5x+1= 2727chia 27
lam on giup minh nhe
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\(\frac{1+2+2^2+...+2^{2008}}{1-2^{2008}}\)
Ta có: Đặt A = 1 + 2 + 22 + ... + 22008
2A = 2 + 22 + 23 + ... + 22009
2A - A = (2 + 22 + 23 + ... + 22009) - (1 + 2 + 22 + ... + 22008)
A = 22009 - 1
=> \(\frac{1+2+2^2+...+2^{2008}}{1-2^{2008}}=\frac{2^{2009}-1}{1-2^{2008}}\)
\(\frac{x^3-x^2-x-2}{x^5-3x^4+4x^3-5x^2+3x-2}\)
\(=\frac{x^3-2x^2+x^2-2x+x-2}{x^5-2x^4-x^4+2x^3+2x^3-4x^2-x^2+2x+x-2}\)
\(=\frac{\left(x^3-2x^2\right)+\left(x^2-2x\right)+\left(x-2\right)}{\left(x^5-2x^4\right)-\left(x^4-2x^3\right)+\left(2x^3-4x^2\right)-\left(x^2-2x\right)+\left(x-2\right)}\)
\(=\frac{x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)}{x^4\left(x-2\right)-x^3\left(x-2\right)+2x^2\left(x-2\right)-x\left(x-2\right)+\left(x-2\right)}\)
\(=\frac{\left(x-2\right)\left(x^2+x+1\right)}{\left(x-2\right)\left(x^4-x^3+2x^2-x+1\right)}=\frac{x^2+x+1}{x^4-x^3+2x^2-x+1}\)
\(\left(1-\frac{1}{15}\right)\left(1-\frac{1}{21}\right)\left(1-\frac{1}{28}\right)...\left(1-\frac{1}{225}\right)\)
\(=\frac{14}{15}.\frac{20}{21}.\frac{27}{28}...\frac{224}{225}\)
\(=\frac{2.7}{3.5}.\frac{5.4}{7.3}.\frac{3.9}{4.7}...\frac{16.14}{15.15}\)
\(=\frac{2}{3}.\frac{14}{15}\) ( rút gọn )
\(=\frac{28}{45}\)
a/ \(\left|5x+\frac{3}{4}\right|-\frac{5}{4}=2\)
\(\left|5x+\frac{3}{4}\right|=\frac{13}{4}\)
\(\Rightarrow x=\left\{\frac{1}{2};-\frac{4}{5}\right\}\)
b/\(\frac{3}{2}-\left|\frac{1}{2}x+1\right|=\frac{1}{4}\)
\(\left|\frac{1}{2}x+1\right|=\frac{5}{4}\)
1/\(\frac{1}{2}x+1=\frac{5}{4}\)
\(\frac{1}{2}x=\frac{1}{4}\)
\(x=\frac{1}{2}\)
2/\(\frac{1}{2}x+1=-\frac{5}{4}\)
\(\frac{1}{2}x=-\frac{9}{4}\)
\(x=-\frac{9}{2}\)
\(\Rightarrow x=\left\{\frac{1}{2};-\frac{9}{2}\right\}\)
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5x + 1 = 2727 : 27
5x + 1 = 101
5x = 101 - 1
5x = 100
x = 100 : 5
x = 20
Vậy x = 20
\(5x+1=\frac{2727}{27}\)
\(5x+1=101\)
\(5x=100\)
\(x=20\)
vậy \(x=20\)