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\(a.32,5-3\cdot0,87=32,5-2,61=29,89\)
\(8,5\cdot\left(1\dfrac{1}{2}+\dfrac{4}{4}\right):5=8,5\cdot\left(\dfrac{3}{2}+\dfrac{4}{4}\right):5\\ =8,5\cdot\left(\dfrac{6}{4}+\dfrac{4}{4}\right):5\\ =8,5\cdot\dfrac{10}{4}:5\\ =\dfrac{85}{4}:5\\ =\dfrac{17}{4}\)
\(b.30,96-6,45+14,4:3=30,96-6,45+4,8\\ =29,31\)
\(\dfrac{2}{5}\cdot\left(\dfrac{4}{5}-\dfrac{1}{2}\right)=\dfrac{2}{5}\cdot\left(\dfrac{8}{10}-\dfrac{5}{10}\right)\\ =\dfrac{2}{5}\cdot\dfrac{3}{10}=\dfrac{3}{25}\)
bài 2
\(a.2,5\cdot12,5\cdot8\cdot0,4=\left(2,5\cdot0,4\right)\left(12,5\cdot8\right)\\ =1\cdot100=100\)
b,\(\dfrac{12}{15}\cdot\dfrac{5}{6}\cdot\dfrac{3}{20}\cdot\dfrac{32}{5}=\dfrac{12\cdot5\cdot3\cdot32}{15\cdot6\cdot20\cdot5}\\ =\dfrac{3\cdot4\cdot5\cdot3\cdot4\cdot8}{3\cdot5\cdot2\cdot3\cdot5\cdot4\cdot5}=\dfrac{16}{25}\)
Bài 1:
a) \(32.5-3\cdot0.87=32.5-2.61=29.89\)
\(8.5\cdot\left(1\dfrac{1}{2}+\dfrac{4}{4}\right):5=8.5\cdot\dfrac{5}{2}:5=\dfrac{17}{2}\cdot\dfrac{5}{2}:5=\dfrac{85}{4}\cdot\dfrac{1}{5}=\dfrac{17}{4}\)
(15/4.4-12/5.5/4)-(7/2:5/2-1/5)-1/5.x=51/5
(15-3)-(7/5-1/5)-1/5.x=51/5
12-6/5-1/5.x=51/5
54/5-1/5.x=51/5
1/5x=54/5-51/5
1/5x=3/5
x=3
\(x+0,9=20,17\Leftrightarrow x=20,17-0,9=19,24\)
\(47,5-x=4,64\Leftrightarrow x=47,5-4,64=42,86\)
\(3x=\dfrac{5}{4}-\dfrac{1}{2}\Leftrightarrow3x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{4}\)
\(a.x=20,17-0,9=19,27\)
\(b.47,5-x=4,56\Leftrightarrow x=42,86\)
\(c.3.x=\dfrac{5}{4}-\dfrac{1}{2}\Leftrightarrow3x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{4}\)
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{x\left(x+2\right)}=\frac{16}{99}\)
\(\Rightarrow\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{x\left(x+2\right)}\right)=\frac{16}{99}\)
\(\Rightarrow\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{x\left(x+2\right)}=\frac{32}{99}\)
=> \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{32}{99}\)
=> \(\frac{1}{3}-\frac{1}{x+2}=\frac{32}{99}\)
=> \(\frac{1}{x+2}=\frac{1}{99}\)
=> x + 2 = 99
=> x = 97
Vậy x = 97 là giá trị cần tìm
\(\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{x\times\left(x+2\right)}\)
\(=\frac{1}{2}\times\left(\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+...+\frac{2}{x\times\left(x+2\right)}\right)\)
\(=\frac{1}{2}\times\left(\frac{5-3}{3\times5}+\frac{7-5}{5\times7}+\frac{9-7}{7\times9}+...+\frac{x+2-x}{x\times\left(x+2\right)}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+2}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{x+2}\right)\)
\(=\frac{1}{6}-\frac{1}{2\times\left(x+2\right)}=\frac{16}{99}\)
\(\Leftrightarrow\frac{1}{2\times\left(x+2\right)}=\frac{1}{6}-\frac{16}{99}=\frac{1}{198}\)
\(\Leftrightarrow2\times\left(x+2\right)=198\)
\(\Leftrightarrow x+2=99\)
\(\Leftrightarrow x=97\)
\(\left(4\times2,5\right)\times\left(0,25\times2\right)\times\left(\dfrac{1}{5}\times\dfrac{1}{2}\right)\)
=\(10\times\dfrac{1}{2}\times\dfrac{1}{10}=\dfrac{1}{2}\)
( 1/2 - 1/3 + 3/4 ) : ( 2/5 - 1/x ) = 2,5
11/12 : ( 2/5 - 1/x ) = 2,5
2/5 - 1/x = 11/12 : 2,5
2/5 - 1/x = 11/30
1/x = 2/5 - 11/30
1/x = 1/30
=> x = 30