3x5+3x7+9x3
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\(\dfrac{1}{1.3}+\dfrac{1}{2.3}+\dfrac{1}{2.5}+\dfrac{1}{3.5}+\dfrac{1}{3.7}+\dfrac{1}{4.7}+\dfrac{1}{4.9}\)
\(=\dfrac{1}{1.3}+\dfrac{1}{3.2}+\dfrac{1}{2.5}+\dfrac{1}{5.3}+\dfrac{1}{3.7}+\dfrac{1}{7.4}+\dfrac{1}{4.9}\)
\(=\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}\right):\dfrac{1}{2}\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right):\dfrac{1}{2}\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{9}\right):\dfrac{1}{2}\)
\(=\dfrac{7}{18}:\dfrac{1}{2}\)
\(=\dfrac{7}{9}\)
\(A=\frac{1}{2.5}+\frac{1}{3.5}+\frac{1}{3.7}+\frac{1}{4.7}+...+\frac{1}{9.19}+\frac{1}{10.19}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{4.5}+\frac{1}{6.5}+\frac{1}{6.7}+\frac{1}{8.7}+...+\frac{1}{18.19}+\frac{1}{20.19}\)
\(\Rightarrow\frac{1}{2}A=\frac{5-4}{4.5}+\frac{6-5}{6.5}+\frac{7-6}{6.7}+...+\frac{20-19}{20.19}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{19}-\frac{1}{20}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{4}-\frac{1}{20}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{5}\)
\(\Rightarrow A=\frac{2}{5}\)
Mình có cách giải khác:
A= \(\frac{1}{2.5}+\frac{1}{3.5}+\frac{1}{3.7}+\frac{1}{4.7}+...+\frac{1}{9.19}+\frac{1}{10.19}\)
A= \(\frac{2.1}{2.2.5}+\frac{2.1}{2.3.5}+\frac{2.1}{2.3.7}+\frac{2.1}{2.4.7}+...+\frac{2.1}{2.9.19}+\frac{2.1}{2.10.19}\)
A= \(\frac{2.1}{4.5}+\frac{2.1}{5.6}+\frac{2.1}{6.7}+\frac{2.1}{7.8}+...+\frac{2.1}{18.19}+\frac{2.1}{19.20}\)
A= \(2.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{18.19}+\frac{1}{19.20}\right)\)
A=\(2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
A= \(2.\left(\frac{1}{4}+0+0+0+...+0+0-\frac{1}{20}\right)\)
A=\(2.\left(\frac{1}{4}-\frac{1}{20}\right)\)
A=\(2.\left(\frac{5}{20}-\frac{1}{20}\right)\)
A= \(2.\frac{1}{5}\)
A=\(\frac{2}{5}\)
Xong rùi đó!!!!! :))
\(A=\dfrac{1}{2.5}+\dfrac{1}{3.5}+\dfrac{1}{3.7}+...+\dfrac{1}{9.19}+\dfrac{1}{10.19}\)
\(A=\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{18.19}+\dfrac{2}{19.20}\)
\(A=2.\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{18.19}+\dfrac{1}{19.20}\right)\)
\(A=2.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\)
\(A=2.\left(\dfrac{1}{4}-\dfrac{1}{20}\right)\)
\(A=2.\dfrac{1}{5}\)
\(A=\dfrac{2}{5}\)
\(A=\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{97.99}\)
\(A=\frac{3}{2}.\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\right)\)
\(A=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(A=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(A=\frac{3}{2}.\frac{32}{99}\)
\(A=\frac{16}{33}\)
96/594=48/297
mình k biết có đúng k nhé tại lâu quá chưa làm dạng này
a/\(\frac{2^3\cdot3^4}{2^2\cdot3^2\cdot5}=\frac{18}{5}\)\(\frac{2^4\cdot5^2\cdot11^2\cdot7}{2^3\cdot5^3\cdot7^2\cdot11}=\frac{2\cdot11}{5\cdot7}=\frac{22}{35}\)
b/\(\frac{121\cdot75\cdot130\cdot169}{39\cdot60\cdot11\cdot198}=\frac{11^2\cdot5^3\cdot13^3\cdot2\cdot3}{2^3\cdot3^4\cdot5\cdot11^2\cdot13}=\frac{5^2\cdot13^2}{2^2\cdot3^3}=\frac{4225}{108}\)
c/\(\frac{1998\cdot1990+3978}{1992\cdot1991-3984}=\frac{2^2\cdot3^3\cdot37\cdot5\cdot199+2\cdot3^2\cdot13\cdot17}{2^3\cdot3\cdot83\cdot11\cdot181-2^4\cdot3\cdot83}=\frac{2\cdot3^2\cdot11\cdot20101}{2^3\cdot3^3\cdot13\cdot17\cdot83}=\frac{11\cdot20101}{2^2\cdot3\cdot13\cdot17\cdot83}\)
\(\frac{3\cdot4+3\cdot7}{6\cdot5+9}=\frac{3\left(4+7\right)}{2\cdot3\cdot5+3\cdot3}=\frac{3\cdot11}{3\left(2\cdot5+3\right)}=\frac{3\cdot11}{3\cdot13}=\frac{11}{13}\)
\(\frac{6\cdot9-2\cdot17}{63\cdot3-119}=\frac{2\cdot3\cdot9-2\cdot17}{7\cdot9\cdot3-7\cdot17}=\frac{2\left(27-17\right)}{7\left(27-17\right)}=\frac{2}{7}\)
\(\frac{21\cdot3-21}{1-24}=\frac{21\left(3-1\right)}{-23}=\frac{21\cdot2}{-23}=\frac{42}{-23}=-\frac{42}{23}\)
\(\frac{9\cdot3-54}{18}=\frac{9\cdot3-9\cdot6}{9\cdot2}=\frac{9\cdot\left(3-6\right)}{9\cdot2}=\frac{3-6}{2}=-\frac{3}{2}\)
a.
\(3x^7+x^4-3x^7+x^5+x+4=x^5+x^4+x+4\)
Đa thức có bậc 5
b.
Đa thức có bậc 0
3 x ( 5 + 7 + 9 )
= 3 x 21
= 63
Hk tốt,
k nhé
63 bạn nhé !