\(B=5\dfrac{9}{10}:\dfrac{3}{2}-\left(2\dfrac{1}{3}\cdot4\dfrac{1}{2}-2\cdot2\dfrac{1}{3}\right):\dfrac{7}{4}\)
Tính giá trị biểu thức :
1. \(A=\dfrac{\dfrac{2}{5}+\dfrac{2}{7}-\dfrac{2}{9}-\dfrac{2}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{9}-\dfrac{4}{11}}\)
2. \(B=\dfrac{1^2}{1\cdot2}\cdot\dfrac{2^2}{2\cdot3}\cdot\dfrac{3^2}{3\cdot4}\cdot\dfrac{4^2}{4\cdot5}\)
3. \(C=\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot\dfrac{5^2}{4\cdot6}\cdot\dfrac{5^2}{4\cdot6}\)
4. \(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right)\cdot\left(\dfrac{2}{3}\cdot\dfrac{1}{4}\right)^2\)
5. Cho \(M=8\dfrac{2}{7}-\left(3\dfrac{4}{9}+4\dfrac{2}{7}\right)\) ; \(N=\left(10\dfrac{2}{9}+2\dfrac{3}{5}\right)-6\dfrac{2}{9}\). Tính \(P=M-N\)
6. \(E=10101\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{3\cdot7\cdot11\cdot13\cdot37}\right)\)
7. \(F=\dfrac{\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{256}+\dfrac{3}{64}}{1-\dfrac{1}{4}+\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
8. \(G=\text{[}\dfrac{\left(6-4\dfrac{1}{2}\right):0,03}{\left(3\dfrac{1}{20}-2,65\right)\cdot4+\dfrac{2}{5}}-\dfrac{\left(0,3-\dfrac{3}{20}\right)\cdot1\dfrac{1}{2}}{\left(1,88+2\dfrac{3}{25}\right)\cdot\dfrac{1}{80}}\text{]}:\dfrac{49}{60}\)
9. \(H=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{4\cdot5\cdot6}+...+\dfrac{1}{98\cdot99\cdot100}\)
10. \(I=\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot\dfrac{24}{25}\cdot...\cdot\dfrac{2499}{2500}\)
11. \(K=\left(-1\dfrac{1}{2}\right)\left(-1\dfrac{1}{3}\right)\left(-1\dfrac{1}{4}\right)...\left(-1\dfrac{1}{999}\right)\)
12. \(L=1\dfrac{1}{3}+1\dfrac{1}{8}+1\dfrac{1}{15}...\) (98 thừa số)
13. \(M=-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{3}}}}\)
14. \(N=\dfrac{155-\dfrac{10}{7}-\dfrac{5}{11}+\dfrac{5}{23}}{403-\dfrac{26}{7}-\dfrac{13}{11}+\dfrac{13}{23}}\)
15. \(P=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{5}-1\right)...\left(\dfrac{1}{2001}-1\right)\)
16. \(Q=\left(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{2005\cdot2006}\right):\left(\dfrac{1}{1004\cdot2006}+\dfrac{1}{1005\cdot2005}+...+\dfrac{1}{2006\cdot1004}\right)\)
1. \(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)
2. \(B=\dfrac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\dfrac{1}{5}\)
3.\(C=\dfrac{2^2.3^2.\text{4^2.5^2}.5^2}{1.2^2.3^2.4^2.5.6^2}=\dfrac{125}{36}\)
4.D=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)
Rút gọn:
a) \(\dfrac{2^{19}\cdot27^3+15\cdot4^9\cdot9^4}{6^9\cdot2^{10}+2^{10}}\)
b) \(\dfrac{\left(-\dfrac{1}{2}\right)^3-\left(\dfrac{3}{4}\right)^3\cdot\left(-2\right)^2}{2\cdot\left(-1\right)+\left(\dfrac{3}{4}\right)^2-\left(\dfrac{3}{8}\right)}\)
a: \(=\dfrac{2^{19}\cdot3^9+3^9\cdot5\cdot2^{18}}{2^{19}\cdot3^9+2^{10}}\)
\(=\dfrac{3^9\cdot2^{18}\cdot\left(2+5\right)}{2^{10}\cdot\left(2^9\cdot3^9+1\right)}=\dfrac{3^9\cdot7\cdot2^8}{6^9+1}\)
b: \(=\dfrac{\dfrac{-1}{8}-\dfrac{27}{64}\cdot4}{-2+\dfrac{9}{16}-\dfrac{3}{8}}=\dfrac{-29}{16}:\dfrac{-29}{16}=1\)
1.tính c = \(\left(1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5+...+18\cdot19\cdot20\cdot21\right)-5-10-15-...-200\)
2. tính d = \(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{18\cdot19\cdot20}-\dfrac{3}{1\cdot2}-\dfrac{3}{2\cdot3}-...-\dfrac{3}{19\cdot20}\)
