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a, \(4\times\left(-\dfrac{1}{2}\right)^3-2\times\left(-\dfrac{1}{2}\right)^2+3\times\left(-\dfrac{1}{2}\right)+1\)
\(=\left(-\dfrac{1}{2}\right)\left[\left(4\times-\dfrac{1}{2}\right)-\left(2\times-\dfrac{1}{2}\right)+3\right]+1\)
\(=\left(-\dfrac{1}{2}\right)\left(-2+1+3\right)+1\)
\(=\left(-\dfrac{1}{2}\right)2+1\)
\(=-1+1\)
\(=0\)
@Trịnh Thị Thảo Nhi
a, 4×(−12)3−2×(−12)2+3×(−12)+14×(−12)3−2×(−12)2+3×(−12)+1
=(−12)[(4×−12)−(2×−12)+3]+1=(−12)[(4×−12)−(2×−12)+3]+1
=(−12)(−2+1+3)+1=(−12)(−2+1+3)+1
=(−12)2+1=(−12)2+1
=−1+1=−1+1
=0=0
a)1.2.3.4...9-1.2.3.4...8-1.2.3.4...8.8
=1.2.3.4...8(9-1-8)
=1.2.3.4...8.0
=0
b)(3.4.216)2/11.123.411-169=(3.22.216)2/11.213.222-236=32.24.232/11.235-236=32.226/235.(11-2)
=32.236/235.9=32.236/235.32=2
c)70.(131313/565656+131313/727272+131313/909090
=70.(13/56+13/72+13/90)
=70.39/70=39
d)1/4.9+1/9.14+1/14.19+...+1/64.69
=4/4.9.4+4/9.4.14+4/14.19.4+...+4/64.69.4.
=1/4.(4/4.9+4/9.14+4/14.19+...+4/64.69)
=1/4.(1/4-1/9+1/9-1/14+1/14-1/19+...+1/64-1/69)
=1/4.(1/4-1/69)
=1/4.65/276=65/1104
~~~~~~~~Chúc bạn học giỏi nhé !~~~~~~~~
\(1\frac{13}{15}\cdot3\cdot(0,5)^2\cdot3+\left[\frac{8}{15}-1\frac{19}{60}:1\frac{23}{24}\right]\)
\(=\frac{28}{15}\cdot3\cdot0,5\cdot0,5\cdot3+\left[\frac{8}{15}-\frac{79}{60}:\frac{47}{24}\right]\)
\(=\frac{28}{5}\cdot0,25\cdot3+\left[\frac{32}{60}-\frac{79}{60}\cdot\frac{24}{47}\right]\)
\(=\frac{28}{5}\cdot\frac{25}{100}\cdot3+\left[\frac{32}{60}-\frac{158}{235}\right]\)
\(=\frac{28}{5}\cdot\frac{1}{4}\cdot3+\frac{-98}{705}=\frac{7}{5}\cdot1\cdot3+\frac{-98}{705}\)
Đến đây là tính dễ rồi :v
\((-3,2)\cdot\frac{-15}{64}+\left[0,8-2\frac{4}{15}\right]:1\frac{23}{24}\)
\(=\frac{-32}{10}\cdot\frac{-15}{64}+\left[\frac{8}{10}-\frac{34}{15}\right]:\frac{47}{24}\)
\(=\frac{-32\cdot(-15)}{10\cdot64}+\left[\frac{4}{5}-\frac{34}{15}\right]:\frac{47}{24}\)
\(=\frac{-1\cdot(-3)}{2\cdot2}+\frac{4\cdot3-34}{15}:\frac{47}{24}\)
\(=\frac{3}{4}+\frac{-22}{15}:\frac{47}{24}\)
\(=\frac{3}{4}+\frac{-517}{180}=\frac{-191}{90}\)
Bài 2 : \(\frac{2\cdot(-13)\cdot9\cdot10}{(-3)\cdot4\cdot(-5)\cdot26}=\frac{1\cdot(-1)\cdot3\cdot2}{(-1)\cdot2\cdot(-1)\cdot2}=\frac{1\cdot3}{-1\cdot2}=\frac{3}{-2}=\frac{-3}{2}\)
\(\frac{15\cdot8+15\cdot4}{12\cdot3}=\frac{15\cdot(8+4)}{12\cdot3}=\frac{15\cdot12}{12\cdot3}=\frac{15}{3}=5\)
Ta có : S = \(\frac{5.2^{30}.6^3.3^{15}-2^3.8^9.3^{17}.21}{21.2^{29}.3^{16}.4-2^{29}.\left(3^4\right)^5}=\frac{5.2^{30}.\left(2.3\right)^3.3^{15}-2^3.\left(2^3\right)^9.3^{17}.3.7}{3.7.2^{29}.3^{16}.2^2-2^{29}.3^{20}}=\frac{5.2^{33}.3^{18}-2^{30}.3^{18}.7}{3^{17}.7.2^{31}-2^{29}.3^{20}}\)
\(=\frac{2^{30}.3^{18}.\left(5.2^3-7\right)}{3^{17}.2^{29}.\left(7.2^2-3^3\right)}=2.3.33=198\)
