Tính hợp lí:
\(\dfrac{-4}{13}.\dfrac{5}{17}+\dfrac{-12}{13}.\dfrac{4}{17}\)
Giải chi tiết dùm mik nha. Thankss
Tính hợp lí:
\(\left(\dfrac{-2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(\dfrac{-1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)
Giải chi tiết dùm mik nha. Thankss
\(\left(-\dfrac{2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{3}+\dfrac{4}{7}\right)+\dfrac{4}{5}\\ =-\dfrac{5}{21}:\dfrac{4}{5}+\dfrac{5}{21}\\ =\left(-\dfrac{5}{21}+\dfrac{5}{21}\right):\dfrac{4}{5}\\ =0:\dfrac{4}{5}\\ =0.\)
Sửa cho mk dòng đầu là :4/5 và dòng tiếp theo mk thiếu :4/5
Tính hợp lí
1) \((\)\(\dfrac{7}{-18}\)+\(\dfrac{-5}{12}\)) -\(\dfrac{13}{-18}\)
2) \(\dfrac{-13}{17}\)+(\(\dfrac{13}{-21}\)+\(\dfrac{-4}{17}\))
3) ( \(\dfrac{13}{-10}\)-\(\dfrac{-4}{13}\))+\(\dfrac{11}{-10}\)
4) \(\dfrac{13}{17}\)\(\times\)\(\dfrac{4}{-5}\)+\(\dfrac{13}{17}\)\(\times\)\(\dfrac{-3}{4}\)
5) \(\dfrac{-5}{12}\)\(\times\)\(\dfrac{7}{-17}\)\(\times\)\(\dfrac{9}{-20}\)
6) \(\dfrac{5}{9}\)\(\times\)\(\dfrac{11}{23}\)+ \(\dfrac{11}{23}\)\(\times\)\(\dfrac{17}{9}\)\(-\)\(\dfrac{13}{9}\)\(\times\)\(\dfrac{11}{23}\)
\(1.\dfrac{-7}{18}+\dfrac{-5}{12}-\dfrac{-13}{18}\text{=}\left(\dfrac{-7}{18}-\dfrac{-13}{18}\right)+\dfrac{-5}{12}\text{=}\dfrac{1}{3}+\dfrac{-5}{12}\text{=}\dfrac{-1}{12}\)
\(2.\dfrac{-13}{17}+\dfrac{-13}{21}+\dfrac{-4}{17}\text{=}\left(\dfrac{-13}{17}+\dfrac{-4}{17}\right)+\dfrac{-13}{21}\text{=}-1+\dfrac{-13}{21}\text{=}\dfrac{-34}{21}\)
\(3.\dfrac{-13}{10}-\dfrac{-4}{13}+\dfrac{-11}{10}\text{=}\dfrac{-12}{5}-\dfrac{-4}{13}\text{=}\dfrac{-136}{65}\)
\(4.\dfrac{13}{17}\times\left(\dfrac{-4}{5}+\dfrac{-3}{4}\right)\text{=}\dfrac{13}{17}\times\dfrac{-31}{20}\text{=}\dfrac{-403}{340}\)
\(5.\left(\dfrac{-5}{12}\times\dfrac{-9}{20}\right)\times\dfrac{-7}{17}\text{=}\dfrac{3}{16}\times\dfrac{-7}{17}\text{=}\dfrac{-21}{272}\)
\(6.\dfrac{11}{23}\times\left(\dfrac{5}{9}+\dfrac{17}{9}-\dfrac{13}{9}\right)\text{=}\dfrac{11}{23}\times1\text{=}\dfrac{11}{23}\)
Tính
\(\dfrac{12^3.18^2}{24^2}\)
Giải chi tiết dùm mik với nha. Thankss
\(\dfrac{12^3.18^2}{24^2}=\dfrac{12^3.6^2.3^2}{6^2.4^2}=\dfrac{4^3.3^3.3^2}{4^2}=4.3^3.3^2=4.3^5=972\)
\(\dfrac{12^3.18^2}{24^2}=\dfrac{1728.324}{576}=\dfrac{559872}{576}=972\)
Vậy giá trị cần tìm là 972
\(=\dfrac{\left(2^2\cdot3\right)^3\cdot\left(2\cdot3^2\right)^2}{\left(2^3\cdot3\right)^2}\)
\(=\dfrac{2^{2\cdot3}\cdot3^3\cdot2^2\cdot3^{2\cdot2}}{2^{3\cdot2}\cdot3^2}\)
\(=\dfrac{2^6\cdot2^2\cdot3^3\cdot3^4}{2^6\cdot3^2}\)
\(=2^2\cdot3^5\) \(=4\cdot243\) \(=486\)
Tính:
\(A=\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+...+\dfrac{1}{44.49}\right).\dfrac{1-3-5-7-...-49}{89}\)
Giải chi tiết dùm mik nha. Thankss
\(A=\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+..+\dfrac{1}{44.49}\right)\left(\dfrac{1-3-5-7-..-49}{89}\right)\\ A=\dfrac{1}{5}\left(\dfrac{5}{4.9}+\dfrac{5}{9.14}+..+\dfrac{5}{44.49}\right)\left(\dfrac{1-3-5-7-...-49}{89}\right)\\ A=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\left(\dfrac{1-3-5-7-...-49}{89}\right)\)
\(A=\dfrac{9}{196}\left(\dfrac{1-3-5-7-...-49}{89}\right)\)
Ta đặt: \(P=1-3-5-7-...-49\\ =1-\left(3+5+7+..+49\right)\\ =1-624\\ =-623\\ \Rightarrow\dfrac{9}{196}.-\dfrac{623}{89}=-\dfrac{9}{28}.\)
\(\dfrac{9}{17}\) x \(\dfrac{21}{13}\) + \(\dfrac{9}{17}\) x \(\dfrac{5}{13}\) - \(\dfrac{9}{17}\) x 2
Các bạn giải giúp mình chi tiết nha! Tks
\(\dfrac{9}{17}\times\dfrac{21}{13}+\dfrac{9}{17}\times\dfrac{5}{13}-\dfrac{9}{17}\times2\)
\(=\dfrac{9}{17}\times\left(\dfrac{21}{13}+\dfrac{5}{13}-2\right)\)
\(=\dfrac{9}{17}\times\left(\dfrac{26}{13}-2\right)=\dfrac{9}{17}\times\left(2-2\right)\)
\(=\dfrac{9}{17}\times0=0\)
= 9/17 x ( 21/13 + 5/13 - 2 )
= 9/17 x 0
= 0
=9/17 x (21/13 + 5/13 - 2)
=9/17 x (26/13 - 2)
=9/17 x (13 - 2)
=9/17 x 11
= 99/17
Tick cho mik nhé!!!
