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Tính M
\(M=\frac{\left(1^3+2^3+3^3\right)\left(2^3+3^3+4^3\right)......\left(98^3+99^3+100^3\right)}{\left(1+2+3\right)\left(2+3+4\right)........\left(98+99+100\right)}\)
Tính nhanh :
A = \(\left(\frac{2}{3}+\frac{3}{4}+....+\frac{99}{100}\right)\cdot\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+....+\frac{98}{99}\right)-\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\cdot\left(\frac{2}{3}+\frac{3}{4}+...+\frac{98}{99}\right)\)
A=(2/3+3/4+...+99/100)x(1/2+2/3+3/4+...+98/99)-(1/2+2/3+...+99/100)x(2/3+3/4+4/5+...98/99)
ta cho nó dài hơn như sau
A=(2/3+3/4+4/5+5/6+....+98/99+99/100)
ta thấy các mẫu số và tử số giống nhau nên chệt tiêu các số
2:3:4:5...99 vậy ta còn các số 2/100
ta làm vậy với(1/2+2/3+3/4+.....+98/99) thi con 1/99
làm vậy với câu (1/2+2/3+...+99/100) thì ra la 1/100
vậy với (2/3+3/4+...+98/99) ra 2/99
xùy ra ta có 2/100.1/99-1/100.2/99=1/50x1/99-1/100x2/99=tự tinh nhe mình ngủ đây
bài 1: tính nhanh
a, (100-13).(100-23).(100-33)..........(100-503)
b, \(\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}+\frac{5^2}{26.31}+...........+\frac{5^2}{56.61}\)
c, \(2^{100}-2^{99}+2^{98}-2^{97}+.......+2^2-2^1\)
d, \(2015^{\left(1.4.9.6\right).\left(1.4.9.7\right).\left(1.4.9.8\right)....\left(1.9.9.9\right)}\)
bài 2: Tính
a, \(A=\left(-1\right).\left(-1\right)^2.\left(-1\right)^3.\left(-1\right)^4.\left(-1\right)^5.........\left(-1\right)^{99}\)
b, \(B=512-\frac{512}{2}-\frac{512}{2^2}-\frac{512}{2^3}-........-\frac{512}{2^{10}}\)
đang bận làm để thông cảm nha có j kiếm lại chất xám mình giải cho
1, Tính \(\frac{1}{2}-\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)-\left(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right)+...+\left(\frac{1}{100}+\frac{2}{100}+\frac{3}{100}+...+\frac{99}{100}\right)\)2,Tính \(\left(1-\frac{1}{2^2}\right)x\left(1-\frac{1}{3^2}\right)x\left(1-\frac{1}{4^2}\right)x...x\left(1-\frac{1}{n^2}\right)\)
Bài 1 : Thực hiện phép tính
(1) D = \(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{16}\left(1+2+...+16\right)\)
(2) M =\(\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
Bài 2 : Tìm x biết
(1) \(x-\left\{x-\left[x-\left(-x+1\right)\right]\right\}=1\)
(2) \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right]\cdot x=\frac{2015}{1}+\frac{2014}{2}+...+\frac{1}{2015}\)
(3) \(\frac{x}{\left(a+5\right)\left(4-a\right)}=\frac{1}{a+5}+\frac{1}{4-a}\)
(4) \(\frac{x+2}{11}+\frac{x+2}{12}+\frac{x+2}{13}=\frac{x+2}{14}+\frac{x+2}{15}\)
(5) \(\frac{x+1}{2015}+\frac{x+2}{2014}+\frac{x+3}{2013}+\frac{x+4}{2012}+4=0\)
Bài 3 :
(1) Cho : A =\(\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{1}{9}\); B =\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\)
CMR : \(\frac{A}{B}\)Là 1 số nguyên
(2) Cho : D =\(\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+...+\frac{1}{2000}\)CMR : \(D< \frac{3}{4}\)
Bài 4 : Ký hiệu [x] là số nguyên lớn nhất không vượt quá x , gọi là phần nguyên của x.
