em xin giải chi tiết bài này em cảm ơn ạaaTT
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\(a,\dfrac{x}{9}=\dfrac{5}{3}\\ \Leftrightarrow x=9\cdot\dfrac{5}{3}\\ \Leftrightarrow x=15\\ b,\dfrac{17}{x}=\dfrac{85}{105}\\ \Leftrightarrow x=17\cdot\dfrac{105}{85}\\ \Leftrightarrow x=21\\ c,\dfrac{x}{8}+\dfrac{2}{3}=\dfrac{7}{6}\\ \Leftrightarrow\dfrac{x}{8}=\dfrac{1}{2}\\ \Leftrightarrow x=4\\ d,\dfrac{3}{x-7}=\dfrac{27}{135}\\ \Leftrightarrow x-7=15\\ \Leftrightarrow x=22\)
\(e,\dfrac{75}{20-x}=\dfrac{3}{2}\times10\\ \Leftrightarrow\dfrac{75}{20-x}=15\\ \Leftrightarrow20-x=5\\ \Leftrightarrow x=15\\ f,\left(x-50\%\right)\times\dfrac{5}{3}=\dfrac{7}{4}-0,5\\ \Leftrightarrow\left(x-\dfrac{1}{2}\right)\times\dfrac{5}{3}=\dfrac{5}{4}\\ \Leftrightarrow x-\dfrac{1}{2}=\dfrac{3}{4}\\ \Leftrightarrow x=\dfrac{5}{4}\\ g,\left(\dfrac{2}{15}+\dfrac{3}{35}+\dfrac{2}{63}\right):x=\dfrac{1}{18}\\ \Leftrightarrow\dfrac{2}{9}:x=\dfrac{1}{18}\\ \Leftrightarrow x=4\)
\(h,\left[\left(x-\dfrac{1}{2}\right):6+4\right]\times\dfrac{2}{3}=0,6\times\dfrac{40}{6}\\ \Leftrightarrow\left[\left(x-\dfrac{1}{2}\right):6+4\right]\times\dfrac{2}{3}=4\\ \Leftrightarrow\left(x-\dfrac{1}{2}\right):6+4=6\\ \Leftrightarrow\left(x-\dfrac{1}{2}\right):6=2\\ \Leftrightarrow x-\dfrac{1}{2}=12\\ \Leftrightarrow x=\dfrac{25}{2}\)
\(\overline{2003ab}\) : 2;5 dư 1 ⇔ b = 1
\(\overline{2003ab}\) : 9 dư 1 ⇔ 2+0+0+3+a+b - 1⋮ 9
4 + a + 1 ⋮ 9
5 + a ⋮ 9 ⇒ a =4;
Thay a = 4; b = 1 vào biểu thức \(\overline{2003ab}\) ta có
\(\overline{2003ab}\) = 200341
Bài 4: Thực hiện phép tính:
a, \(\dfrac{3}{4}\) \(\times\) \(\dfrac{7}{9}\) + 0,25 \(\times\) \(\dfrac{7}{9}\) - \(\dfrac{7}{9}\)
= \(\dfrac{7}{9}\) (0,75 + 0,25 - 1)
= 0
b, \(\dfrac{5}{12}\) - \(\dfrac{1}{12}\): (\(\dfrac{3}{8}\) + \(\dfrac{1}{2}\).[\(\dfrac{9}{4}\) - 2.(19,5 -18,5)])
= \(\dfrac{5}{12}\) - \(\dfrac{1}{12}\): ( \(\dfrac{3}{8}\) + \(\dfrac{1}{2}\). [ \(\dfrac{9}{4}\)- 2])
= \(\dfrac{5}{12}\) - \(\dfrac{1}{12}\): ( \(\dfrac{3}{8}\) + \(\dfrac{9}{8}\)- 1)
= \(\dfrac{5}{12}\) - \(\dfrac{1}{12}\): ( \(\dfrac{1}{2}\))
= \(\dfrac{5}{12}\) - \(\dfrac{1}{6}\)
= \(\dfrac{1}{4}\)
c, [ 50,2 - 0,2 .(12,05 - 10,05)] - 37,38
= [ 50,2 - 0,2. 2 ] - 37,38
= [ 50,2 - 0,4] - 37,38
= 49,8 - 37,38
= 12,42
Bài 3:
B = \(\dfrac{1}{1.4}\)+\(\dfrac{1}{4.7}+\dfrac{1}{7.11}+...+\dfrac{1}{97.100}\)
B = \(\dfrac{1}{3}\).( \(\dfrac{3}{1.4}\)+ \(\dfrac{3}{4.7}\)+ \(\dfrac{3}{7.11}\)+...+ \(\dfrac{3}{97.100}\))
B = \(\dfrac{1}{3}\).( \(\dfrac{1}{1}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{7}\)+ \(\dfrac{1}{7}\) - \(\dfrac{1}{11}\) +...+ \(\dfrac{1}{97}\) - \(\dfrac{1}{100}\))
B = \(\dfrac{1}{3}\).( \(\dfrac{1}{1}\) - \(\dfrac{1}{100}\))
B = \(\dfrac{1}{3}\). \(\dfrac{99}{100}\)
B = \(\dfrac{33}{100}\)
\(\dfrac{1}{5}\sqrt[]{25x+50}-5\sqrt[]{x+2}+\sqrt[]{9x+18}+9=0\)
\(\Leftrightarrow\dfrac{1}{5}\sqrt[]{25\left(x+2\right)}-5\sqrt[]{x+2}+\sqrt[]{9\left(x+2\right)}+9=0\)
\(\Leftrightarrow\dfrac{1}{5}.5\sqrt[]{x+2}-5\sqrt[]{x+2}+3\sqrt[]{x+2}+9=0\)
\(\Leftrightarrow\sqrt[]{x+2}-5\sqrt[]{x+2}+3\sqrt[]{x+2}+9=0\)
\(\Leftrightarrow\sqrt[]{x+2}\left(1-5+3\right)+9=0\)
\(\Leftrightarrow-\sqrt[]{x+2}+9=0\)
\(\Leftrightarrow\sqrt[]{x+2}=9\)
\(\Leftrightarrow x+2=81\)
\(\Leftrightarrow x=79\)
\(a,A=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{99\times100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}\\ =\dfrac{99}{100}\\ b,B=\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+...+\dfrac{1}{97\times100}\\ =\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\\ =\dfrac{1}{3}\cdot\left(1-\dfrac{1}{100}\right)\\ =\dfrac{1}{3}\cdot\dfrac{99}{100}\\ =\dfrac{33}{100}\)
\(c,C=\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{128}\\ =\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^7}\\ \Rightarrow2C=1+\dfrac{1}{2}+...+\dfrac{1}{2^6}\\ \Rightarrow2C-C=1-\dfrac{1}{2^7}\\ \Rightarrow C=\dfrac{127}{128}\)
\(d,D=\dfrac{2}{1\times2}+\dfrac{2}{2\times3}+...+\dfrac{2}{99\times100}\\ =2\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =2\cdot\left(1-\dfrac{1}{100}\right)\\ =\dfrac{99}{55}\)
\(e,E=\dfrac{1}{10}+\dfrac{1}{40}+...+\dfrac{1}{340}\\ =\dfrac{1}{2\times5}+\dfrac{1}{5\times8}+...+\dfrac{1}{17\times20}\\ =\dfrac{1}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{17}-\dfrac{1}{20}\right)\\ =\dfrac{1}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{20}\right)\\ =\dfrac{1}{3}\cdot\dfrac{9}{20}\\ =\dfrac{3}{20}\)