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c) \(3^{11}:3^9-147:7^2\)
= \(3^{\left(11-9\right)}-147:49\)
= \(3^2-3\)
= \(9-3\)
= \(6\)
Đề hình như bị sai ban ơi sửa lại
\(A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{92.95}\)
\(A=3\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}\right)\)
\(A=3.\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}\right)\)
\(A=\dfrac{1}{2}-\dfrac{1}{95}\)
\(A=\dfrac{93}{190}\)
\(B=\dfrac{2}{2.5}+\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{92.95}\)
\(3B=2\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}\right)\)
\(3B=2.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}\right)\)
\(3B=2\left(\dfrac{1}{2}-\dfrac{1}{95}\right)\)
\(3B=2.\dfrac{93}{190}\)
\(3B=\dfrac{93}{95}\)
\(\Rightarrow B=\dfrac{31}{95}\)
\(76-\left\{2\cdot\left[2\cdot5^2-\left(31-2\cdot3\right)\right]\right\}\)
\(=76-\left\{2\cdot\left[2\cdot25-\left(31-6\right)\right]\right\}\)
\(=75-\left[2\cdot\left(50-25\right)\right]\)
\(=76-\left(2\cdot25\right)\)
\(=76-50\)
\(=26\)
____________________
\(6^2\cdot10:\left\{780:\left[10^3-\left(2\cdot5^3+35\cdot14\right)\right]\right\}\)
\(=36\cdot10:\left\{780:\left[1000-\left(2\cdot125+490\right)\right]\right\}\)
\(=360:\left\{780:\left[1000-\left(250+490\right)\right]\right\}\)
\(=360:\left[780:\left(1000-740\right)\right]\)
\(=360:\left(780:260\right)\)
\(=360:3\)
\(=120\)
a: =76-{2*[2*25-31+6]}
=76-{2*[50-31+6]}
=76-2*25
=76-50=26
b: \(=360:\left\{\dfrac{780}{1000-2\cdot125-490}\right\}\)
\(=360:\dfrac{780}{260}\)
=360/3=120
2.25.9+{[2.125-(5x+4).5]:(4.3.5)}=453
50.9+{[350-(5x+4).5]:60}=453
450+{[350-(5x+4).5]:60}=453
[350-(5x+4).5]:60=453-450
[350-(5x+4).5]:60=3
350-(5x+4).5=3.60
350-(5x+4).5=180
350-(5x+4)=180:5
350-(5x+4)=36
5x+4=350-36
5x+4=314
5x=314-4
5x=310
x=310:5
x=62
\(C=...\)
\(=\frac{1}{3^2}+\frac{1}{4^2}-\frac{1}{4^2}+\frac{1}{5^2}-\frac{1}{5^2}+...-\frac{1}{14^2}+\frac{1}{14^2}-\frac{1}{15^2}\)
\(=\frac{1}{3^2}-\frac{1}{15^2}\)
\(=\frac{1}{9}-\frac{1}{225}\)
Do \(\frac{1}{9}-\frac{1}{225}\)<\(\frac{1}{5}\)
\(=>C< \frac{1}{5}\)( ĐPCM )
C = ...
=> C = \(\frac{1}{3^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{5^2}+...+\frac{1}{14^2}-\frac{1}{15^2}\)
C = \(\frac{1}{3^2}-\frac{1}{15^2}\)
Ta thấy : \(\frac{1}{3^2}< \frac{1}{5}\Leftrightarrow\frac{1}{3^2}-\frac{1}{15^2}< \frac{1}{5}\)
=> C < \(\frac{1}{5}\)
98:{29-32.5(130-22.52)}
=98:{29-9.5(130-4.25)}
=98:{29-9.5.30}
=98:(-1321)
=-0,07418622255
Mik ko bt đúng hay sai
Hok tốt!