Tính hợp lý:
\(B=\dfrac{51\times125-51\times42-17\times150}{3+6+9...+99}\)
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1: =5/8*17/13*7/17*8/3*6
=5/3*7/13*6
=5*7*2/13=70/13
2: =-7/9(24/19-1)
=-7/9*5/19
=-35/171
3: =-3/8(7/6-9/6)
=-3/8*(-2/6)
=3/8*1/3=1/8
\(=\dfrac{17\left(375-126-150\right)}{33\cdot\left(99+3\right):2}=\dfrac{17\cdot99}{33\cdot102:2}=\dfrac{17\cdot3\cdot33}{33\cdot51}=1\)
\(\frac{51.125-51.42-17.150}{3+6+9+...+99}\)=\(\frac{51.125-51.42-17.3.50}{1683}\)=\(\frac{51.125-51.42-71.50}{1683}\)=\(\frac{51\left(125-42-50\right)}{1683}=\frac{51.33}{51.33}=1\)
\(\dfrac{33}{38}:\dfrac{11}{19}=\dfrac{33\times19}{38\times11}=\dfrac{3\times1}{2\times1}=\dfrac{3}{2}\)
a) Ta có: \(\dfrac{-5}{18}+\dfrac{32}{45}-\dfrac{9}{10}\)
\(=\dfrac{-25}{90}+\dfrac{64}{90}-\dfrac{81}{90}\)
\(=\dfrac{-42}{90}=-\dfrac{7}{15}\)
b) Ta có: \(\left(-\dfrac{1}{4}+\dfrac{51}{33}-\dfrac{5}{3}\right)-\left(-\dfrac{15}{12}+\dfrac{6}{11}-\dfrac{42}{29}\right)\)
\(=\dfrac{-1}{4}+\dfrac{17}{11}-\dfrac{5}{3}+\dfrac{5}{4}-\dfrac{6}{11}+\dfrac{42}{29}\)
\(=\dfrac{-5}{3}+\dfrac{42}{29}\)
\(=\dfrac{-145}{87}+\dfrac{126}{87}=\dfrac{-19}{87}\)
c) Ta có: \(1-\dfrac{1}{2}+2-\dfrac{2}{3}+3-\dfrac{3}{4}+4-\dfrac{1}{4}-3-\dfrac{1}{3}-2-\dfrac{1}{2}-1\)
\(=\left(1-1\right)-\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(2-2\right)-\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(3-3\right)-\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+4\)
\(=-1-1-1+4\)
=1
a) Ta có: −518+3245−910−518+3245−910
=−2590+6490−8190=−2590+6490−8190
=−4290=−715=−4290=−715
b) Ta có: (−14+5133−53)−(−1512+611−4229)(−14+5133−53)−(−1512+611−4229)
=−14+1711−53+54−611+4229=−14+1711−53+54−611+4229
=−53+4229=−53+4229
=−14587+12687=−1987=−14587+12687=−1987
c) Ta có: 1−12+2−23+3−34+4−14−3−13−2−12−11−12+2−23+3−34+4−14−3−13−2−12−1
=(1−1)−(12+12)+(2−2)−(23+13)+(3−3)−(34+14)+4=(1−1)−(12+12)+(2−2)−(23+13)+(3−3)−(34+14)+4
=−1−1−1+4=−1−1−1+4
=1
\(A=17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)
\(=11-\dfrac{15}{17}=\dfrac{172}{17}\)
\(B=\left(31\dfrac{6}{13}+5\dfrac{9}{41}\right)-36\dfrac{6}{12}=36\dfrac{363}{533}-36\dfrac{6}{12}=\dfrac{193}{1066}\)
\(C=27\dfrac{51}{59}-\left(7\dfrac{51}{59}-\dfrac{1}{3}\right)=27\dfrac{51}{59}-7\dfrac{51}{59}+\dfrac{1}{3}=20+\dfrac{1}{3}=\dfrac{61}{3}\)
\(A=17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)
\(=\left(17\dfrac{2}{31}-6\dfrac{2}{31}\right)-\dfrac{15}{17}=11-\dfrac{15}{17}=\dfrac{172}{17}\)
\(B=\left(31\dfrac{6}{13}+5\dfrac{9}{41}\right)-36\dfrac{6}{12}=36\dfrac{363}{533}-36\dfrac{1}{2}=\dfrac{193}{1066}\) (Casio :>)
\(C=27\dfrac{51}{59}-\left(7\dfrac{51}{59}-\dfrac{1}{3}\right)=27\dfrac{51}{59}-7\dfrac{51}{59}+\dfrac{1}{3}\)
\(=20+\dfrac{1}{3}=\dfrac{61}{3}\)
Ta có: \(D=\left(2\dfrac{2}{15}\cdot\dfrac{9}{17}\cdot\dfrac{3}{32}\right):\left(-\dfrac{3}{17}\right)\)
\(=\dfrac{32}{15}\cdot\dfrac{9}{17}\cdot\dfrac{3}{32}\cdot\dfrac{-17}{3}\)
\(=\dfrac{-9}{15}=-\dfrac{3}{5}\)
Tính tử: \(51.125-51.42-17.150\)
\(=51.125-51.42-17.3.50\)
\(=51.125-51.42-51.50\)
\(=51.\left(125-42-50\right)=51.33\)
Tính mẫu:
Số các số hạng: \(\left(99-3\right):3+1=33\)
Mẫu = \(33.\left(\dfrac{99+3}{2}\right)=33.51\)
\(B=1\)