Tính nhanh: \(\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{96}+\dfrac{2}{192}\)
A = \(\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{96}\)
Tính tổng sau đây bằng cách nhanh nhất
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{96}+\dfrac{1}{192}\)
GIÚP MÌNH VỚI NHÉ MÌNH ĐAG CẦN GẤP.CHO MÌNH GỬI LỜI CẢM ƠN TRC NHÉ!
=\(\dfrac{64}{192}+\dfrac{32}{192}+\dfrac{16}{192}+\dfrac{8}{192}+\dfrac{4}{192}+\dfrac{2}{192}+\dfrac{1}{192}\)
= \(\dfrac{127}{192}\)
Tính nhanh:
a) A= 1 + \(\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}+\dfrac{1}{625}+...+\dfrac{1}{78125}\)
b) B= \(\dfrac{1}{3}+\dfrac{1}{12}+\dfrac{1}{48}+\dfrac{1}{192}+\dfrac{1}{768}+...+\dfrac{1}{36864}\)
c) M= \(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{9900}\)
d) P= \(\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+3+4+...+2018}\)
Bài 1. Tính nhanh giá trị của biểu thức sau:
a, E = \(\dfrac{1}{2}\)+ \(\dfrac{1}{3}\)+ \(\dfrac{1}{6}\)+ \(\dfrac{1}{24}\)+ \(\dfrac{1}{8}\)+ \(\dfrac{1}{2}\)+\(\dfrac{1}{12}\)
\(E=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{8}+\dfrac{1}{2}+\dfrac{1}{12}\)
\(E=\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{1}{24}\right)\)
\(E=\dfrac{2}{2}+\dfrac{3}{6}+\left(\dfrac{1}{8}+\dfrac{3}{24}\right)\)
\(E=1+\dfrac{1}{2}+\left(\dfrac{1}{8}+\dfrac{1}{8}\right)\)
\(E=\left(\dfrac{2}{2}+\dfrac{1}{2}\right)+\dfrac{2}{8}\)
\(E=\dfrac{3}{2}+\dfrac{1}{4}\)
\(E=\dfrac{6}{4}+\dfrac{1}{4}\)
\(E=\dfrac{7}{4}\)
Tính: A= \(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{24}\)+\(\dfrac{1}{48}\)+\(\dfrac{1}{96}\)
A= 1/3+1/6+1/12+1/24+1/48+1/96
= (1/3+1/6)+(1/12+1/24)+(1/48+1/96)
= (2/6+1/6)+(2/24+1/24)+(2/96+1/96)
= 1/2+1/8+1/32
= 16/32+4/32+1/32
= 21/32
Vậy A=21/32
Giải:
A=1/3+1/6+1/12+1/24+1/48+1/96
A=1/3+(1/2.3+1/3.4)+(1/4.6+1/6.8)+1/96
A=1/3+(1/2-1/3+1/3-1/4)+[1/2.(2/4.6+2/6.8)]+1/96
A=1/3+(1/2-1/4)+[1/2.(1/4-1/6+1/6-1/8)]+1/96
A=1/3+1/4+[1/2.(1/4-1/8)]+1/96
A=1/3+1/4+[1/2.1/8]+1/96
A=1/3+1/4+1/16+1/96
A=7/12+7/96
A=21/32
Tính:
