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\(P=\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)...\left(1-\dfrac{1}{1225}\right)\left(1-\dfrac{1}{1275}\right)\)
\(=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}...\dfrac{1224}{1225}.\dfrac{1274}{1275}\)
\(=\dfrac{2.2}{3.2}.\dfrac{5.2}{6.2}.\dfrac{9.2}{10.2}...\dfrac{1224.2}{1225.2}.\dfrac{1274.2}{1275.2}\)
\(=\dfrac{4}{9}.\dfrac{10}{12}.\dfrac{18}{20}...\dfrac{2448}{2450}.\dfrac{2548}{2550}\)
\(=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}...\dfrac{48.51}{49.50}.\dfrac{49.52}{50.51}\)
\(=\dfrac{1.2.3...48.49}{2.3.4...49.50}.\dfrac{4.5.6...51.52}{3.4.5...50.51}\)
\(=\dfrac{1}{50}.\dfrac{52}{3}\)
\(=\dfrac{26}{75}\).
a: =>2x-x=-5/2-1/3
=>x=-17/6
b: =>4(x-2)2=36
=>(x-2)2=9
=>x-2=3 hoặc x-2=-3
hay x=5 hoặc x=-1
c: =>2x+1/2=5/6
=>2x=1/3
hay x=1/6
1) 1 + 2 + 3 + ... + x = 1225
=> (1 + x).x:2 = 1225
=> (1 + x).x = 1225.2
=> (1 + x).x = 2450
=> (1 + x).x = 50.49 = (-49).(-50)
Vậy \(x\in\left\{50;-49\right\}\)
2) 2 + 4 + 6 + ... + 2x = 210
=> 2.(1 + 2 + 3 + ... + x) = 210
=> 2.(1 + x).x:2 = 210
=> (1 + x).x = 15.14 = (-14).(-15)
Vậy \(x\in\left\{15;-14\right\}\)
a)
\(1+2+......+x=1225\)
\(\Rightarrow\frac{\left(x+1\right)x}{2}=1225\)
\(\Rightarrow\left(x+1\right)x=2450\)
\(\Rightarrow\left(x+1\right)x=49.50\)
=> x = 49
Vậy x = 49
b)
\(2+4+....+2x=210\)
\(\Rightarrow2\left(1+2+....+x\right)=210\)
\(\Rightarrow2.\frac{\left(x+1\right)x}{2}=210\)
\(\Rightarrow x\left(x+1\right)=210\)
\(\Rightarrow x\left(x+1\right)=210\)
\(\Rightarrow x\left(x+1\right)=14.15\)
=> x = 14
Vậy x = 14
(x - 1) + (x - 2) + (x - 3) + ⋯ + (x - 50) = 1225
x-1+x-2+x-3+....+x-50=1225
50x-(1+2+3+...+50)=1225
50x-1275=1225
50x=1225+1275
50x=2500
x=2500:50
x=50