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\(44.179+20^2-79.44\\ =44\left(179-79\right)+400\\ =44.100+400\\ =100\left(44+4\right)=4800\\ 3.4^2:\left[500-\left(7.35+125\right)\right]\\ =3.16:\left[500-370\right]\\ =48:130=\dfrac{24}{65}\\ 187+\left[921-\left(921+887\right)\right]\\ =187+921-921-887\\ =187-887=-700\)
44. 179+ 202 - 79.44 3. 42 : [500- (7.35+125)]
= 44. 179+400 - 79.44 = 3. 16: [500- (245+125)]
= 44. (179-79) +400 = 48: [ 500-370 ]
= 44. 100+400 = 48:130
= 4400+400 = 0, 369...
= 4800
187+ [ 921-( 921+887) = 187+ [921- 1808]
= 187+ (-159)
= 28
\(a/(-8978)-\left(2007-8978\right)+\left(-193\right)\)
\(=\left(-8978\right)-\left(-6971\right)+\left(-193\right)\)
\(=\left(-8978\right)+6971+\left(-193\right)\)
\(=-2007+\left(-193\right)\)
\(=-2200\)
\(b/187+\left[923-\left(923+887\right)\right]\)
\(=187+\left[923-1810\right]\)
\(=187+\left(-887\right)\)
\(=-700\)
\(c/39.178+30^2-78.39\)
\(=39.\left(178-78\right)+900\)
\(=39.100+900\)
\(=3900+900\)
\(=4800\)
\(d/3.4^2:\left[500-\left(7.35+125\right)\right]\)
\(=3.16:\left[500-\left(245+125\right)\right]\)
\(=48:\left(500-370\right)\)
\(=48:130\)
\(=\frac{24}{35}\)
\(125\%\cdot\left(\frac{-1}{2}\right)^2:\left(21-1,5\right)+2008^0\)
\(=\frac{5}{4}\cdot\frac{1}{4}:\left(21-\frac{3}{2}\right)+1\)
\(=\frac{5}{16}:\frac{39}{2}+1\)
\(=\frac{5}{312}+1\)
\(=\frac{317}{312}\)
b ) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
= 1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
c ) Đặt A = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)
=> A < \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
=> A < 1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100= 1 - 1/100 = 99/100 < 1
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)< 1
b, \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\)\(\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
c,Ta thấy
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(.....\)
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\left(đpcm\right)\)
a)\(12:\left\{400:\left[500-\left(125+25×7\right)\right]\right\}\)
\(12:\left\{400:\left[500-300\right]\right\}\)
\(12:2\)
\(6\)
b)\(\left[\left(7-3^3:3^2\right):2^2+99\right]-100\)
\(=\left[4:4+99\right]-100\)
\(=100-100\)
\(=0\)
\(c,3^2×\left[\left(5^2-3\right):11\right]-2^4+2×10^3\)
\(=9×2-16+2×10000\)
\(=18-16+20000\)
\(=20002\)
a) 125 + 70 + 375 + 230 = (125 + 375) + (70 + 230) = 500 + 300 = 800
b) 62 : 4 . 3 + 2 . 52 = 62 : 22 . 3 + 2 . 25 = 32 . 3 + 50 = 33 + 50 = 27 + 50 = 77
c) 150 : [25 . (18 - 42)] = 150 : (25 . 2) = 150 : 50 = 3