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a) \(A=\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.10}+\dfrac{1}{143}\)
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)+\dfrac{1}{143}\)
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{100}\right)+\dfrac{1}{143}=\dfrac{1}{2}.\dfrac{99}{100}+\dfrac{1}{143}=\dfrac{99}{200}+\dfrac{1}{143}=\dfrac{99.143+200.1}{200.143}=\dfrac{14157+200}{28600}=\dfrac{14357}{28600}\)
b) \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+99\right)=14950\)
\(\Rightarrow x+x+...+x+\left(1+2+...+99\right)=14950\)
\(\Rightarrow100x+\left(\left(99+1\right):2\right).99:2=14950\)
\(\Rightarrow100x+2475=14950\Rightarrow100x=12475\Rightarrow x=\dfrac{12475}{100}=\dfrac{499}{4}\)
a: \(\Leftrightarrow n+2\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{-1;-3;3;-7\right\}\)
b: \(\Leftrightarrow n-1-2⋮n-1\)
\(\Leftrightarrow n-1\in\left\{1;-1;2;-2\right\}\)
hay \(n\in\left\{2;0;3;-1\right\}\)
c: \(\Leftrightarrow3n-6+8⋮n-2\)
\(\Leftrightarrow n-2\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)
hay \(n\in\left\{3;1;4;0;6;-2;10;-6\right\}\)
a) (−27). 1011 − 27. (−12) + 27. (−1)
= (-27). (1011 - 12 + 1)
= (-27). 1000 = -27 000
b) (−9). (−9). (−9) + 103 + 9.3
= (-9). 3 + 103 + 9.3
= 9.3 - 9.3 + 103
= 0 + 103 = 103
c) (−157). (127 − 316) − 127. (316 − 157)
= (-157). 127 - 316. (-157) - 127.316 - 127. (-157)
= (-157). 127 + 316. 157 - 127. 316 + 127.157
= (127.157 - 157.127) + (316. 157 - 127.316)
= 0 + (157 - 127). 316
= 0 + 30. 316
= 9480
Ta có:
\(3.9^{x+1}+6.9^{x-1}=249\\\Leftrightarrow3.9.9^x+6.9^x:9=249 \\ \Leftrightarrow27.9^x+\dfrac{2}{3}.9^x=249\\ \Leftrightarrow\left(27+\dfrac{2}{3}\right)9^x=249\\ \Leftrightarrow\dfrac{83}{3}.9^x=249\\ \Leftrightarrow9^x=249:\dfrac{83}{3}=9\\ \Rightarrow x=1\)
Đs...