a. \(5\sqrt{2}-2\sqrt{48}+6\sqrt{75}-\sqrt{108}\)
b.\(2\sqrt{147}-\dfrac{3}{32}\sqrt{192}+\dfrac{4}{18}\sqrt{243}-\dfrac{1}{10}\sqrt{300}\)
c. \(-\dfrac{1}{2}\sqrt{108}+\dfrac{1}{15}\sqrt{75}-\dfrac{1}{22}\sqrt{363}+\sqrt{12}\)
d. \(\dfrac{5}{8}\sqrt{48}-\dfrac{1}{33}\sqrt{363}+\dfrac{3}{14}\sqrt{147}-\dfrac{1}{4}\sqrt{192}\)
e. \(\dfrac{3}{2}\sqrt{12}+\dfrac{7}{5}\sqrt{75}-\dfrac{9}{10}\sqrt{300}+\dfrac{11}{6}\sqrt{108}\)
a: \(5\sqrt{2}-8\sqrt{3}+30\sqrt{3}-6\sqrt{3}=5\sqrt{2}+16\sqrt{3}\)
b: \(=14\sqrt{3}-\dfrac{3}{32}\cdot8\sqrt{3}+\dfrac{4}{18}\cdot9\sqrt{3}-\dfrac{1}{10}\cdot10\sqrt{3}\)
\(=14\sqrt{3}-\dfrac{3}{4}\sqrt{3}+2\sqrt{3}-1\sqrt{3}=\dfrac{57}{4}\sqrt{3}\)
c: \(=\dfrac{-1}{2}\cdot6\sqrt{3}+\dfrac{1}{15}\cdot5\sqrt{3}-\dfrac{1}{22}\cdot11\sqrt{3}+2\sqrt{3}\)
\(=-3\sqrt{3}+\dfrac{1}{3}\sqrt{3}-\dfrac{1}{2}\sqrt{3}+2\sqrt{3}=-\dfrac{7}{6}\sqrt{3}\)
d: \(=\dfrac{5}{8}\cdot4\sqrt{3}-\dfrac{1}{33}\cdot11\sqrt{3}+\dfrac{3}{14}\cdot7\sqrt{3}-\dfrac{1}{4}\cdot8\sqrt{3}\)
\(=\dfrac{5}{2}\sqrt{3}-\dfrac{1}{3}\sqrt{3}+\dfrac{3}{2}\sqrt{3}-2\sqrt{3}=\dfrac{5}{3}\sqrt{3}\)
Thực hiện phép tính, tính nhanh nếu có thể
a, \(\dfrac{7}{13}\) + \(\dfrac{12}{13}\) + \(\dfrac{6}{-13}\)
b, ( \(\dfrac{4}{5}\) + \(\dfrac{1}{2}\) ) x ( \(\dfrac{6}{13}\) - 2 )
c, 75% : \(\dfrac{1}{2}\) - \(2\dfrac{1}{2}\) x ( \(-1\dfrac{1}{3}\) )
d, 0,375 x \(1\dfrac{3}{5}\) + 60% x \(\dfrac{2}{7}\) + \(\dfrac{3}{5}\) x \(\dfrac{5}{7}\)
e, \(\dfrac{-5}{6}\) x \(\dfrac{7}{13}\) + \(\dfrac{19}{13}\) : \(\dfrac{6}{-5}\) + \(\dfrac{2}{5}\)
f, 1,25 x \(\dfrac{7}{9}+\dfrac{5}{4}x\dfrac{15}{9}-1\dfrac{1}{4}:\dfrac{19}{3}\)
a: \(=\dfrac{7+12-6}{13}=1\)
b: \(=\dfrac{13}{10}\cdot\dfrac{6-26}{13}=\dfrac{-20}{10}=-2\)
c: \(=\dfrac{3}{4}\cdot2-\dfrac{5}{2}\cdot\dfrac{-4}{3}=\dfrac{3}{2}+\dfrac{20}{6}=\dfrac{3}{2}+\dfrac{10}{3}=\dfrac{29}{6}\)
d: \(=\dfrac{3}{8}\cdot\dfrac{8}{5}+\dfrac{3}{5}\cdot\dfrac{2}{7}+\dfrac{3}{5}\cdot\dfrac{5}{7}=\dfrac{3}{5}+\dfrac{3}{5}=\dfrac{6}{5}\)
Bài 4. Trên tia Ox lấy điểm A và B sao cho OA = 8cm, OB = 12cm.
