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Ta có : A = \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{9.9}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)
\(\Rightarrow A< \frac{8}{9}\)(1)
Lại có : \(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{9.9}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
\(\Rightarrow A>\frac{2}{5}\)(2)
Từ (1);(2) => \(\frac{8}{9}>A>\frac{2}{5}\)
Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mn hiểu đề của bạn hơn.
a) \(A=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right).\left(1-\dfrac{1}{5}\right)...\left(1-\dfrac{1}{2003}\right).\left(1-\dfrac{1}{2004}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}...\dfrac{2002}{2003}.\dfrac{2003}{2004}\)
\(=\dfrac{1}{2004}\)
b) \(B=5\dfrac{9}{10}:\dfrac{3}{2}-\left(2\dfrac{1}{3}.4\dfrac{1}{2}-2.2\dfrac{1}{3}\right):\dfrac{7}{4}\)
\(=\dfrac{59}{10}:\dfrac{3}{2}-\left(\dfrac{7}{3}.\dfrac{9}{2}-2.\dfrac{7}{3}\right).\dfrac{4}{7}\)
\(=\dfrac{59}{15}-\left(\dfrac{21}{2}-\dfrac{14}{3}\right).\dfrac{4}{7}\)
\(=\dfrac{59}{15}-\dfrac{35}{6}.\dfrac{4}{7}\)
\(=\dfrac{59}{15}-\dfrac{10}{3}\)
\(=\dfrac{3}{5}\)
Dễ thế mà không làm được thì bạn nên xem lại nhé,một hai câu thì còn được chứ cả 10 câu thế kia rõ là ỷ lại rồi bạn ạ.Thân!
a, \(\Leftrightarrow2x^2=72\)
\(\Leftrightarrow x^2=36\)
\(\Leftrightarrow x=\pm6\)
Vậy ...
\(b,\Leftrightarrow\dfrac{3}{5}x-0,75=2\dfrac{4}{5}.\dfrac{3}{7}=\dfrac{6}{5}\)
\(\Leftrightarrow\dfrac{3}{5}x=\dfrac{6}{5}+0,75=\dfrac{39}{20}\)
\(\Leftrightarrow x=\dfrac{39}{20}:\dfrac{3}{5}=\dfrac{13}{4}\)
Vậy ...
\(c,\Leftrightarrow2x=1\dfrac{5}{6}.\dfrac{6}{11}-\dfrac{3}{10}=\dfrac{7}{10}\)
\(\Leftrightarrow x=\dfrac{7}{10}:2=\dfrac{7}{20}\)
Vậy ...
\(d,\Leftrightarrow\dfrac{1}{x-7\dfrac{1}{3}}=1.5:2\dfrac{1}{4}=\dfrac{2}{3}\)
\(\Leftrightarrow x-7\dfrac{1}{3}=\dfrac{3}{2}\)
\(\Leftrightarrow x=\dfrac{3}{2}+7\dfrac{1}{3}=\dfrac{53}{6}\)
Vậy ...
a) 2x2 - 72 = 0
\(\Rightarrow\) 2x2 = 72
\(\Rightarrow\) x2 = 36 = 62 = (- 6)2
\(\Rightarrow\) x = 6 hoặc x = - 6
Vậy x = 6 hoặc x = - 6
b) (\(\dfrac{3}{5}\)x - 0,75) : \(\dfrac{3}{7}\) = \(2\dfrac{4}{5}\)
\(\Rightarrow\) (\(\dfrac{3}{5}\)x - 0,75) : \(\dfrac{3}{7}\) = \(\dfrac{14}{5}\)
\(\Rightarrow\) \(\dfrac{3}{5}\)x - \(\dfrac{3}{4}\) = \(\dfrac{6}{5}\)
\(\Rightarrow\) \(\dfrac{3}{5}\)x = \(\dfrac{39}{20}\)
\(\Rightarrow\) x = \(\dfrac{13}{4}\)
Vậy x = \(\dfrac{13}{4}\)
B= 1/2 x 2/3 x 3/4 x ...........x 2002/2003 x 2003/2004
1 x 2 x 3 x 4 x .............x 2002 x 2003
2 x 3 x 4 x .............x 2003 x 2004
1
2004
b: Ta có: \(1992+\left(-53\right)+158+\left(-247\right)+\left(-1592\right)\)
\(=\left(1992-1592\right)+\left(-53-247\right)+158\)
\(=400-300+158=258\)
a)
\(\left(1-\dfrac{1}{5}\right)x\left(1-\dfrac{2}{5}\right)x...x\left(1-\dfrac{9}{5}\right)\\ =\left(1-\dfrac{1}{5}\right)x...x\left(1-\dfrac{5}{5}\right)x...x\left(1-\dfrac{9}{5}\right)\\ =\left(1-\dfrac{1}{5}\right)x...x0x...x\left(1-\dfrac{9}{5}\right)=0\)
x là nhân nhé :))
b)
\(\dfrac{1}{2}x\dfrac{2}{3}x...x\dfrac{9}{10}\\ =\dfrac{1x2x...x9}{2x3x...x10}=\dfrac{2x3x...x9}{2x3x...x9x10}=\dfrac{1}{10}\)