\(-382-\left[125+\left(-382\right)\right]-\left[\left(3210-697\right)+\left(-3210-957\right)\right]\)
tính nhanh:
-385-[1925+(-382)]-[(3210-697)+(-3210+597)]
-385 -[1925 +(-382)]-[(3210-697)+(3210+597)]
= -385 -1925 +(-382)-[3210 -697-3210+597]
= -385 -1925 +(-382)- [3210-3210 +597-697]
= -385 + (-382) -1925 -[0+(-100)
= -385 +(-382)-1925-(-100)
=-767 -1925+100
= -2692 +100
= -2592
Ta có:-385-[ 1925+(-382) ] - [ ( 3210 - 697 )+(-3210+597) ]
=-385 - [ 1925-382 ] - [ 3210 - 697 - 3210+597 ]
=-385 - 1925 + 382 - [ ( 3210 - 3210 )+( 597 - 697 ) ]
= -3 - 1925 - [ 0 - 100 ]
= -1928 + 100 = -1828
Giúp mik với
Tính nhanh:
a. A=\(\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}\left(n\in N\right)\)
b. B=\(\left(10000-1^2\right)\left(10000-2^2\right)\left(10000-3^2\right)..\left(10000-1000^2\right)\)
c. C=\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)\left(\frac{1}{125}-\frac{1}{3^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
d. D=\(1999^{\left(1000-1^3\right)\left(1000-2^3\right)\left(1000-3^3\right)...\left(1000-10^3\right)}\)
a) \(A=\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}=\left(-1\right)^{3n+1}\)
b) \(B=\left(10000-1^2\right)\left(10000-2^2\right).........\left(10000-1000^2\right)\)
\(=\left(10000-1^2\right)\left(10000-2^2\right)......\left(10000-100^2\right)....\left(10000-1000^2\right)\)
\(=\left(10000-1^2\right)\left(10000-2^2\right).....\left(10000-10000\right).....\left(10000-1000^2\right)=0\)
c) \(C=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)..........\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right).....\left(\frac{1}{125}-\frac{1}{5^3}\right)......\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)........\left(\frac{1}{125}-\frac{1}{125}\right).....\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
d) \(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-10^3\right)}\)
\(=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-1000\right)}=1999^0=1\)
Tính nhanh :
a) \(\left(-4\right)\left(+125\right)\left(-25\right)\left(-6\right)\left(-8\right)\)
b) \(\left(-98\right)\left(1-246\right)-246.98\)
Sách Giáo Khoa
a) (-4).(+125).(-25).(-6).(-8) = [(-4).(-25)].[(+125).(-8)].(-6) = 100.(-1000).(-6) = 600000 b) (-98).(1 - 246) – 246.98 = -98 + 98.246 - 246.98 = -98 + 98.(246 - 246) = -98 + 98.0 = -98a) (-4).(+125).(-25).(-6).(-8)
= [(-4).(-25)].[(+125).(-8)].(-6)
= 100.(-1000).(-6)
= 600000
b) (-98).(1 - 246) – 246.98
= -98 + 98.246 - 246.98
= -98 + 98.(246 - 246)
= -98 + 98.0
= -98
a)(-4).(+125).(-25).(-6).(-8)
=4 . 125 . 25. 6 . 8
=(4.25).(125.8).6
=100 . 1000.6
= 600000
b)(-98).(1-246)-246.98
= (-98)-(-245)-246.98
= 98. 245 -246.98
= 98.(245-246)
= 98.(-1)
= -98
Tính nhanh :
a) \(\left(-4\right).\left(+3\right).\left(-125\right).\left(+25\right).\left(-8\right)\)
b) \(\left(-65\right).\left(1-301\right)-301.67\)
a (-4) . (+3) . (-125) . (+25) . (-8)
=[(-4) . (+25)] . [(-125) . (-8)] . (+3)
=(-100) + (1000) . (+3)
= -300 000
b (-67) . (1 - 301) - 301 . 67
=(-67) . 1 + 67 . 301 - 67 . 301
= - 67
A)\(2009^{\left(1000-1^3\right)\cdot\left(1000-2^3\right)\cdot...\cdot\left(1000-15^3\right)}\)
B)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\cdot\left(\frac{1}{125}-\frac{1}{2^3}\right)\cdot...\cdot\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
C)\(\left(\frac{1}{38}-1\right)\cdot\left(\frac{1}{37}-1\right)\cdot\left(\frac{1}{36}-1\right)\cdot...\cdot\left(\frac{1}{2}-1\right)\)
HELP ME!!!!!!!!!!!!!!!!!!!
#)Giải :
a)\(2009^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-15^3\right)}=2009^{\left(1000-1^3\right)...\left(1000-10^3\right)...\left(1000-15^3\right)}=2009^0=1\)
b)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
Tính:
\(P=\frac{\left(81,624:4,8-4,505\right)^2+125\cdot0,75}{\left\{\left[\left(0,44\right)^2:0,88+3,53\right]^2-\left(2,75\right)^2\right\}:0,52}\)
Tính giá trị của biểu thức \(A=\left(\frac{1}{125}-\frac{1}{1^3}\right).\left(\frac{1}{125}-\frac{1}{2^3}\right).\left(\frac{1}{125}-\frac{1}{3^3}\right)...\left(\frac{1}{125}-\frac{1}{19^3}\right).\left(\frac{1}{125}-\frac{1}{20^3}\right)\)
Tính \(C=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)\left(\frac{1}{125}-\frac{1}{3^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(\frac{\left(81,624:4\frac{4}{3}-4,505\right)^2+125\frac{3}{4}}{\left(\left(\left(\frac{11}{25}\right)^2:0,88+3,53\right)^2-\left(2,75\right)^2\right):\frac{13}{25}}\)