Tính hợp lý:
\(a.\dfrac{3}{17}+\dfrac{-5}{13}+\dfrac{-18}{35}+\dfrac{14}{17}+\dfrac{17}{-35}+\dfrac{-8}{13}\)
\(b.\dfrac{-3}{8}\cdot\dfrac{1}{6}+\dfrac{3}{-8}\cdot\dfrac{5}{6}+\dfrac{-10}{16}\)
\(c.\dfrac{-4}{11}\cdot\dfrac{5}{15}\cdot\dfrac{11}{-4}\)
a: \(=\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-5}{13}-\dfrac{8}{13}\right)+\left(\dfrac{-18}{35}-\dfrac{17}{35}\right)\)
=1-1-1
=-1
b: \(=\dfrac{-3}{8}\left(\dfrac{1}{6}+\dfrac{5}{6}\right)+\dfrac{-5}{8}=\dfrac{-3}{8}-\dfrac{5}{8}=-1\)
c: \(=\dfrac{4}{4}\cdot\dfrac{5}{15}\cdot\dfrac{11}{11}=\dfrac{1}{3}\)
a) \(=\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(-\dfrac{5}{13}+-\dfrac{8}{13}\right)+\left(-\dfrac{18}{35}+-\dfrac{17}{35}\right)=1+-1+-1=-1\)
b) \(=-\dfrac{3}{8}\cdot\left(\dfrac{1}{6}+\dfrac{5}{6}\right)-\dfrac{10}{16}=-\dfrac{3}{8}-\dfrac{10}{16}=-1\)
c) \(=\left(-\dfrac{4}{11}\cdot-\dfrac{11}{4}\right)\cdot\dfrac{5}{15}=1\cdot\dfrac{1}{3}=\dfrac{1}{3}\)
a)\(=\left(-\dfrac{5}{13}+\dfrac{-8}{13}\right)+\left(-\dfrac{18}{35}-\dfrac{17}{35}\right)+\left(\dfrac{3}{14}+\dfrac{14}{17}\right)=-1-1+1=-1\)
b)\(=\dfrac{-3}{8}.\left(\dfrac{1}{6}+\dfrac{5}{6}\right)-\dfrac{10}{16}=-\dfrac{3}{8}.1-\dfrac{10}{16}=-\dfrac{6}{16}-\dfrac{10}{16}=-\dfrac{16}{16}=-1\)
c)\(\dfrac{-4.5.11}{11.5.3.-4}=\dfrac{1}{3}\)
Tính giá trị của các biểu thức sau 1) \(A=1+2+2^2+...+2^{2015}\) 2) \(B=\left(\dfrac{1}{4}-1\right)\cdot\left(\dfrac{1}{9}-1\right)\cdot\left(\dfrac{1}{16}-1\right)\cdot\cdot\cdot\cdot\cdot\left(\dfrac{1}{400}-1\right)\) 3) \(C=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\) 4) \(D=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\) 5) \(E=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\) 6) Cho 13+23+...+103=3025 Tính S= 23+43+63+...+203
Bài 2: Tính nhanh:
a) \(\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot\dfrac{24}{25}\cdot.....\cdot\dfrac{120}{121}\cdot\dfrac{143}{144}\)
b)\(\dfrac{5}{9}\cdot\dfrac{21}{25}\cdot\dfrac{45}{49}\cdot\dfrac{77}{81}\cdot.....\dfrac{357}{361}\cdot\dfrac{437}{441}\)
ai làm đúng mk dùng 3nick mk tick cho :))))
a) \(A=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}.....\dfrac{120}{121}.\dfrac{143}{144}\)
= \(\dfrac{1.3.2.4.3.5.4.6....10.12.11.13}{2^2.3^2.4^2.5^2...11^2.12^2}\)
= \(\dfrac{1.2.12.13}{2^2.12^2}=\dfrac{13}{2.12}=\dfrac{13}{24}\)
b) \(B=\dfrac{5}{9}.\dfrac{21}{25}.\dfrac{45}{49}.\dfrac{77}{81}....\dfrac{357}{361}.\dfrac{437}{441}\)
= \(\dfrac{1.5.3.7.5.9.7.11.....17.21.19.23}{3^2.5^2.7^2....19^2.21^2}=\dfrac{1.3.21.23}{3^2.21^2}\)
= \(\dfrac{23}{3.21}=\dfrac{23}{63}\)
c) \(\left(6-\dfrac{14}{5}\right)\cdot\dfrac{25}{8}-\dfrac{18}{5}\cdot\dfrac{1}{4}\)
