Tính giá trị biểu thức sau:{[(3+2]+6-2)+(-2+5)-23+(2-1)}+2-8+(-13+223)
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a/\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}\)
= \(\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\frac{2^{10}.3^8.\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=\frac{1-3}{1+5}=\frac{-2}{6}=-3\)
A = (1- 2) \(\times\) ( 4 - 3) \(\times\) (5 - 6) \(\times\) (8 - 7) \(\times\) (9 - 10) \(\times\) (12 - 11) \(\times\)(13 - 14)
A = (-1) \(\times\) 1 \(\times\) (-1) \(\times\) 1 \(\times\) (-1) \(\times\) 1 \(\times\) (-1)
A = 1
1) 2611
2)(-21)
3)(-44)
4)(-253)
5)82
6(-45)
7)10
8)(-10)
k mk nha
1)2763+152=2915
2)(-7)+(-14)=-21
3)(-35)+(-9)=-44
4)(-5)+(-248)=-253
5)(-23)+105=82
6)78+(-123)=-45
7)23+(-13)=10
8)(-23)+13=-10
B=1-2-3+4+5-6-7+8+..........+21-22-23+24
B=(1-2-3+4)+(5-6-7+8)+.......+(21-22-23+24)
B=0+0+............+0
B=0
\(B=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+\frac{1}{18\cdot19\cdot20}\)
\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+\frac{2}{18\cdot19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{18\cdot19}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\cdot\frac{189}{380}=\frac{189}{760}\)
\(C=\frac{52}{1\cdot6}+\frac{52}{6\cdot11}+\frac{52}{11\cdot16}+...+\frac{52}{31\cdot36}\)
\(C=\frac{52}{5}\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{6}{31\cdot36}\right)\)
\(C=\frac{52}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{31}-\frac{1}{36}\right)\)
\(C=\frac{52}{5}\cdot\left(1-\frac{1}{36}\right)\)
\(C=\frac{91}{9}\)
A= \(49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)
=\(\frac{1135}{32}-\left(\frac{167}{32}+\frac{330}{23}\right)\)
=\(\frac{1135}{23}-\frac{167}{32}-\frac{330}{23}\)
=\(\left(\frac{1135}{23}-\frac{330}{23}\right)-\frac{167}{32}\)
=\(\frac{805}{23}-\frac{167}{32}\)
=\(\frac{953}{32}\)