3.tìm x: \(2\left(x-3\right)-3\left(1-2x\right)=4+4\left(1-x\right)\)
4.tìm x: \(\dfrac{x-2}{2}-\dfrac{1+x}{3}=\dfrac{4-3x}{4}-1\)
giải nhanh giùm tớ nhé. ghi cả lời giả ra kìa
3/ \(2\left(x-3\right)-3\left(1-2x\right)=4+4\left(1-x\right)\)
\(\Leftrightarrow2x-6-3+6x=4+4-4x\)
\(\Leftrightarrow8x-9=8-4x\)
\(\Leftrightarrow8x+4x=8+9\)
\(\Leftrightarrow12x=17\)
\(\Leftrightarrow x=\dfrac{17}{12}\)
Vậy \(x=\dfrac{17}{12}\)
4/ \(\dfrac{x-2}{2}-\dfrac{1+x}{3}=\dfrac{4-3x}{4}-1\)
\(\Leftrightarrow6\left(x-2\right)-4\left(1+x\right)=3\left(4-3x\right)-12\)
\(\Leftrightarrow6x-12-4-4x=12-9x-12\)
\(\Leftrightarrow6x-4-4x=12-9x\)
\(\Leftrightarrow2x-4=12-9x\)
\(\Leftrightarrow2x+9x=12+4\)
\(\Leftrightarrow11x=16\)
\(\Leftrightarrow x=\dfrac{16}{11}\)
Vậy \(x=\dfrac{16}{11}\)
tính tổng
S=\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{n\cdot\left(n+1\right)\cdot\left(n+2\right)}\)
A=1+\(\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{99}\)
B= \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...\dfrac{1}{100}\)
C=\(\dfrac{99}{1}+\dfrac{98}{2}+\dfrac{97}{3}+...+\dfrac{1}{99}\)
a) Ta có
S = \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{n.\left(n+1\right).\left(n+2\right)}\)
2S = \(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{n.\left(n+1\right).\left(n+2\right)}\)
2S = \(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right).\left(n+2\right)}\)2S = \(\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right).\left(n+2\right)}\)
S = \(\dfrac{1}{4}-\dfrac{1}{\left(n+1\right).\left(n+2\right):2}\)
b) A = \(1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{99}\)
A = \(2-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
A = \(2-\dfrac{1}{99}\)
A = \(\dfrac{197}{99}\)
c) Ta có
B = \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\)
B = \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
B = \(1-\dfrac{1}{100}\)
B = \(\dfrac{99}{100}\)
d) Ta có
C = \(\dfrac{99}{1}+\dfrac{98}{2}+\dfrac{97}{3}+...+\dfrac{1}{99}\)
C = \(1+\left(1+\dfrac{98}{2}\right)+\left(1+\dfrac{97}{3}\right)+...+\left(1+\dfrac{1}{99}\right)\)
C = \(1+50+\dfrac{100}{3}+...+\dfrac{100}{99}\)
C = 51 + 100(\(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{99}\))
Đặt D = \(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{99}\)
D = \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{98}-\dfrac{1}{99}\)
D = \(\dfrac{1}{2}-\dfrac{1}{99}\)
D = \(\dfrac{97}{198}\)
=> C = 51 + 100.\(\dfrac{97}{198}\)
C = 51 + \(\dfrac{4850}{99}\)
C = \(\dfrac{9899}{99}\)
Đây là bài làm của mình sai thì nx nha
Câu hỏi today là a, \(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{x+\left(x+1\right)}=\dfrac{101}{1540}\)
b,\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{x+\left(x+1\right)}=\dfrac{2020}{2021}\)
Chúc các bn học tốt ❤
Giải:
a)1/5.8+1/8.11+...+1/x.(x+1)=101/1540
1/3.(3/5.8+3/8.11+...+3/x.(x+1))=101/1540
1/3.(1/5-1/8+1/8-1/11+...+1/x-1/x+1)=101/1540
1/3.(1/5-1/x+1)=101/1540
1/5-1/x+1=101/1540:1/3
1/5-1/x+1=303/1540
1/x+1=1/5-303/1540
1/x+1=1/308
⇒x+1=308
x=308-1
x=307
b)1/1.2+1/2.3+1/3.4+...+1/x.(x+1)=2020/2021
1/1-1/2+1/2-1/3+1/3-1/4+...+1/x-1/x+1=2020/2021
1/1-1/x+1=2020/2021
1/x+1=1/1-2020/2021
1/x+1=1/2021
⇒x+1=2021
x=2021-1
x=2020
Mk thấy đề bài hơi sai là:
1/x+(x+1) ➜ 1/x.(x+1)
mới ra đc kết quả!
