2.3^2-50:5^2
136.8-36.2^3
84:4+3^9:3^7+5^0
5.3^3+7^11:7^9-12^0
pls help mee
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\(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}-\dfrac{4}{5}\right)\)
=\(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}\)
=\(\left(\dfrac{1}{2}+\dfrac{4}{5}\right)+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)\)
=\(\left(\dfrac{5}{10}+\dfrac{8}{10}\right)+0\)
=\(\dfrac{13}{10}\)
\(-\dfrac{7}{25}.\dfrac{11}{13}+\left(-\dfrac{7}{25}\right).\dfrac{2}{13}-\dfrac{18}{25}\)
=\(-\dfrac{7}{25}.\cdot\left(\dfrac{11}{13}+\dfrac{2}{13}\right)-\dfrac{18}{25}\)
=\(-\dfrac{7}{25}.1-\dfrac{18}{25}\)
=\(-\dfrac{7}{25}-\dfrac{18}{25}\)
=\(-\dfrac{25}{25}\) = \(-1\)
a: =(5/7+2/7)+(4/3+5/3)=3+1=4
b: =(17/12+7/12)+(29/7-8/7)
=2+3=5
c: =(2/5+3/5)+(6/9+1/3)+(7/4+1/4)
=1+2+1
=4
d: =(1/3+2/3)+(13/17+4/17)+(29/11+4/11)
=1+1+3=5
\(1)\)\(-\dfrac{10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
\(=\dfrac{10}{11}\left(-\dfrac{8}{9}+\dfrac{7}{18}\right)\)
\(=\dfrac{10}{11}\left(\dfrac{-16}{18}+\dfrac{7}{18}\right)\)
\(=\dfrac{10}{11}.\left(-\dfrac{1}{2}\right)=-\dfrac{5}{11}\)
\(2)\)\(\dfrac{12}{25}.\dfrac{23}{7}-\dfrac{12}{7}.\dfrac{13}{25}\)
\(=\dfrac{12}{7}.\dfrac{23}{25}-\dfrac{12}{7}.\dfrac{13}{25}\)
\(=\dfrac{12}{7}.\left(\dfrac{23}{25}-\dfrac{13}{25}\right)\)
\(=\dfrac{12}{7}.\dfrac{2}{5}=\dfrac{24}{35}\)
\(3)\)\(\dfrac{3}{7}.\dfrac{16}{15}-\dfrac{2}{15}.\dfrac{-3}{7}\)
\(=\dfrac{3}{7}.\dfrac{16}{15}-\dfrac{3}{7}.\dfrac{-2}{15}\)
\(=\dfrac{3}{7}.\left(\dfrac{16}{15}+\dfrac{2}{15}\right)\)
\(=\dfrac{3}{7}.\dfrac{18}{15}=\dfrac{18}{35}\)
\(4)\)\(-\dfrac{4}{13}.\dfrac{5}{17}+\dfrac{-12}{13}.\dfrac{4}{17}\)
\(=-\dfrac{4}{13}.\dfrac{5}{17}+\dfrac{-4}{13}.\dfrac{12}{17}\)
\(=-\dfrac{4}{13}.\left(\dfrac{5}{17}+\dfrac{12}{17}\right)\)
\(=-\dfrac{4}{13}.\dfrac{17}{17}=-\dfrac{4}{13}\)
`#040911`
`1)`
`-10/11 * 8/9 + 7/18 . 10/11`
`= 10/11 * (-8/9 + 7/18)`
`= 10/11 * (-1/2)`
`= -5/11`
`2)`
`12/25 * 23/7 - 12/7 *13/25`
`= 12/7 * 23/25 - 12/7 * 13/25`
`= 12/7 * (23/25 - 13/25)`
`= 12/7 * 2/5`
`= 24/35`
`3)`
`3/7 * 16/15 - 2/15 * (-3)/7`
`= 3/7 * (16/15 + 2/15)`
`= 3/7 * 6/5`
`= 18/35`
`4)`
`-4/13 * 5/17 + (-12)/13 * 4/17`
`= -4/17 * 5/13 + (-12)/13 * 4/17`
`= 4/17 * (-5/13 - 12/13)`
`= 4/17 * (-17)/13`
`= -4/13`
B={[5^10. 7^3 - 25^2 . 49^2] : [(125 . 7)^3+5^9 . 14^3]} - {[2^12-4^6 . 9^2] : [(2^2.3)^6+8^4 . 3^5]
Sửa đề: \(5^9\cdot49^2\)
\(=\dfrac{5^{10}\cdot7^3-5^9\cdot7^4}{5^9\cdot7^3+5^9\cdot14^3}-\dfrac{2^{12}-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}\)
\(=\dfrac{5^9\cdot7^3\left(5-7\right)}{5^9\cdot7^3\left(1+8\right)}-\dfrac{2^{12}\left(1-3^4\right)}{2^{12}\left(3^6+3^5\right)}=\dfrac{-2}{9}+\dfrac{80}{972}\)
