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a) \(3.5^2-27:3^2-5^2.4-18:3^2\)
\(=3.\left(5^2-5^2\right).27:\left(3^2-3^2\right)\)
\(=15.0.27.0\)
\(=0.0=0\)
b)
2x-1 là bội của x+3
=> 2x-1 chia hết cho x+3
hay [2(x+3)-7] chia hết ho x+ 3
=> 7 chia hết cho x+ 3
x+3 εεƯ(7)={1,-1,7,-7}
x+3=1 x+3=-1 x+3=7 x+3= -7
x = 1-3 x = -1-3 x = 7-3 x = -7-3
x = -2 x = -4 x =4 x = -10
Vậy x= -2, x=-4,x= 4, x= -10
c) \(205-\left[1200-\left(4^2-2.3\right)^3\right]:40\)
\(=205-\left[1200-16-6^3\right]:40\)
\(=205-\left[1200-10^3:40\right]\)
\(=205-1200-1000:40\)
\(=205-200:40\)
\(=205-5\)
\(=200\)
=(2-1)*(2+1)+(4-1)*(4+1)+ ...+(2n-1)*(2n+1) =(2^2-1)+(4^2-1)+...+(4n^2-1) =(2^2+4^2+...+4n^2)-(1+1+...+1) =4(1^2+2^2+...n^2)-n n(n+1)(2n+1)/6: 1^2+2^2+3^2+…+n^2=n(n+1)(2n+1)/6n^2=n 1x3+3x5+5x7+7x9+...+17x19 =4(1^2+2^2+...n^2)-n =4*n(n+1)(2n+1)/6-n; n=10,1x3+3x5+5x7+7x9+...+17x19=1530
con Nguyễn Kim Ngân ngu vãi tao ko biết thì tao mới phải hỏi
\(\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)...\left(1+\dfrac{1}{49.51}\right)\)+\(\dfrac{2}{51}\)
=\(\dfrac{4}{1.3}.\dfrac{9}{2.4}.\dfrac{16}{3.5}.....\dfrac{2500}{49.51}\)+\(\dfrac{2}{51}\)
=\(\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.....\dfrac{50^2}{49.51}\)+\(\dfrac{2}{51}\)
=\(\dfrac{\left(2.3.4.....50\right)\left(2.3.4.....50\right)}{\left(1.2.3.....49\right)\left(3.4.....51\right)}\)+\(\dfrac{2}{51}\)
=\(\dfrac{\left(2.3.4.....49\right).50.2.\left(3.4.5.....50\right)}{1.\left(2.3.4.....49\right)\left(3.4.5.....50\right).51}\)+\(\dfrac{2}{51}\)
=\(\dfrac{50.2}{1.51}\)+\(\dfrac{2}{51}\)=\(\dfrac{100}{51}\)+\(\dfrac{2}{51}\)=\(\dfrac{102}{51}\)=2
\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\)
\(S=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(S=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)
\(S=\frac{1}{2}.\frac{2016}{2017}\)
\(S=\frac{1008}{2017}< \frac{1}{2}\)
\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\)
\(2S=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2015.2017}\)
\(2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\)
\(2S=1-\frac{1}{2017}< 1\)
=> 2S < 1
=> S < \(\frac{1}{2}\)(đpcm)
\(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)y=\frac{2}{3}\)
=> \(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)y=\frac{2}{3}\)
=> \(\frac{1}{2}\left(1-\frac{1}{11}\right)y=\frac{2}{3}\)
=> \(\frac{1}{2}.\frac{10}{11}y=\frac{2}{3}\)
=> \(\frac{5}{11}y=\frac{2}{3}\)
=>y = \(\frac{2}{3}:\frac{5}{11}\)
=> y = \(\frac{22}{15}\)
cho mk cái lời giải thích chỗ nhân 1/2 ý mk ko hiểu mong bn thông cảm
\(3x^5-1=2\)
\(3x^5=3\)
\(x^5=1\)
\(x^5=1^5\)
\(x=1\)
x=1