\dfrac{2}{3}-5-\dfrac{1}{3}+\dfrac{6}{5}+6+\dfrac{7}{4}-\dfrac{3}{2}\\ =\left(3-5+6\right)+\left(-\dfrac{1}{4}+\dfrac{7}{4}\right)+\left(\dfrac{2}{3}-\dfrac{1}{3}\right)+\left(\dfrac{6}{5}+\dfrac{7}{4}\right)\\ =4-\dfrac{3}{2}+\dfrac{1}{3}+\dfrac{59}{20}\\ =\dfrac{5}{2}+\dfrac{1}{3}+\dfrac{59}{20}\\ =\dfrac{17}{6}+\dfrac{59}{20}\\ =\dfrac{347}{60}.\)
\(c,\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\\ =\dfrac{1}{3}+\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\\ =\left(\dfrac{1}{3}-\dfrac{2}{9}\right)+\left(\dfrac{3}{4}-\dfrac{1}{36}\right)+\left(\dfrac{3}{5}+\dfrac{1}{15}\right)+\dfrac{1}{64}\\ =\dfrac{1}{9}+\dfrac{13}{18}+\dfrac{2}{3}+\dfrac{1}{64}\\ =\dfrac{3}{2}+\dfrac{1}{64}\\ =\dfrac{65}{64}.\)
a/ \(\dfrac{\left(1+2+.....+100\right)\left(\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(6,3.12-21.36\right)}{\dfrac{1}{2}+\dfrac{1}{3}+.......+\dfrac{1}{100}}\)
\(=\dfrac{\left(1+2+3+.....+100\right)\left(\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}-\dfrac{1}{9}\right).0}{\dfrac{1}{2}+\dfrac{1}{3}+.......+\dfrac{1}{100}}\)
\(=\dfrac{0}{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{100}}\)
\(=0\)
e: \(=\dfrac{5^{30}\cdot3^{20}}{3^{15}\cdot5^{30}}=3^5=243\)
B = .................
Xét thừa số 63.1,2 - 21.3,6 = 0 nên B = 0
\(C=\left|\dfrac{4}{9}-\left(\dfrac{\sqrt{2}}{2}\right)^2\right|+\left|0,4+\dfrac{\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}}{\dfrac{2}{3}-\dfrac{4}{5}-\dfrac{6}{7}}\right|\)
\(C=\left|\dfrac{4}{9}-\dfrac{1}{2}\right|+\left|0,4+\dfrac{\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}}{2\left(\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}\right)}\right|\)
\(C=\left|\dfrac{4}{9}-\dfrac{1}{2}\right|+\left|0,4+\dfrac{1}{2}\right|=\dfrac{1}{18}+\dfrac{9}{10}=\dfrac{43}{45}\)
Mình làm câu 1,2 trước, câu 3 sau
Câu 1:
\(\sqrt{x^2}=0\)
=> \(\left(\sqrt{x^2}\right)^2=0^2\)
\(\Leftrightarrow x^2=0\Leftrightarrow x=0\)
Câu 2:
\(A=\left(0,75-0,6+\dfrac{3}{7}+\dfrac{3}{12}\right)\left(\dfrac{11}{7}+\dfrac{11}{3}+2,75-2,2\right)\)
\(A=\left(\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}\right)\left(\dfrac{11}{7}+\dfrac{11}{3}+\dfrac{11}{4}-\dfrac{11}{5}\right)\)
\(A=3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)\cdot11\left(\dfrac{1}{7}+\dfrac{1}{3}+\dfrac{11}{4}-\dfrac{11}{5}\right)\)
\(A=33\cdot\dfrac{491}{1820}\cdot\dfrac{221}{420}=\dfrac{3580863}{764400}\)
\(1,A=\dfrac{1}{21}\\ 2,B=\dfrac{101}{200}\\ 3,a\in\left\{-14;-13;-12;-11;-10\right\}\\ 4,D=\dfrac{48}{7}\\ 5,E=-\dfrac{1}{3}\\ 6,F=2-\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)
Câu 8:
Ta có: \(A=2+2^2+2^3+2^4+...+2^{50}\)
\(\Leftrightarrow2\cdot A=2^2+2^3+...+2^{51}\)
\(\Leftrightarrow A=2^{51}-2\)