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\(A=\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+...+\dfrac{5^2}{26.31}\)
\(A=5\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+..+\dfrac{5}{26.31}\right)\)
\(A=5\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{31}\right)\)
\(A=5\left(1-\dfrac{1}{31}\right)\)
\(A=5-\dfrac{1}{155}\)
\(A< 5\rightarrowđpcm\)
\(A=\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+.......\dfrac{5^2}{26.31}\)
\(\Leftrightarrow A=5\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+.........+\dfrac{5}{26.31}\right)\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+..........+\dfrac{1}{26}-\dfrac{1}{31}\right)\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{31}\right)\)
\(\Leftrightarrow A=5.\dfrac{30}{31}=\dfrac{150}{31}\)
Bài 1:
a: \(A=\left(-\dfrac{1}{5}\right)^{33}:\left(-\dfrac{1}{5}\right)^{32}=\dfrac{-1}{5}\)
c: \(C=\dfrac{2^{12}\cdot3^{10}+3^9\cdot2^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)
\(=\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\cdot7}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)
a)\(\frac{5}{3}\)+ \(\left(\frac{-2}{7}\right)\)-(-1,2)
=\(\frac{5}{3}+\left(\frac{-2}{7}\right)+\frac{6}{5}\)
=\(\frac{175+\left(-30\right)+126}{105}\)
=\(\frac{271}{105}\)
b) \(\frac{-4}{9}+\frac{-5}{6}-\frac{17}{4}\)
=\(\frac{-16+\left(-30\right)-153}{36}\)
=\(\frac{-199}{36}\)
\(\frac{5}{3}+\left(\frac{-2}{7}\right)-\left(\frac{-6}{5}\right)\)
=\(\frac{-2}{7}-\left(\frac{5}{3}+\frac{-6}{5}\right)\)
=\(\frac{-79}{105}\)\(\frac{-2}{7}-\frac{7}{15}\)
a)\(\left|-0.75\right|+\dfrac{1}{4}-2\dfrac{1}{2}\)
=0.75+0.25-2.5
=1-2.5=-1.5
b)\(15.\dfrac{1}{5}:\left(\dfrac{-5}{7}\right)-2\dfrac{1}{5}.\left(\dfrac{-7}{5}\right)\)
=3.(-1.4)+3.08
=-4.2+3.08=-1.12
c)\(\dfrac{5}{17}+\dfrac{2}{3}-\dfrac{20}{12}+\dfrac{7}{9}+\dfrac{12}{17}\)
=\(\dfrac{49}{51}-\dfrac{5}{3}+\dfrac{7}{9}+\dfrac{12}{17}\)
=\(\dfrac{-12}{17}+\dfrac{7}{9}+\dfrac{12}{17}\)
=\(\dfrac{11}{153}+\dfrac{12}{17}\)
=\(\dfrac{7}{9}\)
d)\(\dfrac{5}{15}+\dfrac{14}{25}-\dfrac{12}{9}+\dfrac{2}{7}+\dfrac{11}{25}\)
=\(\dfrac{67}{75}-\dfrac{4}{3}+\dfrac{2}{7}+\dfrac{11}{25}\)
=-0.44+\(\dfrac{127}{175}\)
=\(\dfrac{2}{7}\)
\(a,\dfrac{1}{2}x=3+2\)
\(\dfrac{1}{2}x=5\)
\(x=5\div\dfrac{1}{2}\)
\(x=10\)
\(b,\dfrac{1}{4}x^2-\sqrt{36}=10\)
\(\dfrac{1}{4}x^2-6=10\)
\(\dfrac{1}{4}x^2=10+6\)
\(\dfrac{1}{4}x^2=16\)
\(x^2=16\div\dfrac{1}{4}\)
\(x^2=64\)
\(x^2=\left(8\right)^2\)
\(\Rightarrow x=8\)
Ta có: \(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+...+\dfrac{5}{96.101}\) \(=1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{96}-\dfrac{1}{101}\) \(=1-\dfrac{1}{101}\) \(\dfrac{100}{101}\)
\(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+.....+\dfrac{5}{96.101}\)
\(=1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+......+\dfrac{1}{96}-\dfrac{1}{101}\)
\(=1-\dfrac{1}{101}\)
\(=\dfrac{101}{101}-\dfrac{1}{101}\)
\(=\dfrac{101-1}{101}\)
\(=\dfrac{100}{101}\)
\(=\dfrac{102}{5}\cdot\dfrac{6}{5}-\dfrac{107}{3}\cdot\dfrac{6}{5}-\dfrac{3}{4}\)
\(=\dfrac{6}{5}\cdot\dfrac{-229}{15}-\dfrac{3}{4}\)
\(=\dfrac{-458}{25}-\dfrac{3}{4}=-\dfrac{1907}{100}\)
= 102/5 x 6/5 - 107/3 x 6/5 - 3/4
= 6/5 x ( 102/5 - 107/3 ) - 3/4
= 6/5 x -229/15 - 3/4
= -458/25 - 3/4
= -1907/100
\(a,\left(31\dfrac{6}{13}+5\dfrac{9}{41}\right)-36\dfrac{6}{13}\\ =\left(31-36\right)+\left(\dfrac{6}{13}-\dfrac{6}{13}\right)+5\dfrac{9}{41}\\ =-5+0+5\dfrac{9}{41}\\ =\left(-5+5\right)+\dfrac{9}{41}=\dfrac{9}{41}\)
\(b,\dfrac{5}{3}+\left(-\dfrac{2}{7}\right)-\left(-1,2\right)\\ =\dfrac{5}{3}-\dfrac{2}{7}+\dfrac{6}{5}\\ =\dfrac{5.35-2.15+6.21}{105}=\dfrac{271}{105}\\ c,0,25+\dfrac{3}{5}-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)=\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{1}{8}+\dfrac{2}{5}-1\dfrac{1}{4}\\ =\left(-1\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(\dfrac{3}{5}+\dfrac{2}{5}\right)-\dfrac{1}{8}=-1+1-\dfrac{1}{8}=-\dfrac{1}{8}\)
a) (31 6/13 + 5 9/41) - 36 6/13
= 409/13 + 214/41 - 474/13
= (409/13 - 474/13) + 214/41
= -5 + 214/41
= 9/41
b) 5/3 + (-2/7) - (-1,2)
= 5/3 - 2/7 + 6/5
= 29/21 + 6/5
= 271/105
c) 0,25 + 3/5 - (1/8 - 2/5 + 1 1/4)
= 1/4 + 3/5 - 1/8 + 2/5 - 5/4
= (1/4 - 5/4) + (3/5 + 2/5) - 1/8
= -1 + 1 - 1/8
= -1/8
Đặt: \(S=\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+\dfrac{5^2}{11.16}+...+\dfrac{5^2}{26.31}\)
\(\Rightarrow\dfrac{S}{5}=\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+...+\dfrac{5}{26.31}\\ =1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{26}-\dfrac{1}{31}\\ =1-\dfrac{1}{31}=\dfrac{30}{31}\\ \Rightarrow S=\dfrac{30}{31}.5=\dfrac{150}{31}\)
Sửa bài:
Đặt biểu thức là A. ta được:
\(\dfrac{A}{5}=\dfrac{5}{1.6}+\dfrac{5}{6.11}+...+\dfrac{5}{26.31}\\ \dfrac{A}{5}=1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{31}\\ \dfrac{A}{5}=\dfrac{30}{31}\\ A=\dfrac{150}{31}.\)