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\(A=\dfrac{3^{10}\cdot11+3^{10}\cdot5}{3^9\cdot2^4}=\dfrac{3^{10}\cdot\left(11+5\right)}{3^9\cdot16}=\dfrac{3^{10}\cdot16}{3^9\cdot16}=3\)
\(B=\dfrac{2^{10}\cdot13+2^{10}\cdot65}{2^8\cdot104}=\dfrac{2^{10}\cdot\left(13+65\right)}{2^8\cdot2^2\cdot26}=\dfrac{2^{10}\cdot78}{2^{10}\cdot26}=3\)
\(C=\dfrac{72^3\cdot54^2}{108^4}=\dfrac{\left(2^3\cdot3^2\right)^3\cdot\left(2\cdot3^3\right)^2}{\left(3^3\cdot2^2\right)^4}\\ =\dfrac{2^9\cdot3^6\cdot2^4\cdot3^6}{3^{12}\cdot2^8}=\dfrac{2^{13}\cdot3^{12}}{3^{12}\cdot2^8}=2^5=32\)
\(D=\dfrac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=\dfrac{11\cdot3^{29}-\left(3^2\right)^{15}}{2^2\cdot3^{28}}=\dfrac{11\cdot3^{29}-3^{30}}{2^2\cdot3^{28}}\\ =\dfrac{3^{29}\cdot\left(11-3\right)}{2^2\cdot3^{28}}=\dfrac{3^{29}\cdot8}{4\cdot3^{28}}=3\cdot2=6\)
a: \(=\dfrac{14-2+9}{32}\cdot\dfrac{4}{5}=\dfrac{21}{5}\cdot\dfrac{1}{8}=\dfrac{21}{40}\)
b: \(=10+\dfrac{2}{9}+2+\dfrac{3}{5}+6+\dfrac{2}{9}=18+\dfrac{47}{45}=\dfrac{857}{45}\)
c: \(=\dfrac{3}{10}-\dfrac{12}{5}+\dfrac{1}{10}=\dfrac{4}{10}-\dfrac{12}{5}=\dfrac{2}{5}-\dfrac{12}{5}=-2\)
d: \(=\dfrac{-25}{30}\left(\dfrac{37}{44}+\dfrac{13}{44}-\dfrac{6}{44}\right)=\dfrac{-25}{30}\cdot1=-\dfrac{5}{6}\)
A=\(\dfrac{2^{10}\left(13+65\right)}{2^8.104}\)
=\(\dfrac{2.78}{104}\)=\(\dfrac{78}{52}\)=\(\dfrac{39}{26}\)
\(\dfrac{2}{3}+\dfrac{1}{5}.\dfrac{10}{7}=\dfrac{2}{3}+\dfrac{10}{35}=\dfrac{70}{105}+\dfrac{30}{105}=\dfrac{100}{105}=\dfrac{50}{21}\)
a) Ta có: \(\dfrac{2}{3}+\dfrac{1}{5}\cdot\dfrac{10}{7}\)
\(=\dfrac{2}{3}+\dfrac{2}{7}\)
\(=\dfrac{14}{21}+\dfrac{6}{21}\)
\(=\dfrac{20}{21}\)
2:
a: 2/9-x=-5/9
=>x=2/9+5/9=7/9
b: x-7/13=1/2
=>x=1/2+7/13=27/26
câu a
\(\dfrac{7}{4}+\dfrac{3}{2}+\dfrac{-9}{16}\\ =\dfrac{28}{16}+\dfrac{24}{16}-\dfrac{9}{16}=\dfrac{43}{16}\)
câu b
\(-\dfrac{2}{7}+\dfrac{3}{5}+\dfrac{9}{7}+\dfrac{-18}{5}\\ =-\dfrac{10}{35}+\dfrac{21}{35}+\dfrac{45}{35}-\dfrac{126}{35}\\ =-\dfrac{70}{35}=-2\)
câu c
\(-\dfrac{5}{13}+\dfrac{11}{10}-\dfrac{-9}{10}+\dfrac{-8}{13}\\ =-\dfrac{5}{13}+\dfrac{11}{10}+\dfrac{9}{10}-\dfrac{8}{13}\\ =-\dfrac{50}{130}+\dfrac{143}{130}+\dfrac{117}{130}-\dfrac{80}{130}\\ =\dfrac{130}{130}=1\)
bài 2
câu a