VD : [1.5] =1 ; [3] =3 ; [-3.5] = -4
(1) Tính :\(\left[\frac{100}{3}\right]+\left[\frac{100}{3^2}\right]+\left[\frac{100}{3^3}\right]+\left[\frac{100}{3^4}\right]\)
(2) So sánh : A =\(\left[X\right]+\left[X+\frac{1}{5}\right]+\left[X+\frac{2}{5}\right]+\left[X+\frac{3}{5}\right]+\left[X+\frac{4}{5}\right]\)và B = [5x]. Biết x=3.7
a)A=\(\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+...+2006}\right)\)
b)\(B=1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}\)
c)C=\(\frac{1}{2!}+\frac{2}{3!}+...+\frac{n-1}{n!}\)
d)D=\(1+2^2+3^2+...+98^2\)
e)E=\(3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1\)
f)F=\(2^{2010}-2^{2009}-...-2-1\)
g)G=\(\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{100}-1\right)\left(\frac{1}{121}-1\right)\)
câu g)
\(G=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{121}-1\right).\)
\(=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}...\cdot\frac{120}{121}\)
\(=\frac{3.\left(2.4\right).\left(3.5\right)...\left(10.12\right)}{2.2.3.3.4.4.5.5....11.11}\)
\(=\frac{12}{3}=4\)
Bài 1:Tính S= \(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{99^2}+\frac{1}{100^2}}\)
Bài 2: Tính S= 1+3+9+27+...+1438907
Bài 3: Cho \(f\left(1\right)=1;f\left(m+n\right)=f\left(m\right)+f\left(n\right)+mn.\)Tính f(10), f(2015) (Với m, n là các số nguyên dương)
Bài 1
Ta có \(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}=\sqrt{\left(1+\frac{1}{2}-\frac{1}{3}\right)^2}\)
Tương tự như trên ta được
S = 1+1/2-1/3+1+1/3-1/4+...+1+1/99-1/100
= 98 + 1/2 - 1/100
= 9849/100
a,\(\frac{3^9-2^3.3^7+2^{10}.3^2-2^{13}}{3^{10}-2^2.3^7+2^{10}.3^3}\)
b,\(\left(-2\frac{1}{3}\right)^{100}.\left(-0,5\right)^{99}:\left(\frac{7}{3}\right)^{98}:\left(\frac{1}{4}\right)^{50}\)
mk doan la` de sai, sua: \(\frac{3^9-2^3.3^7+2^{10}.3^2-2^{13}}{3^{10}-2^2.3^7+2^{10}.3^3-2^{12}}\)
\(=\frac{3^7.\left(3^2-2^3\right)+2^{10}.\left(3^2-2^3\right)}{3^7.\left(3^3-2^2\right)+2^{10}.\left(3^3-2^2\right)}=\frac{3^7+2^{10}}{\left(3^7+2^{10}\right).24}=\frac{1}{24}\)
Bai 1 Tính \(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right).........\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\)
Bài 2 cho M=2+4+6+8+.......+98+100 Hỏi M có bao nhiêu số và tính tổng của M
Bài 2
Số các số hang của M là:
(100-2):2+1=50 số
Tổng của M là:
(2+100)x50:2=2550
Đáp/Số:.........
Cho M = \(\frac{\left(\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}\right)}{\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{100}\right)}\);
N = \(\frac{\left(92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-...-\frac{92}{100}\right)}{\left(\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+....+\frac{1}{500}\right)}\)
Tìm tỉ số phần trăm của M và N
Ta có :
M = \(\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
M = \(\frac{1+\left(\frac{1}{99}+1\right)+\left(\frac{2}{98}+1\right)+\left(\frac{3}{91}+1\right)+...+\left(\frac{98}{2}+1\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
M = \(\frac{\frac{100}{100}+\frac{100}{99}+\frac{100}{98}+\frac{100}{97}+...+\frac{100}{2}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
M = \(\frac{100.\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+...+\frac{1}{2}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
M = \(100\)
N = \(\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-...-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}\)
N = \(\frac{\left(1-\frac{1}{9}\right)+\left(1-\frac{2}{10}\right)+\left(1-\frac{3}{11}\right)+...+\left(1-\frac{92}{100}\right)}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}\)
N = \(\frac{\frac{8}{9}+\frac{8}{10}+\frac{8}{11}+...+\frac{8}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}\)
N = \(\frac{8.\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)}{\frac{1}{5}.\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)}\)
N = \(40\)
\(\Rightarrow\)M : N = \(\frac{100}{40}\%=250\%\)
\(M=\frac{1+(\frac{1}{99}+1)+(\frac{2}{98}+1)+(\frac{3}{97}+1)+...+(\frac{98}{2}+1)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
\(M=\frac{\frac{100}{100}+\frac{100}{99}+\frac{100}{98}+\frac{100}{97}+...+\frac{100}{2}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
\(M=\frac{100\cdot(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+...+\frac{1}{2})}{(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100})}=100\)
\(N=\frac{(1-\frac{1}{9})+(1-\frac{2}{10})+(1-\frac{3}{11})+...+(1-\frac{92}{100})}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}\)
\(N=\frac{\frac{8}{9}+\frac{8}{10}+\frac{8}{11}+...+\frac{8}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}=\frac{8(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100})}{\frac{1}{5}(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100})}=40\)
\(M:N=\frac{100}{40}=250\%\)