1 . tính
A=\(\frac{3}{2.5}\)+ \(\frac{3}{5.8}\)+...........+\(\frac{3}{50.53}\)
B=\(\frac{1}{1.5}\)+\(\frac{1}{5.9}\)+\(\frac{1}{9.13}\)+...........+\(\frac{1}{41.45}\)
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\(B=\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{41.45}\)
\(4B=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{41}-\frac{1}{45}\)
\(4B=\frac{44}{45}\)
\(B=\frac{11}{45}\)
\(B=\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{41.45}\)
\(=\frac{1}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{41.45}\right)\)
\(=\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{41}-\frac{1}{45}\right)\)
\(=\frac{1}{4}.\left(1-\frac{1}{45}\right)\)
\(=\frac{1}{4}.\frac{44}{45}\)
\(=\frac{11}{45}\)
a) \(\frac{x-1}{21}=\frac{3}{x+1}\)( ĐKXĐ : x khác -1 )
<=> ( x - 1 )( x + 1 ) = 21.3
<=> x2 - 1 = 63
<=> x2 = 64
<=> x2 = ( ±8 )2
<=> x = ±8 ( tmđk )
b) \(\frac{7}{x}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+...+\frac{4}{41\cdot45}=\frac{29}{45}\)( ĐKXĐ : x khác 0 )
<=> \(\frac{7}{x}+\left(\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+...+\frac{4}{41\cdot45}\right)=\frac{29}{45}\)
<=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)
<=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)
<=> \(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
<=> \(\frac{7}{x}=\frac{7}{15}\)
<=> x = 15 ( tmđk )
a) \(\frac{x-1}{21}=\frac{3}{x+1}\Leftrightarrow\left(x-1\right)\left(x+1\right)=3.21\)
\(\Leftrightarrow x^2-1=63\Rightarrow x^2=63+1=64\Rightarrow x=\pm8\)
b) \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\Leftrightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}=\frac{29}{45}-\frac{8}{45}=\frac{21}{45}=\frac{7}{15}\Rightarrow x=15\)
\(x+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=1\)
\(x+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=1\)
\(x+\frac{1}{5}-\frac{1}{45}=1\)
\(x+\frac{8}{45}=1\)
\(\Rightarrow x=1-\frac{8}{45}\)
\(\Rightarrow x=\frac{37}{45}\)
\(x+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=1\)
\(x+\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}\right)=1\)
\(x+\left[4\left(\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{41.45}\right)\right]=1\)
\(x+\left[4.\frac{1}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)\right]=1\)
\(x+\left[1\left(\frac{1}{5}-\frac{1}{45}\right)\right]=1\)
\(x+\frac{8}{45}=1\)
\(x=1-\frac{8}{45}\)
\(x=\frac{37}{45}\)
\(x+\frac{3}{5.9}+\frac{3}{9.13}+\frac{3}{13.17}+...+\frac{4}{41.45}=-\frac{37}{45}\)
\(\Leftrightarrow x+3\left(\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{41.45}\right)=-\frac{37}{45}\)
\(\Leftrightarrow x+\frac{3}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)=-\frac{37}{45}\)
\(\Leftrightarrow x+\frac{3}{4}\left(\frac{1}{5}-\frac{1}{45}\right)=-\frac{37}{45}\)
\(\Leftrightarrow x+\frac{3}{4}.\frac{8}{45}=-\frac{37}{45}\)
\(\Leftrightarrow x+\frac{2}{15}=-\frac{37}{45}\)
\(\Leftrightarrow x=-\frac{43}{45}\)
\(a,\frac{x-1}{21}=\frac{3}{x+1}\)
\(\Leftrightarrow\left[x-1\right]\left[x+1\right]=63\)
\(\Leftrightarrow x^2-1=63\)
\(\Leftrightarrow x^2=64\)
\(\Leftrightarrow x^2=8^2\)
\(\Leftrightarrow x=\pm8\)
\(b,\frac{7}{x}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left[\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}\right]=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left[\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right]=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left[\frac{1}{5}-\frac{1}{45}\right]=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}=\frac{21}{45}\)
\(\Leftrightarrow\frac{7}{x}=\frac{7}{15}\)
\(\Leftrightarrow x=15\)
Vậy x = 15
Bài cuối tương tự
\(x-\frac{37}{45}=\frac{4}{5.9}+\frac{4}{9.13}+.....+\frac{4}{41.45}\)
\(\Rightarrow x-\frac{37}{45}=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\)
\(\Rightarrow x-\frac{37}{45}=\frac{1}{5}-\frac{1}{45}\)
\(\Rightarrow x-\frac{37}{45}=\frac{8}{45}\)
\(\Rightarrow x=\frac{37}{45}+\frac{8}{45}\)
\(\Rightarrow x=1\)
Đặt \(B=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\)
\(=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)\)
\(=\frac{100}{99}+\frac{100}{3\times97}+\frac{100}{5\times95}+...+\frac{100}{49\times51}\)
\(=100\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)
Đặt \(C=\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{97\times3}+\frac{1}{99\times1}\)
\(=2\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)
\(A=\frac{B}{6}=\frac{100}{2}=50\)
Vậy \(A=50\)
A=(1/2-1/5+1/5-1/8+1/8-....+1/50-1/51)
= 1/2-1/51
= 51/102 - 2/102
= 49/102
B=1.4/1.5.4+1.4/5.9.4+...+1.4/41.45.4
= 1/4(1-1/5+1/5-1/9+1/9-...+1/41-1/45)
= 1/4(1-1/45)
= 1/4.44/45
= 11/45