\(\dfrac{5\times4^{15}\times9^9-4\times3^{20}\times8^9}{5\times2^{10}\times6^{19}-7\times2^{29}\times27^6}\\ =\dfrac{5\times2^{30}\times3^{18}-2^2\times3^{20}\times2^{27}}{5\times2^{10}\times3^{19}\times2^{19}-7\times2^{29}\times3^{18}}\\ =\dfrac{5\times2^{30}\times3^{18}-2^{29}\times3^{20}}{5\times2^{29}\times3^{19}-7\times2^{29}\times3^{18}}\\ =\dfrac{2^{29}\times3^{18}\times\left(5\times2-3^2\right)}{2^{29}\times3^{18}\times\left(5\times3-7\right)}\\ =\dfrac{10-9}{15-7}\\ =\dfrac{1}{8}\)
Ta có :
\(S=\frac{5\times2^{30}\times6^2\times3^{15}-2^3\times8^9\times3^{17}\times21}{21\times2^{29}\times3^{16}\times4-2^{29}\times\left(3^4\right)^5}\)
\(S=\frac{5\times2^{30}\times2^2\times3^2\times3^{15}-2^3\times2^{27}\times3^{17}\times3\times7}{3\times7\times2^{29}\times3^{16}\times2^2-2^{29}\times3^{20}}\)
\(S=\frac{5\times2^{32}\times3^{17}-2^{30}\times3^{18}\times7}{7\times2^{31}\times3^{17}-2^{29}\times3^{20}}\)
\(S=\frac{2^{30}\times3^{17}\times\left(5\times2^2-3\times7\right)}{2^{29}\times3^{17}\times\left(2^2\times7-3^3\right)}\)
\(S=\frac{2^{30}\times3^{17}\times\left(-1\right)}{2^{29}\times3^{17}\times1}\)
\(\Rightarrow S=-2\)
Ko viết đề :)
\(S=\frac{5\cdot2^{30}\cdot2^2\cdot3^2\cdot3^{15}-2^3\cdot2^{27}\cdot3^{17}\cdot3\cdot7}{3\cdot7\cdot2^{29}\cdot3^{16}\cdot2^2-2^{29}\cdot3^{20}}\)
\(=\frac{5\cdot2^{32}\cdot3^{17}-2^{30}\cdot3^{18}\cdot7}{3^{17}\cdot7\cdot2^{31}-2^{29}\cdot3^{20}}\)
\(=\frac{2^{30}\cdot3^{17}\left(5\cdot2^2-3\cdot7\right)}{2^{29}\cdot3^{17}\left(7\cdot2^2-3^3\right)}\)
\(=\frac{2\left(20-21\right)}{28-27}\)
\(=\frac{40-42}{1}=-\frac{2}{1}=-2\)
Vậy S= -2
\(A=\dfrac{2^{19}\cdot3^9-3\cdot5\cdot2^{18}\cdot3^8}{2^9\cdot2^{10}\cdot3^9+2^{20}\cdot3^{10}}\)
\(=\dfrac{2^{19}\cdot3^9-2^{18}\cdot3^9\cdot5}{2^{19}\cdot3^9+2^{20}\cdot3^{10}}\)
\(=\dfrac{2^{18}\cdot3^9\left(2-5\right)}{2^{19}\cdot3^9\cdot7}=\dfrac{1}{2}\cdot\dfrac{-3}{7}=\dfrac{-3}{14}\)
\(E=\dfrac{11.3^{29}-3^{2^{15}}}{2.3^{14}.2.3^{14}}\)
\(=\dfrac{11.3-3^{30}}{2^2}=\dfrac{33-3^{30}}{4}\)
a: \(=\left(\dfrac{17}{10}+\dfrac{70}{10}-\dfrac{87}{10}\right):\left(\dfrac{23}{4}-\dfrac{11}{4}+\dfrac{9}{25}\right)\cdot\left(12,98\cdot0,25\right)+12,5\)
\(=0:\left(3+\dfrac{9}{25}\right)\cdot\left(12,98+0,25\right)+12,5\)
=12,5
b: \(=\dfrac{13}{12}\cdot\dfrac{27}{5}\cdot2\cdot\dfrac{34}{9}\cdot2\cdot\dfrac{2}{17}\)
\(=\dfrac{13}{12}\cdot2\cdot\dfrac{27}{5}\cdot\dfrac{34}{9}\cdot\dfrac{4}{17}\)
\(=\dfrac{13}{6}\cdot\dfrac{27}{5}\cdot\dfrac{8}{9}=\dfrac{8}{6}\cdot3\cdot\dfrac{13}{5}=4\cdot\dfrac{13}{5}=\dfrac{52}{5}\)
\(C=\dfrac{5\times2^{12}\times3^8-3^9\times2^{12}}{2^2\times2^{13}\times3^8+2\times2^{12}\times\left(-3^9\right)}=\dfrac{3^8\times2^{12}\times\left(5-3\right)}{2^{15}\times3^8+2^{13}\times\left(-3\right)^9}\)
\(=\dfrac{3^8\times2^{12}\times2}{2^{13}\times3^8\times\left(4-3\right)}=\dfrac{1}{1}=1\)
\(#PaooNqoccc\)