Bài 1: Tính gái trị biểu thức sau:
1) \(\dfrac{-5}{2}:\left(\dfrac{3}{4}-\dfrac{1}{2}\right)\)
2) \(\dfrac{298}{719}:\left(\dfrac{1}{4}+\dfrac{1}{12}-\dfrac{1}{3}\right)-\dfrac{2011}{2012}\)
3) \(\dfrac{27.18+27.103-120.27}{15.33+33.12}\)
Tính giá trị biểu thức:
\(\dfrac{298}{719}:\left(\dfrac{1}{4}+\dfrac{1}{12}-\dfrac{1}{3}\right)-\dfrac{2011}{2012}\)
\(\dfrac{298}{719}:\left(\dfrac{1}{4}+\dfrac{1}{12}-\dfrac{1}{3}\right)-\dfrac{2011}{2012}\)
\(=\dfrac{298}{719}.0-\dfrac{2011}{2012}\)
\(=0-\dfrac{2011}{2012}\)
\(=-\dfrac{2011}{2012}\)
\(\dfrac{298}{719}:\left(\dfrac{1}{4}+\dfrac{1}{12}-\dfrac{1}{3}\right)-\dfrac{2011}{2012}\)
\(=\dfrac{298}{719}.0-\dfrac{2011}{2012}\)
\(=0-\dfrac{2011}{2012}\)
\(=-\dfrac{2011}{2012}\)
cho \(f\left(x\right)=\dfrac{x^3}{1-3x-3x^2}\). hãy tính giá trị biểu thức sau: \(A=f\left(\dfrac{1}{2012}\right)+f\left(\dfrac{2}{2012}\right)+...+f\left(\dfrac{2010}{2012}\right)+f\left(\dfrac{2011}{2012}\right)\)
Bạn kiểm tra lại đề, \(f\left(x\right)=\dfrac{x^3}{1-3x-3x^2}\) hay \(f\left(x\right)=\dfrac{x^3}{1-3x+3x^2}\)
Tính giá trị của các biểu thức sau :
a)\(\left(7+3\dfrac{1}{4}-\dfrac{3}{5}\right)\)+(0,4 - 5) - \(\left(4\dfrac{1}{4}-1\right)\)
b)\(\dfrac{2}{3}\) - \(\left[\left(-\dfrac{7}{4}\right)-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)
c)\(\left(9-\dfrac{1}{2}-\dfrac{3}{4}\right)\):\(\left(7-\dfrac{1}{4}-\dfrac{5}{8}\right)\)
d)3 - \(\dfrac{1-\dfrac{1}{7}}{1+\dfrac{1}{7}}\)
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\(a,\left(7+3\dfrac{1}{4}-\dfrac{3}{5}\right)+\left(0,4-5\right)-\left(4\dfrac{1}{4}-1\right)\)
\(=\left(7+\dfrac{13}{4}-\dfrac{3}{5}\right)-\dfrac{23}{5}-\left(\dfrac{17}{4}-1\right)\)
\(=7+\dfrac{13}{4}-\dfrac{3}{5}-\dfrac{23}{5}-\dfrac{17}{4}+1\)
\(=\left(7+1\right)+\left(\dfrac{13}{4}-\dfrac{17}{4}\right)-\left(\dfrac{3}{5}+\dfrac{23}{5}\right)\)
\(=8-\dfrac{4}{4}-\dfrac{26}{5}\)
\(=7-\dfrac{26}{5}\)
\(=\dfrac{9}{5}\)
\(b,\dfrac{2}{3}-\left[\left(-\dfrac{7}{4}\right)-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)
\(=\dfrac{2}{3}-\left(-\dfrac{7}{4}-\dfrac{1}{2}-\dfrac{3}{8}\right)\)
\(=\dfrac{2}{3}-\left(-\dfrac{14}{8}-\dfrac{4}{8}-\dfrac{3}{8}\right)\)
\(=\dfrac{2}{3}-\left(-\dfrac{21}{8}\right)\)
\(=\dfrac{2}{3}+\dfrac{21}{8}\)
\(=\dfrac{79}{24}\)
\(c,\left(9-\dfrac{1}{2}-\dfrac{3}{4}\right):\left(7-\dfrac{1}{4}-\dfrac{5}{8}\right)\)
\(=\left(\dfrac{36}{4}-\dfrac{2}{4}-\dfrac{3}{4}\right):\left(\dfrac{56}{8}-\dfrac{2}{8}-\dfrac{5}{8}\right)\)
\(=\dfrac{31}{4}:\dfrac{49}{8}\)
\(=\dfrac{62}{49}\)
\(d,3-\dfrac{1-\dfrac{1}{7}}{1+\dfrac{1}{7}}=3-\dfrac{\dfrac{7}{7}-\dfrac{1}{7}}{\dfrac{7}{7}+\dfrac{1}{7}}=3-\left(\dfrac{6}{7}:\dfrac{8}{7}\right)=3-\dfrac{3}{4}=\dfrac{9}{4}\)
1/ Chứng tỏ rằng : B=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{8^2}< 1\)
2/ Rút gọn: B=\(\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{20}\right)\)
3/ Tính giá trị của biểu thức: A= \(\dfrac{7}{4}\cdot\left(\dfrac{3333}{1212}+\dfrac{3333}{2020}+\dfrac{3333}{3030}+\dfrac{3333}{4242}\right)\)
4/ So sánh : A= \(\dfrac{2011+2012}{2010+2013}\) và B= \(\dfrac{2011}{2012}+\dfrac{2012}{2013}\)
