Bài 1 : Tính : \(100:\left\{250:\left[450-\left(4.5^2-2^2.25\right)\right]\right\}\)
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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) 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, \(x:\left[\left(1800+600\right):30\right]=560:\left(315-35\right)\)
\(\Rightarrow\) \(x:\left[2400:30\right]=560:280\)
\(\Rightarrow\) \(x:80=2\)
\(\Rightarrow\) \(x=160\)
b, \(\left[\left(250-25\right):15\right]:x=\left(450-60\right):130\)
\(\Rightarrow\) \(\left[225:15\right]:x=390:130\)
\(\Rightarrow\) \(15:x=3\)
\(\Rightarrow\) \(x=5\)
Gọi tổng trên là A
A=1/1.2.3+1/2.3.4+1/3.4.5+...1/98.99.100
Ta xét :
1/1.2 ‐ 1/2.3 = 2/1.2.3; 1/2.3 ‐ 1/3.4 = 2/2.3.4;...; 1/98.99 ‐ 1/99.100 = 2/98.99.100
tổng quát: 1/n﴾n+1﴿ ‐ 1/﴾n+1﴿﴾n+2﴿ = 2/n﴾n+1﴿﴾n+2﴿.
Do đó: 2A = 2/1.2.3 + 2/2.3.4 + 2/3.4.5 +...+ 2/98.99.100
= ﴾1/1.2 ‐ 1/2.3﴿ + ﴾1/2.3 ‐ 1/3.4﴿ +...+ ﴾1/98.99 ‐ 1/99.100﴿
= 1/1.2 ‐ 1/2.3 + 1/2.3 ‐ 1/3.4 + ... + 1/98.99 ‐ 1/99.100
= 1/1.2 ‐ 1/99.100
= 1/2 ‐ 1/9900
= 4950/9900 ‐ 1/9900
= 4949/9900.
Vậy A = 4949 / 9900
Bn làm sai r . kết quả là \(\frac{101}{297}\) nhưng mik ko bt cách giải thôi
\(A=\left(1-\frac{2}{2\cdot3}\right)\cdot\left(1-\frac{2}{3\cdot4}\right)\cdot\left(1-\frac{2}{4\cdot5}\right)\cdot...\cdot1-\frac{2}{99\cdot100}\)
\(2A=1-\left(\frac{1}{2\cdot3}\cdot\frac{1}{3\cdot4}\cdot\frac{1}{4\cdot5}\cdot...\cdot\frac{1}{99\cdot100}\right)\)
\(2A=1-\left(\frac{1}{2}-\frac{1}{3}\cdot\frac{1}{3}-\frac{1}{4}\cdot\frac{1}{4}-\frac{1}{5}\cdot...\cdot\frac{1}{99}\cdot\frac{1}{100}\right)\)
\(2A=1-\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(2A=1-\frac{49}{100}\)
\(2A=\frac{51}{100}\)
\(A=\frac{51}{100}:2\)
\(A=\frac{51}{200}\)
\(\left(1-\frac{2}{2.3}\right)\left(1-\frac{2}{3.4}\right)\left(1-\frac{2}{4.5}\right)...\left(1-\frac{2}{99.100}\right)\)
\(=\frac{4}{2.3}.\frac{10}{3.4}.\frac{18}{4.5}...\frac{9898}{99.100}\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{98.101}{99.100}\)
\(=\frac{1.2.3...98}{2.3.4...99}.\frac{4.5.6...101}{3.4.5..100}\)
\(=\frac{1}{99}.\frac{101}{3}=\frac{101}{297}\)
\(=2\left(\frac{1}{2}-\frac{1}{2.3}\right).2\left(\frac{1}{2}-\frac{1}{3.4}\right)...2\left(\frac{1}{2}-\frac{2}{99.100}\right)\)
\(=2^{89}.\left(\frac{1}{2}.98-\frac{1}{2}+\frac{1}{100}\right)\)
\(=2^{98}.\left(49-\frac{49}{100}\right)\)
= \(\frac{2^{98}.4851}{100}\)
\(B=\left(\frac{2}{2.3}-1\right)\left(\frac{2}{3.4}-1\right)...\left(\frac{2}{2008.2009}-1\right)\)
\(B=\left(\frac{2}{2.3}-\frac{6}{2.3}\right)\left(\frac{2}{3.4}-\frac{12}{3.4}\right)...\left(\frac{2}{2008.2009}-\frac{2008.2009}{2008.2009}\right)\)
\(B=\left(-\frac{4}{2.3}\right)\left(-\frac{10}{3.4}\right)...\left(\frac{2-2008.2009}{2008.2009}\right)\)
\(B=\left(-\frac{1.4}{2.3}\right)\left(-\frac{2.5}{3.4}\right)...\left(-\frac{2007.2010}{2008.2009}\right)\)
Biểu thức B có (2008 - 2) : 1 + 1 = 2007 (thừa số)
Vì cả 2007 thừa số của biểu thức B đều mang dấu (-)
Nên biểu thức B mang dấu (-)
\(B=-\frac{1.2....2007}{2.3...2008}.\frac{4.5...2010}{3.4...2009}\)
\(B=-\frac{1}{2008}.\frac{2010}{3}\)
\(B=-\frac{1.2010}{2008.3}=-\frac{1.1005}{1004.3}=-\frac{1.335}{1004.1}\)
\(B=-\frac{335}{1004}\)
Vậy\(B=-\frac{335}{1004}\)
=100;(250;[450-(22.52-22.52)])
=100;(250;450)=100;(5/9)=180
nhớ **** nha