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Câu 2:
b: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{n\left(n+1\right)}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(=1-\dfrac{1}{n+1}=\dfrac{n}{n+1}\)
c: \(\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{110}\)
\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-...+\dfrac{1}{10}-\dfrac{1}{11}\)
\(=\dfrac{1}{4}-\dfrac{1}{11}=\dfrac{7}{44}\)
b1
a) \(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{5}-\dfrac{1}{10}\)
\(=\dfrac{2}{10}-\dfrac{1}{10}\)
\(=\dfrac{1}{10}\)
b) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{1}-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
c) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\)
\(=\dfrac{1}{3}-\dfrac{1}{11}\)
\(=\dfrac{8}{33}\)
d) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(=\dfrac{1}{3}-\dfrac{1}{101}\)
\(=\dfrac{98}{303}\)
Bài 1:
a) x + \(\frac{1}{9}\) - \(\frac{3}{5}\) = \(\frac{3}{6}\) b) \(\frac{3}{4}\) - x + \(\frac{6}{11}\) = \(\frac{5}{6}\)
x + \(\frac{1}{9}\) = \(\frac{3}{6}\) + \(\frac{3}{5}\) \(\frac{3}{4}\) - x = \(\frac{5}{6}\) - \(\frac{6}{11}\)
x + \(\frac{1}{9}\) = \(\frac{11}{10}\) \(\frac{3}{4}\) - x = \(\frac{19}{66}\)
x = \(\frac{11}{10}\) - \(\frac{1}{9}\) x = \(\frac{19}{66}\) + \(\frac{3}{4}\)
x = \(\frac{89}{90}\) x = \(\frac{137}{132}\)
Bài 2
a) x : 13/16 = 5/8 b)x - 14/28 = 6/9 + 8/25
x = 5/8 * 13/16 x - 14/28 = 74/75
x = 65/128 x = 74/75 + 14/28
x = 223/150
Bài 3
a)62/7 * x = 29/9 : 3/56 b)1/5 : x = 1/5 + 1/7
62/7 * x = 1624/27 1/5 : x = 12/35
x = 1624/27 : 62/7 x = 12/35 * 1/5
x = 5684/637 x = 12/175
,mai thi rùi mà đề khó quá
a) \(3.5^2-16:2^3.2\)
\(=3.25-16:8.2\)
\(=75-2.2\)
\(=75-4\)
\(=71\)
b) \(168+\left\{\left[2\left(2^4+3^2\right)-256^0\right]:7^2\right\}\)
\(=168+\left\{\left[2\left(16+9\right)-256^0\right]:7^2\right\}\)
\(=168+\left[\left(2.25-256^0\right):7^2\right]\)
\(=168+\left[\left(50-1\right):7^2\right]\)
\(=168+\left(49:7^2\right)\)
\(=168+\left(49:49\right)\)
\(=168+1\)
\(=169\)
c) \(9^{20}:9^{18}-\left(4^2-7\right)^2+8.5^2+5600:\left(3^3+1^8\right)\)
\(=9^{20}:9^{18}-\left(16-7\right)^2+8.5^2+5600:\left(27+1\right)\)
\(=9^{20}:9^{18}-9^2+8.5^2+5600:28\)
\(=9^{20-18}-9^2+8.25+5600:28\)
\(=9^2-9^2+200+200\)
\(=81-81+200+200\)
\(=200+200\)
\(=400\)
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các bạn làm rồi cho mik xem thử nhá tại mik cũng đang ôn mí dạng này
a; - \(\dfrac{10}{13}\) + \(\dfrac{5}{17}\) - \(\dfrac{3}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{11}{20}\)
= - (\(\dfrac{10}{13}\) + \(\dfrac{3}{13}\)) + (\(\dfrac{5}{17}\) + \(\dfrac{12}{17}\)) - \(\dfrac{11}{20}\)
= - 1 + 1 - \(\dfrac{11}{20}\)
= 0 - \(\dfrac{11}{20}\)
= - \(\dfrac{11}{20}\)
b; \(\dfrac{3}{4}\) + \(\dfrac{-5}{6}\) - \(\dfrac{11}{-12}\)
