23.42.53
9.33.\(\dfrac{1}{81}\).32
32.52.\((\dfrac{2}{3})\)2
\((\dfrac{1}{3})\)2.\(\dfrac{1}{3}\).92
tính H = \(\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}}:\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{500}}\)
Tính hợp lý
\(A= (\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\) B= \(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}}{\dfrac{1}{9}+\dfrac{2}{8}+\dfrac{3}{7}+...+\dfrac{8}{2}+\dfrac{9}{1}})\)
Tính
a)\((\dfrac{3}{4}-81).(\dfrac{3^2}{5}-81).(\dfrac{3^3}{6}-81)...(\dfrac{3^{2000}}{2003}-81)\)
b)\(\dfrac{69}{157}-(2+(3+(4+5^{-1})^{-1})^{-1}\)
a: \(\left(\dfrac{3}{4}-81\right)\left(\dfrac{3^2}{5}-81\right)\cdot...\cdot\left(\dfrac{3^{2000}}{2003}-81\right)\)
\(=\left(\dfrac{3^6}{9}-81\right)\left(\dfrac{3}{4}-81\right)\cdot\left(\dfrac{3^2}{5}-81\right)\cdot...\cdot\left(\dfrac{3^{2000}}{2003}-81\right)\)
\(=\left(81-81\right)\left(\dfrac{3}{4}-81\right)\cdot\left(\dfrac{3^2}{5}-81\right)\cdot...\cdot\left(\dfrac{3^{2000}}{2003}-81\right)\)
=0
b: \(\dfrac{69}{157}-\left(2+\left(3+4+5^{-1}\right)^{-1}\right)^{-1}\)
\(=\dfrac{69}{157}-\left(2+\left(3+4+\dfrac{1}{5}\right)^{-1}\right)^{-1}\)
\(=\dfrac{69}{157}-\left(2+1:\dfrac{36}{5}\right)^{-1}\)
\(=\dfrac{69}{157}-\left(2+\dfrac{5}{36}\right)^{-1}\)
\(=\dfrac{69}{157}-\left(\dfrac{77}{36}\right)^{-1}\)
\(=\dfrac{69}{157}-\dfrac{36}{77}=\dfrac{-339}{12089}\)
Tính hợp lí:
A= \(\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
B= \(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}}{\dfrac{1}{9}+\dfrac{2}{8}+\dfrac{3}{7}+...+\dfrac{8}{2}+\dfrac{9}{1}}\)
Cái này mk từng làm nhưng có chút sai sót vậy nên bn sữa cho mk chút nhé ! Thay vì N = ... thì bn thay bằng A = ... nha
Ta có :
N = 40 ( A = 40 )
Cho M = \(\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+..........+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+..........+\dfrac{1}{100}}\) ; N = \(\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-.........-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+......+\dfrac{1}{500}}\)
Tìm tỉ số phần trăm của M và N
+)Đặt A= \(\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\dfrac{99}{1}\)
A= \(\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\left(1+1+1+...+1\right)\) (99 chữ số 1)
A= \(\left(\dfrac{1}{99}+1\right)+\left(\dfrac{2}{98}+1\right)+...+\left(\dfrac{98}{2}+1\right)+1\)
A= \(\dfrac{100}{99}+\dfrac{100}{98}+...+\dfrac{100}{2}+1\)
A= \(100.\left(\dfrac{1}{99}+\dfrac{1}{98}+...+\dfrac{1}{2}+\dfrac{1}{100}\right)\)
⇒ M= \(\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+...+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}\)
M= \(\dfrac{100.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{100}}\)
M= 100 (1)
+) Đặt B= \(92-\dfrac{1}{9}-\dfrac{2}{10}-...-\dfrac{92}{100}\)
B= \(\left(1+1+1+...+1\right)-\dfrac{1}{9}-\dfrac{2}{10}-...-\dfrac{92}{100}\) ( 92 chữ số 1)
B= \(\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+...+\left(1-\dfrac{92}{100}\right)\)
B= \(\dfrac{8}{9}+\dfrac{8}{10}+...+\dfrac{8}{100}\)
B= \(8.\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{100}\right)\)
⇒ N= \(\dfrac{8.\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{100}\right)}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{500}}\)
N= 8 (2)
Từ (1) và (2)⇒ \(\dfrac{M}{N}\) = \(\dfrac{100}{8}\)= \(\dfrac{25}{2}\)
Vậy \(\dfrac{M}{N}=\dfrac{25}{2}\)
Tính \(H=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...........+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+.............+\dfrac{1}{100}}:\dfrac{92-\dfrac{1}{9}-\dfrac{1}{10}-..............\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+........+\dfrac{1}{500}}\)
Help me!!!