=-34/243
a) \(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}-\dfrac{3}{4}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{72}\)
\(=\dfrac{5+9+1}{15}-\dfrac{27+8+1}{36}+\dfrac{1}{72}=1-1+\dfrac{1}{72}=\dfrac{1}{72}\)
b) \(=\dfrac{1}{5}-\dfrac{1}{5}-\dfrac{3}{7}+\dfrac{3}{7}+\dfrac{5}{9}-\dfrac{5}{9}-\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{7}{13}-\dfrac{7}{13}-\dfrac{9}{16}\)
\(=\dfrac{9}{16}\)
a)=\(\frac{-2.4}{5.7}+\frac{-3.2}{5.7}+\frac{-3}{5}\)
=\(\frac{-2.4}{5.7}+\frac{-2.3}{5.7}+\frac{-3}{5}\)
=\(\frac{-2}{5}\left(\frac{4+3}{7}\right)+\frac{-3}{5}\)
=\(\frac{-2}{5}.1+\frac{-3}{5}\)
=-1
b)
a: \(=\dfrac{5}{7}\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}\cdot\dfrac{-7}{11}=-\dfrac{5}{11}\)
b: \(=\dfrac{12}{7}\left(19+\dfrac{5}{8}-15-\dfrac{1}{4}\right)=\dfrac{12}{7}\cdot\left(4+\dfrac{3}{8}\right)\)
\(=\dfrac{12}{7}\cdot\dfrac{35}{8}=\dfrac{3}{2}\cdot5=\dfrac{15}{2}\)
c: \(=\dfrac{2}{15}-\dfrac{2}{15}\cdot5+\dfrac{3}{15}=\dfrac{2}{15}\cdot\left(-4\right)+\dfrac{3}{15}=\dfrac{-8+3}{15}=\dfrac{-5}{15}=-\dfrac{1}{3}\)
d: \(=\dfrac{4}{9}\left(19+\dfrac{1}{3}-39-\dfrac{1}{3}\right)=\dfrac{4}{9}\cdot\left(-20\right)=-\dfrac{80}{9}\)
\(\dfrac{8}{9}+\dfrac{3}{18}=\dfrac{16}{18}+\dfrac{3}{18}=\dfrac{19}{18}\) \(\dfrac{5}{7}-\dfrac{4}{11}=\dfrac{55}{77}-\dfrac{28}{77}=\dfrac{3}{11}\)
\(\dfrac{3}{4}+\dfrac{4}{5}=\dfrac{15}{20}+\dfrac{16}{20}=\dfrac{21}{20}\) \(\dfrac{5}{7}-\dfrac{3}{5}=\dfrac{25}{35}-\dfrac{21}{35}=\dfrac{4}{35}\)
\(\dfrac{7}{9}X\dfrac{3}{4}=\dfrac{7}{9}X\dfrac{3}{4}=\dfrac{7}{12}\) \(\dfrac{8}{3}X\dfrac{21}{4}=\dfrac{8}{3}X\dfrac{21}{4}=14\)
\(\dfrac{12}{7}:\dfrac{1}{4}=\dfrac{12}{7}X\dfrac{4}{1}=\dfrac{48}{7}\) \(\dfrac{7}{2}:\dfrac{9}{4}=\dfrac{7}{2}X\dfrac{4}{9}=\dfrac{14}{9}\)
2 . 3 ^2 - 50 :5 ^ 2
= 2 . 6 - 50 : 25
= 12 - 2
=10
136. 8 - 36.2^3
= 136 . 8 - 36 . 8
= 1088 - 288
= 800
84:4+3^9:3^7+5^0
=84 : 4 + 3^9: 3^7 + 1
= 84:4 + 3^9-^7+1 ( số 9 và 7 đểu là mũ của số 3 nhé)
= 84:4 + 3^2+1
= 84:4+ 9 +1
= 41 +9+1
=50+1
= 51
5.3^3+7^11: 7^9-12^0
= 5 .3^3+7^11: 7^9-1
= 5.3^3+ 7^11- ^ 9 - 1 ( số 11 và 9 đều là mũ của số 11)
= 5.3^3+ 7 ^2 - 1
= 5 . 27 + 49 - 1
= 135 + 48
= 183.
a, Ta có: AM = AN = \(\frac{MN}{2}\) (A là Trung điểm của MN)
BP = BQ = PQ (B là trung điểm của PQ)
mà MN = PQ(MNPQ là hình bình hành)
nên AM = BP = AN = BQ
Xét tam giác AMBP có:
AM =PB (cmt)
AM =PB(A \(\in\) PQ)
Vậy tứ giácAMBP là hình bình hành.
b, HÌnh bình hành MNPQ có O là trung điểm của đường chéo QN(1)
nên O là trung điểm của đường chéo MP(2)
Mà hình bình hành AMBP có O là trung điểm của đường chéo MP (cmt) nên O cũng là trung điểm của đường chéo AB(3)
Từ (1), (2), (3) suy ra QN, MP, AB đồng quy.