Bài 1 :Tìm a,b biết :
\(a+b=3.\left(a-b\right)=\)\(2\frac{a}{b}\)
Bài 2 :
\(A=50+\frac{50}{3}+\frac{25}{3}+\frac{20}{4}+\frac{10}{5}+\frac{100}{67}+...+\frac{100}{98.99}+\frac{1}{99}\)
Tìm X:
\(\frac{50+\frac{50}{3}+\frac{25}{3}+\frac{20}{4}+\frac{10}{3}+\frac{100}{6.7}+...+\frac{100}{98.99}+\frac{1}{99}}{\left(2x-1\right)}\)=11
Tính
\(B=50+\frac{50}{3}+\frac{25}{3}+\frac{20}{4}+\frac{10}{3}+\frac{100}{6.7}+...+\frac{100}{98.99}+\frac{1}{99}\)
Tính tổng :
\(B=50+\frac{50}{3}+\frac{25}{3}+\frac{20}{4}+\frac{10}{3}+\frac{100}{6.7}+...+\frac{100}{98.99}+\frac{1}{99}\)
\(B=50+\frac{50}{3}+\frac{25}{3}+\frac{20}{4}+\frac{10}{3}+\frac{100}{6.7}+...+\frac{100}{98.99}+\frac{1}{99}\)
\(B=\frac{100}{1.2}+\frac{100}{2.3}+\frac{100}{3.4}+\frac{100}{4.5}+\frac{100}{5.6}+\frac{100}{6.7}+...+\frac{100}{98.99}+\frac{100}{99.100}\)
\(B=100\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(B=100\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(B=100\left(1-\frac{1}{100}\right)\)
\(B=100.\frac{99}{100}=99\)
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
bài 1:
tìm n biết: 5n+7 chia hết 3n+2
bài 2:
1, tìm chữ số tận cùng của:
a,57^1999
b,93^1999
2, Cho A= 999993^1999 - 555557^1997
chứng minh rằng: A chia hết cho 5
bài 3:chứng minh rằng:
a) \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
b)\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
Bài 5:Tìm x biết:
a)11.(x-6)=4.x+11
b)\(4\frac{1}{3}.\left(\frac{1}{6}-\frac{1}{2}\right)\le x\le\frac{2}{3}.\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)\)với x\(\in\)Z
c)|x-3|+1=x
Bài 3:
a,Đặt A = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)
A = \(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\)
2A = \(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\)
2A + A = \(\left(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\right)+\left(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\right)\)
3A = \(1-\frac{1}{2^6}\)
=> 3A < 1
=> A < \(\frac{1}{3}\)(đpcm)
b, Đặt A = \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
3A = \(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
3A + A = \(\left(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\right)-\left(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\right)\)
4A = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
=> 4A < \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\) (1)
Đặt B = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
3B = \(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
3B + B = \(\left(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\right)+\left(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\right)\)
4B = \(3-\frac{1}{3^{99}}\)
=> 4B < 3
=> B < \(\frac{3}{4}\) (2)
Từ (1) và (2) suy ra 4A < B < \(\frac{3}{4}\)=> A < \(\frac{3}{16}\)(đpcm)
bài 1:
5n+7 chia hết cho 3n+2
=> [3(5n+7) - 5(3n + 2)] chia hết cho 3n+2
=> (15n + 21 - 15n - 10) chia hết cho 3n+2
=> 11 chia hết cho 3n + 2
=> 3n + 2 thuộc Ư(11) = {1;-1;11;-11}
Ta có bảng:
3n + 2 | 1 | -1 | 11 | -11 |
n | -1/3 (loại) | -1 (chọn) | 3 (chọn) | -13/3 (loại) |
Vậy n = {-1;3}
Bài 2:
1, chữ số tận cùng
a, Xét 71999
Ta có: 71999 = 71996.73 = (74)499.343 = (...1)499.343 = (....1).343 = ....3 (1)
Vậy số 571999 có tận cùng là 3
b, Xét 31999
Ta có: 31999 = 31996.33 = (34)499.27 = (...1)499.27 = (...1) . 27 = ....7 (2)
Vậy số 931999 có chữ số tận cùng là 7
2,
Từ (1) và (2) suy ra A = 9999931999 + 5555571999 = ...7 + ...3 = ....0
Vì A có chữ số tận cùng là 0 nên A chia hết cho 5.
\(A=\frac{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+......+\frac{1}{999}}{\frac{1}{1.999}+\frac{1}{3.997}+\frac{1}{5.995}+......+\frac{1}{999.1}}\)
\(B=\frac{1+\left(1+2\right)+\left(1+2+3\right)+\left(1+2+3+4\right)+......+\left(1+2+3+...+98\right)}{1.2+2.3+3.4+4.5+......+98.99}\)
\(C=\frac{\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+......+\frac{1}{100.400}}{\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+......+\frac{1}{299.400}}\)
\(D=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+......+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{100}}:\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{97}-......-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+......+\frac{1}{500}}\)
Giup tui voi !!!!!!!!!!!!!!!!!!!!!!!!!!! Mai phai nop roi !!!!!!!!!!!!!!!!!!!
Tìm tỉ số phần trăm của A và B biết:
\(A=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+.....+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{99}+\frac{1}{100}}\) \(B=\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-....-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+....+\frac{1}{500}}\)
1.Chứng tỏ rằng:
A=75.(42004+42003+...+42+4+1)+25 chia hết cho 100
2.tính nhanh:
\(A=\frac{\left(1+2+3+...+99+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)
\(B=\frac{\left(\frac{1}{14}-\frac{\sqrt{2}}{7}+\frac{\sqrt[3]{2}}{35}\right).\left(-\frac{4}{15}\right)}{\left(\frac{1}{10}+\frac{\sqrt[3]{2}}{25}-\frac{\sqrt{2}}{5}\right).\frac{5}{7}}\)
3.a)tính giá trị của biểu thức A=3x2-2x+1 với |x|=\(\frac{1}{2}\)
b)Tìm x nguyên để \(\sqrt{x+1}\)chia hết cho \(\sqrt{x-3}\)
1. A = 75(42004 + 42003 +...+ 42 + 4 + 1) + 25
A = 25 . [3 . (42004 + 42003 +...+ 42 + 4 + 1) + 1]
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 3 + 1)
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 4)
A = 25 . 4 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1)
A =100 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1) \(⋮\) 100
3a) |x| = 1/2
=> x = 1/2 hoặc x = -1/2
với x = 1/2:
A = \(3.\left(\frac{1}{2}\right)^2-2.\frac{1}{2}+1\)
\(A=\frac{3}{4}-1+1=\frac{3}{4}\)
với x = -1/2
A = \(3.\left(-\frac{1}{2}\right)^2-2\left(-\frac{1}{2}\right)+1\)
\(A=\frac{3}{4}+1+1=\frac{3}{4}+2=\frac{11}{4}\)
2.
A=\(\frac{\left(1+2+3+.....+99+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+......+99-100}\)\
A=0
1/ Tính
a) \(P=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{16}\left(1+2+3+...+16\right)\)
b) Cho \(a+b+c=2010\)và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{3}\)
Tính \(S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
2/ Tìm x biết
\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}...\frac{30}{62}\cdot\frac{31}{64}=2^x\)
3/ Tìm \(a_1;a_2;a_3;...;a_{100}\)biết \(\frac{a_1-1}{100}=\frac{a_2-2}{99}=\frac{a_3-3}{98}=...=\frac{a_{100}-100}{1}\)và \(a_1+a_2+a_3+...+a_{100}=10100\)