Cho \(A=\left(\dfrac{1}{4}-1\right).\left(\dfrac{1}{9}-1\right)...\left(\dfrac{1}{100}-1\right)\)
So sánh \(A\) với \(\dfrac{-11}{21}\)
Giải chi tiết dùm mik nha. Thankss
\(A=\left(\dfrac{1}{4}-1\right).\left(\dfrac{1}{9}-1\right)....\left(\dfrac{1}{100}-1\right).\)
\(\Rightarrow A=\left(-\dfrac{3}{4}\right).\left(-\dfrac{8}{9}\right)....\left(-\dfrac{99}{100}\right)\)
mà A có 9 dấu - \(\left(4;9;16;25;36;49;64;81;100\right)\)
\(\Rightarrow0>A=\left(-\dfrac{3}{4}\right).\left(-\dfrac{8}{9}\right)....\left(-\dfrac{99}{100}\right)=-\dfrac{1}{2}\)
Ta lại có \(\left\{{}\begin{matrix}\dfrac{1}{2}=\dfrac{21}{42}\\\dfrac{11}{21}=\dfrac{22}{42}\end{matrix}\right.\) \(\Rightarrow\dfrac{1}{2}< \dfrac{11}{21}\Rightarrow-\dfrac{1}{2}>-\dfrac{11}{21}\)
\(\Rightarrow A>-\dfrac{11}{21}\)
\(A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)...\left(\dfrac{1}{100}-1\right)\)
\(A=\left(-\dfrac{2^2-1}{2^2}\right)\left(-\dfrac{3^2-1}{3^2}\right)...\left(-\dfrac{10^2-1}{10^2}\right)\)
\(A=\left[-\dfrac{1\cdot3}{2\cdot2}\right]\left[-\dfrac{2\cdot4}{3\cdot3}\right]...\left[-\dfrac{9\cdot11}{10\cdot10}\right]\)
Dễ thấy A có 9 thừa số, suy ra
\(A=-\dfrac{1\cdot3\cdot2\cdot4\cdot...\cdot9\cdot11}{2\cdot2\cdot3\cdot3\cdot...\cdot10.10}=-\dfrac{1\cdot11}{2\cdot10}=\dfrac{-11}{20}\)
Vì 20 < 21 nên \(\dfrac{11}{20}>\dfrac{11}{21}\), suy ra \(\dfrac{-11}{20}< \dfrac{-11}{21}\)
Vậy \(A< \dfrac{-11}{21}\)
Tính
\(\dfrac{6^3+2.6^2+2^3}{37}\)
Giải chi tiết dùm mik nha. Thankss
\(\dfrac{6^3+2.6^2+2^3}{37}=\dfrac{2^3.3^3+2.2^2.3^2+2^3}{37}\\ \\ \\ \\ \\ \\ \\ \\ \\=\dfrac{2^3.3^3+2^3.3^2+2^3}{37}\\ \\ \\ \\ \\ \\ \\ \\ \\ =\dfrac{2^3.\left(3^3+3^2+1\right)}{37}=\dfrac{2^3.37}{37}=2^3=8\)
\(\dfrac{6^3+2.6^2+2^3}{37}=\dfrac{216+72+8}{37}=\dfrac{296}{37}=8\)
\(\dfrac{-5}{17}\) x \(\dfrac{11}{25}\) + \(\dfrac{-5}{17}\) x \(\dfrac{26}{25}\) - \(\dfrac{-5}{17}\) x \(\dfrac{3}{25}\)
Các bạn giải chi tiết dùm mình nha, tks
Tính (tính nhanh nếu có thể):
a) \(\dfrac{-13}{4}+\dfrac{19}{12}+\dfrac{5}{6}\)
b) \(\dfrac{7.6^{10}.2^{20}.3^6-2^{19}.6^{15}}{9.6^{19}.2^9-4.3^{17}.2^{26}}\)
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a: \(=\dfrac{-39+19+10}{12}=\dfrac{-10}{12}=\dfrac{-5}{6}\)
b: \(=\dfrac{2^{30}\cdot3^{16}\cdot7-2^{34}\cdot3^{15}}{2^{28}\cdot3^{21}-2^{28}\cdot3^{17}}\)
\(=\dfrac{2^{30}\cdot3^{15}\left(3\cdot7-2^4\right)}{2^{28}\cdot3^{17}\left(3^4-1\right)}=\dfrac{2^2}{3^2}\cdot\dfrac{21-16}{80}=\dfrac{4}{9}\cdot\dfrac{5}{80}\)
\(=\dfrac{20}{720}=\dfrac{1}{36}\)