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Ta có:
12 . 12 . 12 . 2. 6 = 12 . 12 . 12 . 12 = 124
25 . 5 . 4 . 2 . 10 = 25 . 4 . 2 . 10 . 5 = 100 . 100 = 1002
2. 10 . 10 . 3 . 5. 10 = 10 . 10 . 10 . 2 . 5 . 3 = 10 . 10 . 10 . 10 . 3 = 104 . 3
a,12.12.2.12.6=12.12.12.(2.6)=12.12.12.12=124
b,25.5.4.2.10=(25.4)(5.2.10)=100.100=1002
c,2.10.10.3.5.10=10.10.10.2.5.3=10.10.10.10.3=104.3
1) \(12\cdot12\cdot2\cdot12\cdot6=12\cdot12\cdot12\cdot12=12^4\)
2) \(25\cdot5\cdot4\cdot2\cdot10=\left(25\cdot4\right)\cdot\left(5\cdot2\cdot10\right)=100\cdot100=100^2=10^4\)
3) \(2\cdot10\cdot10\cdot3\cdot5\cdot10=3\cdot10\cdot10\cdot10\cdot10=3\cdot10^4\)
4) \(a\cdot a\cdot a+b\cdot b\cdot b\cdot b=a^3+b^4\)
Viết gọn các biểu thức sau bằng cách dùng lũy thừa
1) 12.12.2.12.6
=> 12 . 12 . 12 . (2 . 6)
= 12 . 12 . 12 . 12 . 12 = 125
2) 25.5.4.2.10
=> ( 25 . 4 ) . ( 2 . 5 . 10 )
= 100 . 100 = 1002
3) 2.10.10.3.5.10
=> ( 2 . 5 ) . 10 . 10 . 10 . 3
= 10 . 10 . 10 . 10 . 3 = 104 . 3
4) a.a.a cộng b.b.b.b
=> a3 + b4
\(7\cdot35\cdot7\cdot25=7^2\cdot7\cdot5\cdot5^2=7^3\cdot5^3\)
\(2\cdot3\cdot8\cdot12\cdot24=2\cdot3\cdot2^3\cdot2^2\cdot3\cdot2^3\cdot3=2^9\cdot3^3\)
\(25\cdot5\cdot4\cdot2\cdot10=5^2\cdot5\cdot2^2\cdot2\cdot5\cdot2=5^4\cdot2^4\)
\(x\cdot x\cdot y\cdot y\cdot x\cdot y\cdot x=x^4\cdot y^3\)
\(7.35.7.25=7.7.5.7.5^2=7^3.5^3=\left(7.5\right)^3=35^3\\ 2.3.8.12.24=2.3.2^3.2^2.3.2^3.3=2^9.3^3\\ 25.5.4.2.10=5^2.5.2.2.2.2.5=5^4.2^4=\left(5.2\right)^4=10^4\\ x.x.y.y.x.y.x=x^4.y^3\)
7.7.7 = 7^3
7.35.7.25 = 7^2*35^2
2.3.8.12.24=2*3*2*2*2*3*2*2*2*2*2*3 = 2^9*3^3
x.x.y.y.x.y.x = x^3*y^3
b1
a.7^3
b.7^3*5^3
c.2^5*3^3*4^2
dx^3 *y^3
mình nha mình làm b1 thui
a) 7.7.7 = 73
b) 7.35.7.25 = 53 x 73
c) 2.3.8.12.24 = 29 x 33
d) x.x.y.y.x.y.x = x4 . y3
Bài 2:Viết kết quả dưới dạng lũy thừa.
a) 35 : 37 = 3-2
b) 125 : 53 = 11
c) 75 : 343 = 72
d) a12 : a8 = a4
A B C D E K
a) Dễ dàng c/m được ABED là hình chữ nhật => AB = DE
Ta có : \(S_{ADE}=\frac{1}{2}.AD.DE=\frac{1}{2}.\frac{1}{4}\left(AB+CD\right).DE=\frac{1}{8}.\left(8+12\right).8=20\left(cm^2\right)\)
\(S_{AEC}=S_{ADC}-S_{ADE}=\frac{1}{2}.AD.CD-20=\frac{1}{2}.\frac{1}{4}\left(AB+CD\right).CD-20=\frac{1}{8}\left(8+12\right).12-20=10\left(cm^2\right)\)
b) Ta có : \(S_{ABE}=S_{ABC}=\frac{1}{2}.AB.BE\)
Mà \(S_{ABE}=S_{ABK}+S_{AKE}\) ; \(S_{ACB}=S_{BKC}+S_{ABK}\)
=> \(S_{AEK}=S_{BKC}\)
b) 7.35.7 .25 = 42875
c)2.3.8.12.24 = 13824
d)12.12.2.12.6 = 20736
e)25.5.4.2.10 = 10000
f)2.10.10.3.5.10 = 30000
\(b,7x35x7x25\)
\(=7x5x7x7x5x5\)
\(=7^3\)\(x5^3\)\(=\left(7x5\right)^3\)
\(=35^3\)\(=42875\)
\(c,2x3x2^3\)\(x2^2\)\(x3\)\(x2^3\)\(x3\)
\(=2^9\)\(x3^3\)\(=512x27=13824\)
\(d,12^3\)\(x2x6\)
\(=12^3\)\(x12\)\(=12^4\)\(=20736\)
\(25x5x4x2x10\)
\(=5^2\)\(x5x2^2\)\(x2x10\)
\(=5^3\)\(x2^3\)\(x10\)
\(=\left(5x2\right)^3\)\(10=10^3\)\(x10\)\(=10^4\)\(=10000\)
\(f,10^3\)\(x5x2x3\)
\(=10^3\)\(x10x3\)
\(=10^4\)\(x3=1000x3=3000\)