\(\dfrac{2}{9}-x=-\dfrac{5}{9}\\ x=\dfrac{2}{9}-\dfrac{-5}{9}\\ x=\dfrac{7}{9}\)
câu b
\(x+\dfrac{-7}{13}=\dfrac{1}{2}\\ x=\dfrac{1}{2}-\dfrac{-7}{13}\\ x=\dfrac{13}{26}+\dfrac{14}{26}\\ x=\dfrac{17}{26}\)
1) = (8/31 + 23/31)+ ( -12/25+-13/25) = 1 + (-1) = 0
2) = 1/2 +3/4 -3/4 + 4/5 = 1/2 +4/5 = 13/10
3)= ( 7/3 x 15/21) x ( -5/2 x 4/-5 ) = ( 7/3 x 5/7 ) x 2 = 5/3 x 2 = 10/3
4) = 1/4 + 3/4 x -7/6 = 1/4 + -7/8 = -5/8
5)= 3/29 x 29/3 - 1/5 x 29/3 = 1 x 29/15 = 29/15
6)= 5/7 x ( 5/11 + 2/11 - 14/11) = 5/7 x -7/11 = -5 /11
7) = 11/12 x -1/8 + 11/12 x -3/16 - 11/12 = 11/12 x ( -1/8 + -3/16 - 1) = 11/12 x -21/16 = -77/64 ( mk ko chắc , bạn ấn máy tính để thử lại )
8) = 203/20 - 7/4 + 41/13 = 198/20 + 41/13 = 99/10 + 41/13 = 1697/130 ( câu này cứ lỏ lỏ kiểu gì ý :v )
1) = (8/31 + 23/31)+ ( -12/25+-13/25) = 1 + (-1) = 0
2) = 1/2 +3/4 -3/4 + 4/5 = 1/2 +4/5 = 13/10
3)= ( 7/3 x 15/21) x ( -5/2 x 4/-5 ) = ( 7/3 x 5/7 ) x 2 = 5/3 x 2 = 10/3
4) = 1/4 + 3/4 x -7/6 = 1/4 + -7/8 = -5/8
5)= 3/29 x 29/3 - 1/5 x 29/3 = 1 x 29/15 = 29/15
6)= 5/7 x ( 5/11 + 2/11 - 14/11) = 5/7 x -7/11 = -5 /11
7) = 11/12 x -1/8 + 11/12 x -3/16 - 11/12 = 11/12 x ( -1/8 + -3/16 - 1) = 11/12 x -21/16 = -77/64 ( mk ko chắc , bạn ấn máy tính để thử lại )
8) = 203/20 - 7/4 + 41/13 = 198/20 + 41/13 = 99/10 + 41/13 = 1697/130 ( câu này cứ lỏ lỏ kiểu gì ý :v )
a: \(=\dfrac{7+12-6}{13}=1\)
b: \(=\dfrac{13}{10}\cdot\dfrac{6-26}{13}=\dfrac{-20}{10}=-2\)
c: \(=\dfrac{3}{4}\cdot2-\dfrac{5}{2}\cdot\dfrac{-4}{3}=\dfrac{3}{2}+\dfrac{20}{6}=\dfrac{3}{2}+\dfrac{10}{3}=\dfrac{29}{6}\)
d: \(=\dfrac{3}{8}\cdot\dfrac{8}{5}+\dfrac{3}{5}\cdot\dfrac{2}{7}+\dfrac{3}{5}\cdot\dfrac{5}{7}=\dfrac{3}{5}+\dfrac{3}{5}=\dfrac{6}{5}\)
cảm ơn bn, mình đặt câu hỏi, bn thườg xuyên trả lời câu hỏi của mình. Thank you very much.
Sửa đề: \(C=1+3^1+3^2+...+3^{100}\)
b) Ta có: \(C=1+3^1+3^2+...+3^{100}\)
\(\Leftrightarrow3\cdot C=3+3^2+...+3^{101}\)
\(\Leftrightarrow C-3\cdot C=1+3+3^2+...+3^{100}-3-3^2-...-3^{100}-3^{101}\)
\(\Leftrightarrow-2\cdot C=1-3^{101}\)
hay \(C=\dfrac{3^{101}-1}{2}\)
b) Ta có: C=1+31+32+...+3100C=1+31+32+...+3100
⇔3⋅C=3+32+...+3101⇔3⋅C=3+32+...+3101
⇔C−3⋅C=1+3+32+...+3100−3−32−...−3100−3101⇔C−3⋅C=1+3+32+...+3100−3−32−...−3100−3101
⇔−2⋅C=1−3101