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\(160-\left(2^3.5^2-6.25\right)\)
\(=160-\left(8.25-6.25\right)\)
\(=160-\left[\left(8-6\right).25\right]\)
\(=160-\left[2.25\right]\)
\(=160-50\)
\(=110\)
~ Hok tốt ~
a: \(160-\left(2^3\cdot5^2-6\cdot25\right)\)
\(=160-\left(8\cdot25-150\right)\)
=160-(200-150)
=160-50
=110
\(=2\cdot5\cdot11\)
b: \(4\cdot5^2-32:2^4=4\cdot25-32:16=100-2=98=2\cdot7^2\)
c: \(5871:\left\lbrack928-\left(247-82\cdot5\right)\right\rbrack\)
\(=5871:\left\lbrack928-\left(247-410\right)\right\rbrack\)
=5871:[928-247+410]
=5871:1091
d: \(777:7+1331:11^3=111+1=112=2^4\cdot7\)
e: \(6^2:4\cdot3+2\cdot5^2=36:4\cdot3+2\cdot25\)
=27+50
=77
\(=7\cdot11\)
g: \(5\cdot4^2-18:3^2=5\cdot16-18:9=80-2=78=2\cdot3\cdot13\)
h: \(80-\left(4\cdot5^2-3\cdot2^3\right)=80-\left(4\cdot25-3\cdot8\right)\)
=80-(100-24)
=80-76
=4
=2^2
i: \(23\cdot75+25\cdot23+180\)
\(=23\left(75+25\right)+180\)
=2300+180
=2480
\(=2^4\cdot5\cdot31\)
k: \(2^4\cdot5-\left\lbrack131-\left(13-4\right)^2\right\rbrack\)
\(=16\cdot5-\left\lbrack131-81\right\rbrack\)
=80-50
=30
=2*3*5
m: 100:{250:[450-\(\left(4\cdot5^3-2^2\cdot25\right)\) ]}
=100:{250:[450-\(\left(4\cdot125-4\cdot25\right)\) ]}
=100:{250:[450-500+100]}
=100:{250:50}
=100:5
=20
=2^2*5
a)160-(23.52-6.25)
160-(1196-150)
160-1046
-886
b)4.52-32:24
208-4/3
620/3
c)5871:(928-(247-82).5)
5871:(928-165.5)
5871:(928-825)
5871:103
57
A, 160-(2^3.5^2-6.25)
160-50
=110
=> 110= 2.5.7
B,4.5^2-32:2^4
=4.25-32:16
=100-2
=98
=>98=2.7^2
Cau c mk giong kq vs bn tren
a)(1000+1):11
=1001:11
=91
=7x13
b)=\(14^2+5^2+2^2\)
=196+25+4
=196+4+25
=200+25
=225
=\(3^2\times5^2\)
c)29.31+144:\(12^2\)
=899+144:144
=899+1
=900
=\(2^2\times3^2\times5^2\)
d)333:3+225:\(15^2\)
=111+225:225
=111+1
=112
=\(2^4\times7\)
a) (1000 + 1) :11 = 91 = 7 . 13,
b) 142 + 52 + 22 = 225 = 32 .52
c) 29.31 + 144 : 122 = 900 = 22. 32 . 52
d) 333: 3 + 225 : 152 = 112 =24 . 7
160-(23.52-6.25)
=160-(8*25-6*25)
=160-25*(8-6)
=160-25*2
=160-50
=110
Phân tích ra thừa số nguyên tố là:2x5x11
a) \(160-\left(2^3.5^2-6.25\right)\)
\(=160-\left(8.25-6.25\right)\)
\(=160-\left(200-150\right)\)
\(=160-50\)
\(=110\)
110 2 55 5 11 11 1
\(110=2.5.11\)
b) \(4.5^2-32:2^4\)
\(=4.25-32:16\)
\(=100-2\)
\(=98\)
98 2 49 7 7 7 1
\(98=2.7^2\)
c) \(5371:\left[928-\left(247-82\right).5\right]\)
\(=5371:\left[928-165.5\right]\)
\(=5371:\left[928-825\right]\)
\(=5371:103\)
\(=\frac{5371}{103}\)
Vì đây là phân số nên ko thể phân tích ra thừa số nguyên tố đc
a. \(160 - \left( {{2^3}{{.5}^2} - 6.25} \right)\)
\(\begin{array}{l} = 160 - \left( {8.25 - 6.25} \right)\\ = 160 - 25.\left( {8 - 6} \right)\\ = 160 - 25.2\\ = 160 - 50\\ = 110\end{array}\)
Ta có: 110 = 2.5.11
b. \(37.3 + 225:{15^2}\)
\(\begin{array}{l} = 37.3 + 225:225\\ = 37.3 + 1\\ = 111 + 1\\ = 112\end{array}\)
Ta có: \(112 = 2^4.7\)
c. \(5871:103 - 64:{2^5}\)
\(\begin{array}{l} = 5871:103 - 64:32\\ = 57 - 2 = 55\end{array}\)
Ta có: 55 = 5. 11
d. \(\left( {1 + 2 + 3 + 4 + 5 + 6 + 7 + 8} \right){.5^2} - 850:2\)
\(\begin{array}{l} = \left[ {\left( {1 + 8} \right) + \left( {2 + 7} \right) + \left( {3 + 6} \right) + \left( {4 + 5} \right)} \right]{.5^2} - 850:2\\ = \left( {9 + 9 + 9 + 9} \right){.5^2} - 850:2\\ = {9.4.5^2} - 850:2\\ = {36.5^2} - 425\\ = {36.5^2} - {5^2}.17\\ = {5^2}.\left( {36 - 17} \right)\\ = {5^2}.19=475\end{array}\)
Ta có: \(475 = 5^2.19\)
a: \(160-\left(2^3\cdot5^2-6\cdot25\right)\)
\(=160-\left(8\cdot25-150\right)\)
\(=160-200+150=10=2\cdot5\)
b: \(=111+225:225=112=2^4\cdot7\)
c: \(=57-64:32=57-2=55=5\cdot11\)
d: \(=\left(9\cdot\dfrac{8}{2}\right)\cdot25-425=36\cdot25-425=25=5^2\)