cho A = 1.9+5.13+9.17+...+97+105 . Tính A
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a/\(C=\dfrac{2}{1.7}+\dfrac{2}{7.13}+\dfrac{2}{13.19}+...+\dfrac{2}{1013.1019}\)
\(=\dfrac{1}{3}\left(\dfrac{6}{1.7}+\dfrac{6}{7.13}+\dfrac{6}{13.19}+...+\dfrac{6}{1013.1019}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{19}+...+\dfrac{1}{1013}-\dfrac{1}{1019}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{1019}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{1018}{1019}\)
\(=\dfrac{1018}{3057}\)
b/\(D=\dfrac{7}{1.9}+\dfrac{7}{9.17}+\dfrac{7}{17.25}+...+\dfrac{7}{2011.2019}\)
\(=\dfrac{7}{8}\left(\dfrac{8}{1.9}+\dfrac{8}{9.17}+\dfrac{8}{17.25}+...+\dfrac{8}{2011.2019}\right)\)
\(=\dfrac{7}{8}\left(1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{25}+...+\dfrac{1}{2011}-\dfrac{1}{2019}\right)\)
\(=\dfrac{7}{8}\left(1-\dfrac{1}{2019}\right)\)
\(=\dfrac{7}{8}\cdot\dfrac{2018}{2019}\)
\(=\dfrac{7063}{8076}\)
Ta có: \(\left(\frac{8}{1.9}+\frac{8}{9.17}+\frac{8}{17.25}+...+\frac{8}{49.57}\right)+2\left(x-1\right)=\frac{2x+7}{3}+\frac{5x-8}{4}\)
\(\Leftrightarrow1-\frac{1}{9}+\frac{1}{9}-\frac{1}{17}+\frac{1}{17}-\frac{1}{25}+....+\frac{1}{49}-\frac{1}{57}+2x-2=\frac{8x+28+15x-24}{12}\)
\(\Leftrightarrow1-\frac{1}{57}+2x-2=\frac{23x+4}{12}\)
\(\Leftrightarrow2x-\frac{58}{57}=\frac{23x+4}{12}\)
\(\Leftrightarrow24x-\frac{232}{19}=23x+4\)
\(\Leftrightarrow x=\frac{308}{19}\)
Ta có: \(\frac{8}{1.9}+\frac{8}{9.17}+...+\frac{8}{49.57}=1-\frac{1}{9}+\frac{1}{9}-\frac{1}{17}+...+\frac{1}{49}-\frac{1}{57}=1-\frac{1}{57}=\frac{56}{57}\)
Vậy: 56/57 + 58/57 + 2x - 2 = 2x + 7/3 + 5x - 8/4
2 + 2x - 2 = 2x + 7/3 + 5x - 8/4
2x = 2x + 7/3 + 5x - 8/4
=> 7/3 + 5x - 8/4 = 0
1/3 + 5x = 0
=> 5x = -1/3
=> x = -1/3 : 5=-1/15
a: \(\Leftrightarrow\left(1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+...+\dfrac{1}{49}-\dfrac{1}{57}\right)+2x-2=\dfrac{2}{3}x+\dfrac{7}{3}+\dfrac{5}{4}x-2\)
\(\Leftrightarrow\dfrac{56}{57}+2x-2=\dfrac{23}{12}x+\dfrac{1}{3}\)
=>1/12x=77/57
=>x=308/19
b: =>(x^2-4)(x^2-10)=72
=>x^4-14x^2+40-72=0
=>x^4-14x^2-32=0
=>(x^2-16)(x^2+2)=0
=>x^2-16=0
=>x^2=16
=>x=4 hoặc x=-4
a.\(^{101^2}\)= 10201
b.97 x 103 = 9991
c.\(^{77^2}+^{23^2}+77\times46\)=10000
d.\(^{105^5}-^{5^2}\)= 12762815600
a; (-2,5.0,38.0,4) - {0,125.3,15.(-8)}
= (-2.5.0,4).0,38 - {-8.0,125.3,15}
= - 1.0,38 - {-1.3,15}
= -0,38 + 3,15
= 2,77
b; {(-20,83).0,2 + (-9,17).0,2 : {2,47.0,5 - (-3,53.0,5}
= {0,2.(-20,83 - 9,17)}: {0,5.(2,47 + 3,53)}
= {- 0,2.30}:{0,5.6}
= - 6:6 : 0,5
= -1 : 0,5
= - 2
\(a.\)
\(-\left(-315\right)+\left(-115\right)-105+25\)
\(315-115-105+25\)
\(=215\)
\(b.\)
\(888-\left(-333\right)-222+111\)
\(=888+333-222+111\)
\(=1110\)
\(c.\)
\(-97-15+44-35-12+98\)
\(=-17\)