1/ \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}\)
\(B< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}\)
\(B< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}\)
\(B< \dfrac{1}{1}-\dfrac{1}{8}< 1\)
\(B< 1\)
2/ \(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{20}\right)\)
\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{19}{20}\)
\(B=\dfrac{1\times2\times3\times...\times19}{2\times3\times4\times...\times20}\)
\(B=\dfrac{1}{20}\)
3/ \(A=\dfrac{7}{4}\cdot\left(\dfrac{3333}{1212}+\dfrac{3333}{2020}+\dfrac{3333}{3030}+\dfrac{3333}{4242}\right)\)
\(A=\dfrac{7}{4}\cdot\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)
\(A=\dfrac{7}{4}\cdot\left(\dfrac{33}{3.4}+\dfrac{33}{4.5}+\dfrac{33}{5.6}+\dfrac{33}{6.7}\right)\)
\(A=\dfrac{7}{4}.33.\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)
\(A=\dfrac{231}{4}.\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)
\(A=\dfrac{231}{4}\cdot\dfrac{4}{21}\)
\(A=11\)
4/ A phải là \(\dfrac{2011+2012}{2012+2013}\)
Ta có : \(B=\dfrac{2011}{2012}+\dfrac{2012}{2013}>\dfrac{2011}{2013}+\dfrac{2012}{2013}=\dfrac{2011+2012}{2013}>\dfrac{2011+2012}{2012+2013}=A\)
\(\Rightarrow B>A\)
Tính giá trị của biểu thức sau:
\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2\)
\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}\)
\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.4=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{5}{56}\)
\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)
Tính giá trị biểu thức:
\(e,\dfrac{18}{37}+\dfrac{8}{24}+\dfrac{19}{37}-1\dfrac{23}{24}+\dfrac{2}{3}\)
\(f,\left(-2\right)^3.\left(\dfrac{3}{4}-0,25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)
\(g,\left(\dfrac{2}{5}\right)^2+5\dfrac{1}{2}.\left(4,5-2\right)+\dfrac{2^3}{\left(-4\right)}\)
\(h,\dfrac{4}{9}.19\dfrac{1}{3}-\dfrac{4}{9}.39\dfrac{1}{3}\)
\(i,\left(-\dfrac{1}{2}\right)^2:\dfrac{1}{4}-2\left(-\dfrac{1}{2}\right)^2\)
\(j,125\%.\left(\dfrac{-1}{2}\right)^2:\left(1\dfrac{5}{16}-1,5\right)+2008^0\)
\(k,\left(-2\right)^3.\dfrac{-1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right):\dfrac{5}{12}\)
e: \(=\left(\dfrac{18}{37}+\dfrac{19}{37}\right)+\left(\dfrac{8}{24}+\dfrac{2}{3}\right)-\dfrac{47}{24}=2-\dfrac{47}{24}=\dfrac{1}{24}\)
f: \(=-8\cdot\dfrac{1}{2}:\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=-4:\dfrac{13}{12}=\dfrac{-48}{13}\)
g: \(=\dfrac{4}{25}+\dfrac{11}{2}\cdot\dfrac{5}{2}-\dfrac{8}{4}=\dfrac{4}{25}+\dfrac{55}{4}-2=\dfrac{1191}{100}\)
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
Tính giá trị của biểu thức:
a) (-7,05 - \(\dfrac{1}{4}\)) : (-5) + \(\dfrac{1}{15}\) - \(\left(-\dfrac{1}{5}\right)\) : (-3)
b) \(\left(\dfrac{3}{25}-1,12\right)\) : \(\dfrac{3}{7}\) : [\(\left(3\dfrac{1}{2}\right)\) - \(\left(3\dfrac{2}{3}\right)\) : \(\dfrac{1}{14}\)]
a) = (\(-\dfrac{141}{20}\)- \(\dfrac{1}{4}\)) : (-5) + \(\dfrac{1}{15}\) - \(\dfrac{1}{15}\)
= \(-\dfrac{73}{10}\) : - 5
= \(\dfrac{73}{50}\)
b) = \(\left(\dfrac{3}{25}-\dfrac{28}{25}\right)\). \(\dfrac{7}{3}\) : \(\left(\dfrac{7}{2}-\dfrac{11}{3}.14\right)\)
= \(-\dfrac{7}{3}\) . \(-\dfrac{6}{287}\)
= \(\dfrac{2}{41}\)