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\(a,\dfrac{3^{43}+3^4}{3^{39}+1}\)
\(=\dfrac{3^4\cdot\left(3^{39}+1\right)}{3^{39}+1}\)
\(=3^4\)
\(=81\)
\(A=\left(1+2+3+...+2023\right)\left(1^2+2^2+...+2023^2\right)\left(65\cdot111-13\cdot15\cdot37\right)\)
\(=\left(1+2+3+...+2023\right)\cdot\left(1^2+2^2+...+2023^2\right)\cdot\left(13\cdot5\cdot3\cdot37-13\cdot5\cdot3\cdot37\right)\)
=0
B=C*[13*37*(5*3-15)]=0
\(A=\dfrac{2^{10}\cdot78}{2^8\cdot26\cdot4}=\dfrac{78}{26}=3\)
Ta có : (1+2+3+.....+100).(12+ 22 +.....+102 ).(65.111 - 13.15.37)
=A .B.(65.111-13.5.3.37)
=A.B.(65.111-65.111)
=A.B.0
=0
Tính nhanh:
a) (1+2+3+..+100) x (12 +22+32+...+...102) x ( 65.111-13.15.37)
b)19x64+76x34
c)12x35+65x13
a Ta có
(1+2+3....+100).(1^2+2^2+..+10^2).(13.5.3.37-13.15.37)
=(1+2...+100).(1^2+2^2+...+10^2).0=0
b Ta co
19.64+76.34=19.4.16+76.34=76.16+76.34=76(16+34)=76.50=3100 ( cau c tuong tu )
(1+2+3+.....+100).(12+22+32+.....+102).(65.111-13.15.37)
=(1+2+3+.....+100).(12+22+32+.....+102)(65.111-13.555)
=(1+2+3+.....+100).(12+22+32+.....+102)(65.111-13.5.111)
=(1+2+3+.....+100).(12+22+32+.....+102)[111(65-65)]
=(1+2+3+.....+100).(12+22+32+.....+102)(100.0)
=(1+2+3+.....+100).(12+22+32+.....+102)0
=0
b, ( mik cx ko viết đề bài lun ^_^ )
= ( 1 + 2 + 3 + ... + 100 ) . ( 12 + 22 + 32 + ... + 102 ) . ( 65 . 111 - 13 . 555 )
= ( 1 + 2 + 3 + ... + 100 ) . ( 12 + 22 + 32 + ... + 102 ) . ( 65 . 111 - 13 . 5 . 111 )
= ( 1 + 2 + 3 + ... + 100 ) . ( 12 + 22 + 32 + ... + 102 ) . ( 65 . 111 - 65 . 111 )
= ( 1 + 2 + 3 + ... + 100 ) . ( 12 + 22 + 32 + ... + 102 ) . 0
= 0
Chúc hok tốt !
Mk nhầm sr nhé :
A = \(\frac{2^{10}\left(13+65\right)}{2^8.2^3.13}\)
A = \(\frac{2^{10}.78}{2^{11}.13}\)
A = \(\frac{2^{10}.13.3.2}{2^{11}.13}\)
A = \(\frac{2^{11}.13.3}{2^{11}.13}\)
A = \(3\)
Trả lời
a)(1+3+2+...10).(12+23+32+...+1002).(65.111-13.15.37)
=(1+3+2+...10).(12+22+33+...+1002).0
=0