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\(\frac{17}{23}.\frac{8}{6}.\frac{23}{17}.\left(-80\right).\frac{3}{4}\)
\(=\left(\frac{17}{23}.\frac{23}{17}\right).\left(\frac{8}{6}.\frac{3}{4}\right).\left(-80\right)\)
\(=1.1.\left(-80\right)\)
\(=-80\)
a) \(\frac{17}{23}.\frac{18}{16}.\frac{23}{17}-\left(-80\right)-\frac{3}{4}\)
= \(\frac{17.18.23}{23.16.17}+80-\frac{3}{4}\)
= \(\frac{9}{8}+80-\frac{3}{4}\)
= \(\frac{3}{8}+80=\frac{643}{8}\)
b) \(\frac{5}{11}.\frac{18}{29}-\frac{5}{11}.\frac{8}{29}+\frac{5}{11}.\frac{19}{29}\)
= \(\frac{5}{11}.\left(\frac{18}{29}-\frac{8}{29}+\frac{19}{29}\right)\)
= \(\frac{5}{11}.1=\frac{5}{11}\)
c) \(\left(\frac{13}{23}+\frac{1313}{2323}-\frac{131313}{232323}\right).\left(\frac{1}{3}+\frac{1}{4}-\frac{7}{12}\right)\)
= \(\left(\frac{13}{23}+\frac{1313}{2323}-\frac{131313}{232323}\right).0\)
= \(0\)
\(=\dfrac{17}{23}\cdot\dfrac{23}{17}\cdot\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\left(-80\right)=-80\cdot\dfrac{3}{8}=-30\)
Bài 1:
a; \(\dfrac{5}{18}\) + \(\dfrac{8}{19}\) - \(\dfrac{7}{21}\) + (- \(\dfrac{10}{36}\) + \(\dfrac{11}{19}\) + \(\dfrac{1}{3}\)) - \(\dfrac{5}{8}\)
= \(\dfrac{5}{18}\) + \(\dfrac{8}{19}\) - \(\dfrac{1}{3}\) -\(\dfrac{10}{36}\) + \(\dfrac{11}{19}\) + \(\dfrac{1}{3}\) - \(\dfrac{5}{8}\)
= (\(\dfrac{5}{18}\) - \(\dfrac{10}{36}\)) + (\(\dfrac{8}{19}\) + \(\dfrac{11}{19}\)) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\)) - \(\dfrac{5}{8}\)
= (\(\dfrac{5}{18}\) - \(\dfrac{5}{18}\)) + \(\dfrac{19}{19}\) - 0 - \(\dfrac{5}{8}\)
= 0 + 1 - \(\dfrac{5}{8}\)
= \(\dfrac{3}{8}\)
b; \(\dfrac{1}{13}\) + (\(\dfrac{-5}{18}\) - \(\dfrac{1}{13}\) + \(\dfrac{12}{17}\)) - (\(\dfrac{12}{17}\) - \(\dfrac{5}{18}\) + \(\dfrac{7}{5}\))
= \(\dfrac{1}{13}\) - \(\dfrac{5}{18}\) - \(\dfrac{1}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{12}{17}\) + \(\dfrac{5}{18}\) - \(\dfrac{7}{5}\)
= (\(\dfrac{1}{13}\) - \(\dfrac{1}{13}\)) + (\(\dfrac{12}{17}\) - \(\dfrac{12}{17}\)) + (-\(\dfrac{5}{18}\) + \(\dfrac{5}{18}\)) - \(\dfrac{7}{5}\)
= 0 + 0 + 0 - \(\dfrac{7}{5}\)
= - \(\dfrac{7}{5}\)
Bài 1 c;
\(\dfrac{15}{14}\) - (\(\dfrac{17}{23}\) - \(\dfrac{80}{87}\) + \(\dfrac{5}{4}\)) + (\(\dfrac{17}{23}\) - \(\dfrac{15}{14}\) + \(\dfrac{1}{4}\))
= \(\dfrac{15}{14}\) - \(\dfrac{17}{23}\) + \(\dfrac{80}{87}\) - \(\dfrac{5}{4}\) + \(\dfrac{17}{23}\) - \(\dfrac{15}{14}\) + \(\dfrac{1}{4}\)
= (\(\dfrac{15}{14}-\dfrac{15}{14}\)) + (\(-\dfrac{17}{23}+\dfrac{17}{23}\)) - (\(\dfrac{5}{4}\) - \(\dfrac{1}{4}\)) + \(\dfrac{80}{87}\)
= 0 + 0 - 1 + \(\dfrac{80}{87}\)
= - \(\dfrac{7}{87}\)
\(A=98.42-\left\{50.\left[\left(18-2^3\right):2+3^2\right]\right\}\)
\(=98.42-\left\{50.\left[\left(18-8\right):2+9\right]\right\}\)
\(=98.42-\left[50\left(10:2+9\right)\right]\)
\(=98.42-\left(50.14\right)\)
\(=4116-700=3416\)
\(B=-80-\left[-130-\left(12-4\right)^2\right]+2008^0\)
\(=-80-\left(-130-8^2\right)+1\)
\(=-80-\left(-130-64\right)+1\)
\(=-80+130+64+1\)
\(=115\)
\(C=1024:2^4+140:\left(38+2^5\right)-7^{23}:7^{21}\)
\(=1024:16+140:\left(38+32\right)-7^2\)
\(=64+140:70-49\)
\(=64+2-49=17\)
\(D=\left(2^{17}+15^4\right).\left(3^{19}-2^{17}\right).\left(2^4-4^2\right)\)
\(=\left(2^{17}+15^4\right).\left(3^{19}-2^{17}\right).\left(16-16\right)\)
\(=\left(2^{17}+15^4\right).\left(3^{19}-2^{17}\right).0\)
\(=0\)
\(E=100+98+96+....+4+2-97-95-....-3-1\)
\(=100+\left(98-97\right)+\left(96-95\right)+.....+\left(2-1\right)+\left(1-0\right)\)
\(=100+1+1+...+1+1\)
Vì lập được 49 cặp nên sẽ có 49 số 1
\(\Rightarrow E=100+1.49=100+49=149\)
a) \(17+188+183=205+183=388\)
b) \(80-\left[130-\left(12-4\right)^2\right]\)
\(=80-\left(130-64\right)\)
\(80-66=14\)
c) \(2^3.17-2^3.14\)
\(8.17-8.14\)
\(=8.\left(17-14\right)\)
\(=8.3=24\)
\(\frac{17.8.23.\left(-80\right).3}{23.16.17.4}=\frac{8.\left(-80\right).3}{16.4.17}=\frac{1.\left(-20\right).3}{2.1.17}=\frac{\left(-10\right).3}{1.17}=\frac{-30}{17}=-1\frac{13}{17}\)
Bằng -30 nha bạn chắc chắn 100%