\(\left(6-\dfrac{14}{5}\right).\dfrac{25}{8}-\dfrac{18}{5}.\dfrac{1}{4}\)
\(=\left(\dfrac{30}{5}-\dfrac{14}{5}\right).\dfrac{25}{8}-\dfrac{18}{5}.\dfrac{1}{4}\)
\(=\dfrac{16}{5}.\dfrac{25}{8}-\dfrac{18}{20}\)
\(=\dfrac{16.25}{5.8}-\dfrac{18}{20}\)
\(=\dfrac{2.5}{1}-\dfrac{9}{10}\)
\(=10-\dfrac{9}{10}\)
\(=\dfrac{100}{10}-\dfrac{9}{10}\)
\(=\dfrac{91}{10}=9\dfrac{1}{10}\)
\(\left(6-\dfrac{14}{5}\right)\) . \(\dfrac{25}{8}\) - \(\dfrac{18}{5}\) . \(\dfrac{1}{4}\)
=\(\left(\dfrac{30}{5}-\dfrac{14}{5}\right)\) . \(\dfrac{25}{8}\) - \(\dfrac{18}{5}\) .\(\dfrac{1}{4}\)
= \(\dfrac{16}{5}\) . \(\dfrac{25}{8}\) - \(\dfrac{9}{10}\)
= \(\dfrac{16.25}{5.8}\) - \(\dfrac{9}{10}\)
= \(\dfrac{2.5}{1}\) - \(\dfrac{9}{10}\)
= 10 -\(\dfrac{9}{10}\)
= \(\dfrac{100}{10}\) - \(\dfrac{9}{10}\)
= \(\dfrac{91}{10}\) = 9,1
Tính giá trị của các biểu thức sau
1) \(A=1+2+2^2+...+2^{2015}\)
2) \(B=\left(\dfrac{1}{4}-1\right)\cdot\left(\dfrac{1}{9}-1\right)\cdot\left(\dfrac{1}{16}-1\right)\cdot\cdot\cdot\cdot\cdot\left(\dfrac{1}{400}-1\right)\)
3) \(C=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)
4) \(D=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
5) \(E=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\)
6) Cho 13+23+...+103=3025
Tính S= 23+43+63+...+203
Tìm x, biết:
\(x=\dfrac{19}{11}\cdot\dfrac{5}{14}+\dfrac{1}{11}\cdot\dfrac{5}{7}-\sqrt{\dfrac{25}{4}}\cdot\dfrac{2}{11}\)
Tính một cách hợp lí.
a) \(A=\dfrac{-3}{14}+\dfrac{2}{13}+\dfrac{-25}{14}+\dfrac{-15}{13};\)
b) \(B=\dfrac{5}{3}\cdot\dfrac{7}{25}+\dfrac{5}{3}\cdot\dfrac{21}{25}-\dfrac{5}{3}\cdot\dfrac{7}{25}\).
a, \(\begin{array}{l}A = \dfrac{{ - 3}}{{14}} + \dfrac{2}{{13}} + \dfrac{{ - 25}}{{14}} + \dfrac{{ - 15}}{{13}}\\A = \left( {\dfrac{{ - 3}}{{14}} + \dfrac{{ - 25}}{{14}}} \right) + \left( {\dfrac{2}{{13}} + \dfrac{{ - 15}}{{13}}} \right)\\A = \dfrac{{ - 3 + \left( { - 25} \right)}}{{14}} + \dfrac{{2 + \left( { - 15} \right)}}{{13}}\\A = \dfrac{{ - 28}}{{14}} + \dfrac{{ - 13}}{{13}}\\A = - 2 + (-1)\\A = - 3\end{array}\)
b,
Cách 1:
\(\begin{array}{l}B = \dfrac{5}{3}.\dfrac{7}{{25}} + \dfrac{5}{3}.\dfrac{{21}}{{25}} - \dfrac{5}{3}.\dfrac{7}{{25}}\\B = \left( {\dfrac{5}{3}.\dfrac{7}{{25}} - \dfrac{5}{3}.\dfrac{7}{{25}}} \right) + \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = 0 + \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = \dfrac{{5.21}}{{3.25}}\\B = \dfrac{7}{5}\end{array}\)
Cách 2:
\(B = \dfrac{5}{3}.\dfrac{7}{{25}} + \dfrac{5}{3}.\dfrac{{21}}{{25}} - \dfrac{5}{3}.\dfrac{7}{{25}}\\B = \dfrac{5}{3}.({\dfrac{7}{{25}} -\dfrac{7}{{25}} + \dfrac{{21}}{{25}}})\\B = \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = \dfrac{{5.21}}{{3.25}}\\B = \dfrac{7}{5}\)
