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a) 24.5 - [ 131. ( 13 - 4 )2 ]
=120 - [ 131 . 92 ]
=120 - [ 131 . 81 ]
=120 - 10611
= - 10491
b) 100 : {230:[450−(4−53−52.25)]}
= 100 : \(\left\{230:\left[450-\left(4-125-25.25\right)\right]\right\}\)
= \(100:\left\{230:\left[450-\left(4-125-625\right)\right]\right\}\)
= \(100:\left\{230:\left[450-\left(-746\right)\right]\right\}\)
=\(100:\left\{230:1196\right\}\)
= 100 : \(\dfrac{5}{26}\)= 520
a) \(26+173+74+27\)
\(=\left(26+74\right)+\left(173+27\right)\)
\(=100+200\)
\(=300\)
b) \(75\cdot37+89\cdot46+75\cdot52-89\cdot21\)
\(=75\cdot\left(37+52\right)+89\cdot\left(46-21\right)\)
\(=75\cdot89+89\cdot25\)
\(=89\cdot\left(75+25\right)\)
\(=89\cdot100\)
\(=8900\)
c) \(2^7:2^2+5^4:5^3\cdot2^4-3\cdot2^5\)
\(=2^{7-2}+5^{4-3}\cdot2^4-3\cdot2^5\)
\(=2^5+5\cdot2^4-3\cdot2^5\)
\(=2^4\cdot\left(2+5-3\cdot2\right)\)
\(=2^4\cdot\left(7-6\right)\)
\(=2^4\)
\(=16\)
d) \(100:\left\{250:\left[450-\left(4\cdot5^3-2^2\cdot25\right)\right]\right\}\)
\(=100:\left\{250:\left[450-\left(4\cdot5^3-4\cdot5^2\right)\right]\right\}\)
\(=100:\left[250:\left(450-4\cdot5^2\cdot4\right)\right]\)
\(=100:\left[250:\left(450-400\right)\right]\)
\(=100:\left(250:50\right)\)
\(=100:5\)
\(=20\)
a, \(x:\left[\left(1800+600\right):30\right]=560:\left(315-35\right)\)
\(\Rightarrow\) \(x:\left[2400:30\right]=560:280\)
\(\Rightarrow\) \(x:80=2\)
\(\Rightarrow\) \(x=160\)
b, \(\left[\left(250-25\right):15\right]:x=\left(450-60\right):130\)
\(\Rightarrow\) \(\left[225:15\right]:x=390:130\)
\(\Rightarrow\) \(15:x=3\)
\(\Rightarrow\) \(x=5\)
Xét: \(1-\frac{2}{n\left(n+1\right)}=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
Khi đó:
\(1-\frac{2}{2.3}=\frac{1.4}{2.3}\) ; \(1-\frac{2}{3.4}=\frac{2.5}{3.4}\) ; ... ; \(1-\frac{2}{101.102}=\frac{100.103}{101.102}\)
\(\Rightarrow M=\frac{1.4}{2.3}\cdot\frac{2.5}{3.4}\cdot\cdot\cdot\frac{100.103}{101.102}\)
\(M=\frac{\left(1.2...100\right).\left(4.5...103\right)}{\left(2.3...101\right).\left(3.4...102\right)}=\frac{103}{101.3}=\frac{103}{303}\)
Vậy \(M=\frac{103}{303}\)
Bài 1:
\(\left(-\frac{1}{2}\right)^3=\frac{-1}{8}\)
\(\left(-\frac{1}{2}\right)^2=\frac{1}{4}\)
\(\left(-\frac{1}{3}\right)^4=\frac{1}{81}\)
\(\left(-\frac{1}{3}\right)^5=\frac{-1}{243}\)
Bài 2:
\(\left(-\frac{1}{4}\right)^0=1\)
\(\left(-2\frac{1}{3}\right)^2=\left(-\frac{7}{3}\right)^2=\frac{14}{9}\)
\(\left(-1\frac{1}{3}\right)^4=\left(-\frac{4}{3}\right)^4=\frac{256}{81}\)
Với số mũ lẻ, kết quả luôn là âm nếu giá trị trong ngoặc là âm, kết quả luôn là dương với số mũ chẵn.
Đặc biệt số mũ là 0 thì kết quả luôn bằng 1.
\(S=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)......\left(1-\frac{1}{100^2}\right)\)
\(=\frac{3}{4}.\frac{8}{9}.....\frac{9999}{10000}\)
\(=\frac{\left(1.3\right).\left(2.4\right).........\left(99.101\right)}{\left(2.2\right).\left(3.3\right).......\left(100.100\right)}\)
\(=\frac{\left(1.2....99\right)\left(3.4....101\right)}{\left(2.3.....100\right)\left(2.3....100\right)}\)
\(=\frac{1.101}{100.2}=\frac{101}{200}\)
=100;(250;[450-(22.52-22.52)])
=100;(250;450)=100;(5/9)=180
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