= \(\dfrac{9}{12}\) - \(\dfrac{10}{12}\) + \(\dfrac{11}{12}\)
= \(\dfrac{10}{12}\)
= \(\dfrac{5}{6}\)
c; [13.\(\dfrac{4}{9}\) + 2.\(\dfrac{1}{9}\)] - 3.\(\dfrac{4}{9}\)
= [\(\dfrac{52}{9}\) + \(\dfrac{2}{9}\)] - \(\dfrac{4}{3}\)
= \(\dfrac{54}{9}\) - \(\dfrac{4}{3}\)
= \(\dfrac{14}{3}\)
Câu 1:
a)\(\dfrac{12}{25}-\dfrac{7}{25}=\dfrac{5}{25}=\dfrac{1}{5}\)
b)\(\left(-\dfrac{1}{4}+\dfrac{5}{6}\right):\dfrac{2}{3}-\dfrac{4}{5}=\left(\dfrac{-6}{24}+\dfrac{20}{24}\right)\cdot\dfrac{3}{2}-\dfrac{4}{5}\)\(=\dfrac{14}{24}\cdot\dfrac{3}{2}-\dfrac{4}{5}=\dfrac{2\cdot7\cdot3}{3\cdot8\cdot2}-\dfrac{4}{5}=\dfrac{7}{8}-\dfrac{4}{5}=\dfrac{35-32}{40}=\dfrac{3}{40}\)
c)\(\dfrac{2}{9}\cdot\dfrac{6}{7}+\dfrac{2}{7}\cdot\dfrac{1}{9}-\dfrac{2}{9}=\dfrac{2}{9}\cdot\dfrac{6}{7}+\dfrac{2}{9}\cdot\dfrac{1}{7}-\dfrac{2}{9}=\dfrac{2}{9}\cdot\left(\dfrac{6}{7}+\dfrac{1}{7}-1\right)=\dfrac{2}{9}\cdot0=0\)
Câu 2:
a)\(\dfrac{1}{2}x=2\Leftrightarrow x=2\cdot2\Leftrightarrow x=4\)
Vậy tập nghiệm của phương trình là S={4}
b)\(x+\dfrac{2}{3}=\dfrac{6}{5}-\dfrac{1}{5}\Leftrightarrow x+\dfrac{2}{3}=1\Leftrightarrow x=1-\dfrac{2}{3}\Leftrightarrow x=\dfrac{1}{3}\)
Vậy tập nghiệm của phương trình là S={\(\dfrac{1}{3}\)}
c)\(\left(2,8x-23\right):\dfrac{2}{3}=-90\Leftrightarrow2,8x-23=-90\cdot\dfrac{2}{3}\Leftrightarrow2,8x-23=-60\Leftrightarrow2,8x=-60+23\Leftrightarrow2,8x=-37\Leftrightarrow x=-37:2,8\Leftrightarrow x=-\dfrac{185}{14}\)
Vậy tập nghiệm của phương trình là S={\(-\dfrac{185}{14}\)}
d)\(\left(2x-1\right)^2=9\Leftrightarrow2x-1=_-^+3\)
+)2x-1=3
<=>2x=4
<=>x=2
+)2x-1=-3
<=>2x=-2
<=>x=-1
Vậy tập nghiệm của phương trình là S={-1;2}
Câu 3:
a)\(\dfrac{1}{2}\cdot\left(\dfrac{1}{6}+\dfrac{1}{2}\right)\le x\le\left(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{1}{3}\right):\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{1}{2}\cdot\dfrac{2}{3}\le x\le\dfrac{11}{12}\cdot\dfrac{4}{3}\)
\(\Leftrightarrow\dfrac{1}{3}\le x\le\dfrac{11}{9}\)
\(\Leftrightarrow\dfrac{3}{9}\le x\le\dfrac{11}{9}\)
Do x nguyên => x=\(\dfrac{9}{9}=1\)
Vậy x=1
b)\(\dfrac{a}{b}=\dfrac{18}{27}=\dfrac{2}{3}\)
ƯCLN(a;b)={13}
=> \(\dfrac{a}{b}=\dfrac{2.13}{3.13}=\dfrac{26}{39}\)
Vậy phân số cần tìm là \(\dfrac{26}{39}\)
c)Ta có:
\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}>\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{4}=1\)
\(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{8}>4\cdot\dfrac{1}{8}=\dfrac{1}{2}\)
\(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{16}>8\cdot\dfrac{1}{16}=\dfrac{1}{2}\)
\(\dfrac{1}{17}+\dfrac{1}{18}+...+\dfrac{1}{32}>16\cdot\dfrac{1}{32}=\dfrac{1}{2}\)
\(\dfrac{1}{33}+\dfrac{1}{34}+...+\dfrac{1}{64}>32\cdot\dfrac{1}{64}=\dfrac{1}{2}\)
=>\(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{64}>1+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=3\)(đpcm)\(\)