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\(a,97\times327+327\times3\)
\(=327\times\left(97+3\right)\)
\(=327\times100\)
\(=32700\)
\(b,\frac{1}{7}\times\frac{4}{23}+\frac{1}{7}\times\frac{25}{23}+\frac{1}{7}\times\frac{17}{23}+5\times\frac{1}{7}\)
\(=\frac{1}{7}\times\left(\frac{4}{23}+\frac{25}{23}+\frac{17}{23}+5\right)\)
\(=\frac{1}{7}\times7\)
\(=1\)
a) 97 x 327 + 327 x 3
= (97 + 3) x 327
= 100 x 327 = 32700
b) \(\frac{1}{7}\times\frac{4}{23}+\frac{1}{7}\times\frac{25}{23}+\frac{1}{7}\times\frac{17}{23}+5\times\frac{1}{7}\)
\(=\frac{1}{7}\times\left(\frac{4}{23}+\frac{25}{23}+\frac{17}{23}+5\right)\)
\(=\frac{1}{7}\times7=\frac{7}{7}=1\)
\(\frac{2}{3}x+\frac{10}{9}=\frac{23}{18}\)
\(\frac{2}{3}x=\frac{23}{18}-\frac{10}{9}\)
\(\frac{2}{3}x=\frac{1}{6}\)
\(x=\frac{1}{6}:\frac{2}{3}\)
\(x=\frac{1}{4}\)
a) \(\frac{13+x}{20}=\frac{3}{4}\)
\(\Rightarrow\frac{13+x}{20}=\frac{15}{20}\)
\(\Rightarrow13+x=15\)
\(\Rightarrow\) \(x=15-13\)
\(\Rightarrow\)\(x=2\)
b)\(\frac{23-x}{25}=\frac{4}{5}\)
\(\Rightarrow\frac{23-x}{25}=\frac{20}{25}\)
\(\Rightarrow23-x=20\)
\(\Rightarrow\) \(x=23-20\)
\(\Rightarrow\) \(x=3\)
a, 13+x/20= 3/4
Suy ra: ( 13+x). 4= 20.3
Suy ra: (13+x).4=60
Suy ra: 13+x=60:4
Suy ra: 13+x=15
Suy ra: x=15-13
Suy ra: x=2
Vậy x=2
\(\left(X+\frac{1}{1.3}\right)+\left(X+\frac{1}{3.5}\right)+...+\left(X+\frac{1}{23.25}\right)=11.X+\)\(\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(\Leftrightarrow12X+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)+11X\)\(+\frac{\left(1+\frac{1}{3}+...+\frac{1}{81}\right)-\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{243}\right)}{2}\)
\(\Leftrightarrow X+\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{23}+\frac{1}{23}-\frac{1}{25}\right)=\frac{242}{243}:2\)
\(\Leftrightarrow X+\frac{12}{25}=\frac{121}{243}\)
\(\Leftrightarrow X=\frac{109}{6075}\)
Vậy X=109/6075
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Đặt:
\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{23.25}=\frac{3-1}{1.3}+\frac{5-3}{3.5}+...+\frac{25-23}{23.25}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}=1-\frac{1}{25}=\frac{24}{25}\)
=> \(A=\frac{12}{25}\)
Đặt \(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\)
=> \(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)=1-\frac{1}{3^5}=\frac{242}{243}\)
=> \(2B=\frac{242}{243}\Rightarrow B=\frac{121}{243}\)
Giải phương trình:
\(\left(x+\frac{1}{1.3}\right)+\left(x+\frac{1}{3.5}\right)+...+\left(x+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{243}\right)\)
\(12x+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{242}\right)\)
\(12x+\frac{12}{25}=11x+\frac{121}{243}\)
\(12x-11x=\frac{121}{243}-\frac{12}{25}\)
\(x=\frac{109}{6075}\)
a) 15 + 3 ( x - 1 ) = 36
3 ( x - 1 ) = 36 - 15
3 ( x - 1 ) = 21
x - 1 = 21 : 3
x - 1 = 7
x = 7+1
x = 8
- 1a, Ta co:\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}\)
= \(\frac{8}{16}-\frac{4}{16}+\frac{2}{16}-\frac{1}{16}\)
= \(\frac{8-4+2-1}{16}\)
= \(\frac{5}{16}\)
Bài 1: b. 23.(75 + 25) - 108 = 23 x 100 - 108 = 2192
a. 1/2 - 1/4 +1/8 - 1/16 = 1/2 x (1 - 1/2) + 1/8 x (1 - 1/2) = (1 - 1/2) x (1/2 + 1/8) = 1/2 x 5/8 = 5/16
c 31,2x12 - 89,68:8 + 35,16 = 398,35
Bài 2: x + 54 : 6 = 15
\(\Rightarrow\)x + 9 = 15
\(\Rightarrow\)x = 6
211 - (x - 8) = 5628 : 28
\(\Rightarrow\)211 - (x - 8) = 201
\(\Rightarrow\)x - 8 = 10
\(\Rightarrow\)x = 18
75 : x = 160 x 23 : 184
75 : x = 20
x = 75 : 20
x = 3,75
\(\frac{600:x}{184}=\frac{160}{184}\Rightarrow600:x=160\Rightarrow x=3,75\)