Bài 1. Tính
A= \(\left(8\dfrac{2}{7}-4\dfrac{2}{7}\right)-3\dfrac{4}{9}\)
B= \(\left(10\dfrac{2}{9}-6\dfrac{2}{9}\right)+2\dfrac{3}{5}\)
Bài 2. Tính
a) \(5\dfrac{1}{2}.3\dfrac{1}{4}\) b) \(6\dfrac{1}{3}:4\dfrac{2}{9}\) c) \(4\dfrac{3}{7}.2\)
`A=(8 2/7-4 2/7)-3 4/9`
`=8+2/7-4-2/7-3-4/9`
`=4-3-4/9`
`=1-4/9=5/9`
`B=(10 2/9-6 2/9)+2 3/5`
`=10+2/9-6-2/9+2+3/5`
`=4+2+3/5`
`=6+3/5=33/5`
Bài 2:
`a)5 1/2*3 1/4`
`=11/2*13/4`
`=143/8`
`b)6 1/3:4 2/9`
`=19/3:38/9`
`=19/3*9/38=3/2`
`c)4 3/7*2`
`=31/7*2`
`=62/7`
Bài 1:
\(A=\left(8\dfrac{2}{7}-4\dfrac{2}{7}\right)-3\dfrac{4}{9}\)
\(A=\left(\dfrac{58}{7}-\dfrac{30}{7}\right)-\dfrac{31}{9}\)
\(A=4-\dfrac{31}{9}\)
\(A=\dfrac{5}{9}\)
\(B=\left(10\dfrac{2}{9}-6\dfrac{2}{9}\right)+2\dfrac{3}{5}\)
\(B=\left(\dfrac{92}{9}-\dfrac{56}{9}\right)+\dfrac{13}{5}\)
\(B=4+\dfrac{13}{5}\)
\(B=\dfrac{33}{5}\)
Bài 2:
a) \(5\dfrac{1}{2}.3\dfrac{1}{4}=\dfrac{11}{2}.\dfrac{13}{4}=\dfrac{11.13}{2.4}=\dfrac{143}{8}\)
b) \(6\dfrac{1}{3}:4\dfrac{2}{9}=\dfrac{19}{3}:\dfrac{38}{9}=\dfrac{19}{3}.\dfrac{9}{38}=\dfrac{3}{2}\)
c) \(4\dfrac{3}{7}.2=\dfrac{31}{7}.2=\dfrac{31.2}{7}=\dfrac{62}{7}\)
\(\left[18\dfrac{1}{6}-\left(0,06:7\dfrac{1}{2}+3\dfrac{2}{5}\cdot0,38\right)\right]:\left(19-2\dfrac{2}{3}\cdot4\dfrac{3}{4}\right).Tính\)
tìm số tự nhiên thỏa mãn điều kiện
\(2\cdot2^2+3\cdot2^3+4\cdot2^4+........+n\cdot2^n=2^{n+11}\)
rút gọn : \(A=\left(\dfrac{2}{5}-\dfrac{5}{2}+\dfrac{1}{10}\right):\left(\dfrac{5}{2}-\dfrac{2}{3}+\dfrac{1}{12}\right)\)
tính:\(B=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+......+\dfrac{1}{2017}}{\dfrac{2016}{1}+\dfrac{2003}{2}+\dfrac{2002}{3}+.......+\dfrac{1}{2016}}\)
CMR :\(5a+2b⋮13\Leftrightarrow9a+b⋮13\left(a,b\in Z\right)\)
tính
a) \(\left[\dfrac{0.8\div\left(\dfrac{4}{5}\cdot1025\right)}{0.64-1}+\dfrac{\left(1.08-\dfrac{2}{25}\right)\div\dfrac{4}{7}}{\left(6\dfrac{5}{7}-3\dfrac{1}{4}\right)\cdot2\dfrac{2}{17}}+\left(1.2\cdot0.5\right)\div\dfrac{4}{5}\right]\)
b) \(\left(0.2\right)^{-3}\left[\left(-\dfrac{1}{5}\right)^{-2}\right]^{-1}+\left[\left(\dfrac{1}{2}\right)^{-3}\right]^{-2}\div\left(2^{-3}\right)^{-1}-\left(0.175\right)^{-2}\)
c) \(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)
d) \(\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{3}\)
e) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2\div2\)
f) \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
g) \(\dfrac{1}{-\left(2017\right)\left(-2015\right)}+\dfrac{1}{\left(-2015\right)\left(-2013\right)}+...+\dfrac{1}{\left(-3\right)\cdot\left(-1\right)}\)
h) \(\left(1-\dfrac{1}{1\cdot2}\right)+\left(1-\dfrac{1}{2\cdot3}+...+\left(1-\dfrac{1}{2017\cdot2018}\right)\right)\)
c)
Ta có :\(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)
\(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{\dfrac{8}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{3}{8}}\) \(=2+\dfrac{1}{\dfrac{11}{8}}\) \(=2+\dfrac{8}{11}\) \(=\dfrac{30}{11}\)
d) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\dfrac{1}{4}:2\)
\(=3-1+\dfrac{1}{8}\)
\(=\dfrac{17}{8}\)
f) \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}.\dfrac{\dfrac{3}{4}\left(1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}\right)}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}.\dfrac{3}{4}+\dfrac{5}{8}\)
\(=\dfrac{3}{8}+\dfrac{5}{8}\)
\(=1\)