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=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)
Tính giá trị của các biểu thức sau
1) \(A=1+2+2^2+...+2^{2015}\)
2) \(B=\left(\dfrac{1}{4}-1\right)\cdot\left(\dfrac{1}{9}-1\right)\cdot\left(\dfrac{1}{16}-1\right)\cdot\cdot\cdot\cdot\cdot\left(\dfrac{1}{400}-1\right)\)
3) \(C=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)
4) \(D=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
5) \(E=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\)
6) Cho 13+23+...+103=3025
Tính S= 23+43+63+...+203
1) \(A=1+2+2^2+2^3+......+2^{2015}\)
\(\Leftrightarrow2A=2+2^2+2^3+......+2^{2016}\)
\(\Leftrightarrow2A-A=\left(2+2^2+2^3+......+2^{2016}\right)-\left(1+2+2^2+2^3+......+2^{2015}\right)\)
\(\Leftrightarrow A=2^{2016}-1\)
Vậy \(A=2^{2016}-1\)
6)Ta có: \(13+23+33+43+.......+103=3025\)
\(\Leftrightarrow2.13+2.23+2.33+2.43+.......+2.103=2.3025\)
\(\Leftrightarrow26+46+66+86+.......+206=6050\)
\(\Leftrightarrow\left(23+3\right)+\left(43+3\right)+\left(63+3\right)+\left(83+3\right)+.......+\left(203+3\right)=6050\)
\(\Leftrightarrow23+43+63+83+.......+203+3.10=6050\)
\(\Leftrightarrow23+43+63+83+.......+203+=6050-30\)
\(\Leftrightarrow23+43+63+83+.......+203+=6020\)
Vậy S=6020
b, B có 19 thừa số
=> \(-B=(1-\frac{1}{4})(1-\frac{1}{9})(1-\frac{1}{16})...(1-\frac{1}{400}) \)
<=>\(-B=\frac{(2-1)(2+1)(3-1)(3+1)(4-1)(4+1)...(20-1)(20+1)}{4.9.16...400} \)
<=>\(-B=\frac{(1.2.3.4...19)(3.4.5...21)}{(2.3.4.5.6...20)(2.3.4.5...20)} \)
<=>\(-B=\frac{21}{20.2} =\frac{21}{40} \)
<=>\(B=\frac{-21}{40} \)
a) \(\left(7\dfrac{4}{9}+3\dfrac{7}{11}\right)-3\dfrac{4}{9}\)
b) \(5\dfrac{2}{7}\cdot\dfrac{8}{11}+5\dfrac{2}{7}\cdot\dfrac{5}{11}-5\dfrac{2}{7}\cdot\dfrac{2}{11}\)
c) \(1\dfrac{13}{15}\cdot0,75-\left(\dfrac{11}{20}+25\%\right):\dfrac{7}{5}\)
Giúp mình nha
Đang cần gấp
A,
\(\left(7\dfrac{4}{9}+3\dfrac{7}{11}\right)-3\dfrac{4}{9}=7\dfrac{4}{9}+3\dfrac{7}{11}-3\dfrac{4}{9}\)
\(=7\dfrac{4}{9}-3\dfrac{4}{9}+3\dfrac{7}{11}=4+3\dfrac{7}{11}=7\dfrac{7}{11}\)
B,
\(5\dfrac{2}{7}.\dfrac{8}{11}+5\dfrac{2}{7}.\dfrac{5}{11}-5\dfrac{2}{7}.\dfrac{2}{11}=5\dfrac{2}{7}.\left(\dfrac{8}{11}+\dfrac{5}{11}-\dfrac{2}{11}\right)\)
\(=5\dfrac{2}{7}.1=5\dfrac{2}{7}\)
C,
\(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+25\%\right):\dfrac{7}{5}=\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{11}{20}+\dfrac{1}{4}\right).\dfrac{5}{7}\)
\(=\dfrac{7}{5}-\dfrac{3}{5}.\dfrac{5}{7}=\dfrac{7}{5}-\dfrac{3}{7}=\dfrac{34}{35}\)