a, Điểm A có nằm giữa hai điểm O và B không? Vì sao? Tính AB.
b, Gọi M, N lần lượt là trung điểm của OA, OB. Điểm M có nằm giữa hai điểm O và N không? Vì sao? Tính MN.
c, Điểm N có là trung điểm của đoạn thẳng AM không? Vì sao?
cho mìn hỏi câu b nhoa
Câu 1)
1) \(\dfrac{11}{24}\)−\(\dfrac{5}{41}\)+\(\dfrac{13}{24}\)+0,5−\(\dfrac{36}{41}\)=
2)12÷\(\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)=
3) (\(1+\dfrac{2}{3}-\dfrac{1}{4}\))\(\left(0,8-\dfrac{3}{4}\right)^2\) =
4)\(16\dfrac{2}{7}\)÷(\(\dfrac{-3}{5}\))+\(28\dfrac{2}{7}\)÷\(\dfrac{3}{5}\)
5)\(\left(2^2\div\dfrac{4}{3}-\dfrac{1}{2}\right)\times\dfrac{6}{5}-17\)
6)\(\left(\dfrac{1}{3}\right)^{50}\times\left(-9\right)^{25}-\dfrac{2}{3}\div4\)
1: \(\dfrac{11}{24}-\dfrac{5}{41}+\dfrac{13}{24}+0,5-\dfrac{36}{41}\)
\(=\left(\dfrac{11}{24}+\dfrac{13}{24}\right)-\left(\dfrac{5}{41}+\dfrac{36}{41}\right)+\dfrac{1}{2}\)
\(=1-1+\dfrac{1}{2}=\dfrac{1}{2}\)
2: \(12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)
\(=12:\left(\dfrac{9}{12}-\dfrac{10}{12}\right)^2\)
\(=12:\left(-\dfrac{1}{12}\right)^2=12:\dfrac{1}{144}=12\cdot144=1368\)
3: \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\cdot\left(0,8-\dfrac{3}{4}\right)^2\)
\(=\dfrac{12+8-3}{12}\cdot\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)
\(=\dfrac{17}{12}\cdot\left(\dfrac{16-15}{20}\right)^2\)
\(=\dfrac{17}{12}\cdot\dfrac{1}{400}=\dfrac{17}{4800}\)
4: \(16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)+28\dfrac{2}{7}:\dfrac{3}{5}\)
\(=\dfrac{5}{3}\cdot\left(-16-\dfrac{2}{7}\right)+\dfrac{5}{3}\cdot\left(28+\dfrac{2}{7}\right)\)
\(=\dfrac{5}{3}\left(-16-\dfrac{2}{7}+28+\dfrac{2}{7}\right)\)
\(=12\cdot\dfrac{5}{3}=20\)
5: \(\left(2^2:\dfrac{4}{3}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)
\(=\left(4\cdot\dfrac{3}{4}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)
\(=\dfrac{5}{2}\cdot\dfrac{6}{5}-17=3-17=-14\)
6: \(\left(\dfrac{1}{3}\right)^{50}\cdot\left(-9\right)^{25}-\dfrac{2}{3}:4\)
\(=\left(\dfrac{1}{3}\right)^{50}\cdot\left(-1\right)\cdot3^{50}-\dfrac{2}{3\cdot4}\)
\(=-1-\dfrac{2}{12}=-1-\dfrac{1}{6}=-\dfrac{7}{6}\)
tìm x
\([\dfrac{6:\dfrac{3}{5}-1\dfrac{1}{16}.\dfrac{16}{7}}{4\dfrac{1}{5}-\dfrac{10}{11}+5\dfrac{2}{11}}-\dfrac{\left(\dfrac{3}{20}+\dfrac{1}{2}-\dfrac{1}{5}\right).\dfrac{12}{49}}{3\dfrac{1}{3}+\dfrac{2}{9}}].x=2\dfrac{23}{96}\)