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Ta có: \(\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Leftrightarrow100\cdot\dfrac{9}{10}-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Leftrightarrow\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=1\)
\(\Leftrightarrow\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)=2\)
\(\Leftrightarrow x=-\dfrac{81}{100}\)
Đề sai nhá bạn
Đề P = 1.2 + 2.3 + 3.4 + ...... + 19.20 (đúng)
Ta có ; P = 1.2 + 2.3 + 3.4 + ...... + 19.20
=> 3P = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ....... + 19.20.21
=> 3P = 19.20.21
=> P = 19.20.21 /3
=> P = 2660
\(\dfrac{1}{2}=1-\dfrac{1}{2}\)
\(\dfrac{2}{3}=1-\dfrac{1}{3}\)
\(\dfrac{3}{4}=1-\dfrac{1}{4}\)
\(\dfrac{4}{5}=1-\dfrac{1}{5}\)
\(\dfrac{1}{2}>\dfrac{1}{3}>\dfrac{1}{4}>\dfrac{1}{5}\)
=>\(-\dfrac{1}{2}< -\dfrac{1}{3}< -\dfrac{1}{4}< -\dfrac{1}{5}\)
=>\(1-\dfrac{1}{2}< 1-\dfrac{1}{3}< 1-\dfrac{1}{4}< 1-\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{1}{2}< \dfrac{2}{3}< \dfrac{3}{4}< \dfrac{4}{5}\)
=>\(-\dfrac{1}{2}>-\dfrac{2}{3}>-\dfrac{3}{4}>-\dfrac{4}{5}\)
Bài 1:
a) Ta có: \(A=-1.7\cdot2.3+1.7\cdot\left(-3.7\right)-1.7\cdot3-0.17:0.1\)
\(=1.7\cdot\left(-2.3\right)+1.7\cdot\left(-3.7\right)+1.7\cdot\left(-3\right)+1.7\cdot\left(-1\right)\)
\(=1.7\cdot\left(-2.3-3.7-3-1\right)\)
\(=-10\cdot1.7=-17\)
b) Ta có: \(B=2\dfrac{3}{4}\cdot\left(-0.4\right)-1\dfrac{2}{3}\cdot2.75+\left(-1.2\right):\dfrac{4}{11}\)
\(=\dfrac{11}{4}\cdot\left(-0.4\right)-\dfrac{5}{3}\cdot\dfrac{11}{4}+\left(-1.2\right)\cdot\dfrac{11}{4}\)
\(=\dfrac{11}{4}\left(-0.4-\dfrac{5}{3}-1.2\right)\)
\(=-\dfrac{539}{60}\)
c) Ta có: \(C=\dfrac{\left(2^3\cdot5\cdot7\right)\cdot\left(5^2\cdot7^3\right)}{\left(2\cdot5\cdot7^2\right)^2}\)
\(=\dfrac{2^3\cdot5^3\cdot7^4}{2^2\cdot5^2\cdot7^4}\)
\(=10\)
\(\text{Ta thấy:}\text{ }\)
\(2^5-2^4-2^3-2^2-2-2=0\)
\(\Rightarrow\text{ Giá trị của biểu thức trên là 0}\)