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\(\Leftrightarrow x+1,5x+3,5x=252\)
<=>6x=252
<=>x=42
ủng hộ nhe
Câu b:
\(\frac{21}{8}:\frac{5}{6}+\frac{1}{2}:\frac{5}{6}\)
= \(\frac{63}{20}+\frac{3}{5}\)
= \(\frac{15}{4}\)
\(\left(\frac{21}{8}+\frac{1}{2}\right):\frac{5}{6}\)
\(\frac{25}{8}:\frac{5}{6}\)
\(\frac{25}{8}.\frac{6}{5}\)
\(\frac{30}{8}\)
a ) 1 + 2 + 3 + 4 + ... + x = 1275 ( có x số tự nhiên )
( x + 1 ) . x : 2 = 1275
( x + 1 ) . x = 1275 x 2
( x + 1 ) . x = 2550
( x + 1 ) . x = 50 . 51
Mà x , x + 1 là hai số tự nhiên liên tiếp => x = 50
Vậy x = 50
1+2+3+4+...+x=1275
\(\frac{x.\left(x+1\right)}{2}=1275\)
x(x+1)=1275x2=2550
x(x+1)=50.51
x=50
Ta có:
\(A=\left(x-\frac{1}{2}\right).\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1}-\frac{1}{10}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\frac{9}{10}=\frac{1}{3}\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}.\frac{10}{9}\Leftrightarrow x=\frac{47}{54}\)
\(B=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{96.101}=\frac{1}{10.x}\)
\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{96.101}\right)=\frac{1}{10}-\frac{1}{x}\)
\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{96}-\frac{1}{101}\right)=\frac{1}{10}-\frac{1}{x}\)
\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{1}{10}-\frac{1}{x}\Leftrightarrow B=\frac{1}{5}.\frac{100}{101}=\frac{1}{10}-\frac{1}{x}\)
\(\Leftrightarrow B=\frac{1}{x}=\frac{1}{10}-\frac{20}{101}=-\frac{99}{1010}\Leftrightarrow x=-\frac{1010}{99}\)
c) Sai đề nhé bạn vì không có kết quả nên không tìm được x.
d) \(\left(x-5\right).\left(10-9\frac{40}{41}\right):\left(1-\frac{81}{82}\right):\left(1-\frac{204}{205}\right)=2050\)
\(\Rightarrow\left(x-5\right).\frac{1}{41}.82.205=2050\)
\(\Rightarrow\left(x-5\right).2.205=2050\Leftrightarrow x-5=2050:410=5\Leftrightarrow x=10\)
=) \(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{255}{256}.x=\frac{108}{7}\)
=) \(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{15.17}{16.16}.x=\frac{108}{7}\)
=) \(\frac{1.2.3.4.....15}{2.3.4.....16}.\frac{3.4.5.....17}{2.3.4.....16}.x=\frac{108}{7}\)
=) \(\frac{1}{16}.\frac{17}{2}.x=\frac{108}{7}\)=) \(\frac{17}{32}.x=\frac{108}{7}\)=) \(x=\frac{108}{7}:\frac{17}{32}\)
=) \(x=\frac{3456}{119}\)
\(\left(X+\frac{1}{1.3}\right)+\left(X+\frac{1}{3.5}\right)+...+\left(X+\frac{1}{23.25}\right)=11.X+\)\(\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(\Leftrightarrow12X+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)+11X\)\(+\frac{\left(1+\frac{1}{3}+...+\frac{1}{81}\right)-\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{243}\right)}{2}\)
\(\Leftrightarrow X+\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{23}+\frac{1}{23}-\frac{1}{25}\right)=\frac{242}{243}:2\)
\(\Leftrightarrow X+\frac{12}{25}=\frac{121}{243}\)
\(\Leftrightarrow X=\frac{109}{6075}\)
Vậy X=109/6075
Chắc Sai kết quả chứ công thức đúng nha!!!...
Fighting!!!...
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)
\(x.\frac{7}{9}=\frac{2}{3}+2\frac{1}{2}\)
\(x.\frac{7}{9}=\frac{19}{6}\)
\(x=\frac{19}{6}:\frac{7}{9}\)
\(x=\frac{57}{14}\)
b) \(\frac{5}{7}+x:\frac{9}{4}=\frac{4}{3}\)
\(x:\frac{9}{4}=\frac{4}{3}-\frac{5}{7}\)
\(x:\frac{9}{4}=\frac{13}{21}\)
\(x=\frac{13}{21}.\frac{9}{4}\)
\(x=\frac{39}{28}\)
a,
x.2/7.3/4=5/21
x.3/14=5/21
x=5/21:3/14
x=10/9
b,
x.1/2=1/3
x=1/3:1/2
x=2/3
c,
x:4/5=25/8:5/4
x:4/5=5/2
x=5/2.4/5=2
A, x= 5/25 : 3/4 : 2/7 = 14/15
B, x=1/3 : 1/2 = 2/3
C, x=(25/8 : 5/4)x4/5 = 5/2 x 4/5 = 2
a ) \(x+x\times\frac{1}{3}\div\frac{2}{9}+x\div\frac{2}{7}=252\)
\(x\times1+x\times\frac{1}{3}\div\frac{2}{9}+x\div\frac{2}{7}=252\)
\(x\times1+x\times\frac{1}{3}\times\frac{9}{2}+x\times\frac{7}{2}=252\)
\(x\times1+x\times\frac{9}{6}+x\times\frac{7}{2}=252\)
\(x\times\left(1+\frac{9}{6}+\frac{7}{2}\right)=252\)
\(x\times6=252\)
\(x=252\div6\)
\(x=42\)
b ) \(\left(3\times x-0,8\right)\div x+14,5=15\)
\(\left(3\times x-0,8\right)\div x=15-14,5\)
\(\left(3\times x-0,8\right)\div x=0,5\)
\(0,8\div x=3-0,5\)
\(0,8\div x=2,5\)
\(x=0,8\div2,5\)
\(x=0,32\)