Đặt vế đầu là A, vế sau là B.
Vế A:
- Tử:
\(\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\dfrac{99}{1}\)
\(=100\left(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+...+\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{1}{100}\right)\)
\(=100\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{98}+\dfrac{1}{99}+\dfrac{1}{100}\right)\)
Vậy:
\(A=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}}\\ =\dfrac{50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+..+\dfrac{1}{100}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}}\\ \Rightarrow A=50\)
Vế B:
- Tử:
\(92-\dfrac{1}{9}-\dfrac{1}{10}-...-\dfrac{92}{100}\\ =\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+...+\left(1-\dfrac{92}{100}\right)\\ =\dfrac{8}{9}+\dfrac{8}{10}+...+\dfrac{8}{100}\\ =\dfrac{40}{45}+\dfrac{40}{50}+...+\dfrac{40}{500}\\ =40\left(\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{500}\right)\)
Vậy:
\(B=\dfrac{92-\dfrac{1}{9}-\dfrac{1}{10}-...-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{500}}\\ =\dfrac{40\left(\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{500}\right)}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{500}}\\ \Rightarrow B=40\)
Từ 2 vế trên ta tính được \(\dfrac{A}{B}=\dfrac{50}{40}=\dfrac{5}{4}\)
1. \(\left(y+\dfrac{1}{3}\right)\)+\(\left(y+\dfrac{1}{9}\right)\)+\(\left(y+\dfrac{1}{27}\right)\)+\(\left(y+\dfrac{1}{81}\right)\)=\(\dfrac{56}{81}\)
2. 18:\(\dfrac{Xx0,4+0,32}{X}\)+5=14
3. \(\dfrac{3xX}{2}\)=\(\dfrac{2}{5}+\)X\(+\dfrac{1}{3}\)
4. X-\(\dfrac{11}{15}\)=\(\dfrac{3+X}{5}\)
Bài 1:
$(y+\frac{1}{3})+(y+\frac{1}{9})+(y+\frac{1}{27})+(y+\frac{1}{81})=\frac{56}{81}$
$(y+y+y+y)+(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81})=\frac{56}{81}$
$4\times y+\frac{40}{81}=\frac{56}{81}$
$4\times y=\frac{56}{81}-\frac{40}{81}=\frac{16}{81}$
$y=\frac{16}{81}:4=\frac{4}{81}$
Bài 2:
$18: \frac{x\times 0,4+0,32}{x}+5=14$
$18: \frac{x\times 0,4+0,32}{x}=14-5=9$
$\frac{x\times 0,4+0,32}{x}=18:9=2$
$x\times 0,4+0,32=2\times x$
$2\times x-x\times 0,4=0,32$
$x\times (2-0,4)=0,32$
$x\times 1,6=0,32$
$x=0,32:1,6=0,2$
Bài 3:
$\frac{3\times x}{2}=\frac{2}{5}+x+\frac{1}{3}$
$1,5\times x=x+\frac{11}{15}$
$1,5\times x-x=\frac{11}{15}$
$x\times (1,5-1)=\frac{11}{15}$
$x\times 0,5=\frac{11}{15}$
$x=\frac{11}{15}: 0,5=\frac{22}{15}$
Giúp mk với
Câu 1:
Cho A = \(\dfrac{1}{\dfrac{99}{\dfrac{1}{2}+}}+\dfrac{2}{\dfrac{98}{\dfrac{1}{3}+}}+\dfrac{3}{\dfrac{97}{\dfrac{1}{4}+....}}+...+\dfrac{99}{\dfrac{1}{\dfrac{1}{100}}}\).
B =\(\dfrac{92}{\dfrac{1}{45}+}-\dfrac{1}{\dfrac{9}{\dfrac{1}{50}+}}-\dfrac{2}{\dfrac{10}{\dfrac{1}{55}+}}-\dfrac{3}{\dfrac{11}{\dfrac{1}{60}+....}}-...\dfrac{92}{\dfrac{100}{\dfrac{1}{500}}}\). Tính \(\dfrac{A}{B}\)
a,2.(\(\dfrac{1}{4}\)+x)\(^3\)=(\(-\dfrac{27}{4}\))
b,(x+\(\dfrac{1}{2}\))\(^3\):3=\(\dfrac{-1}{81}\)
c,(\(\dfrac{2}{3}\)-x)\(^2\)=1:\(\dfrac{4}{9}\)
d,(2x-\(\dfrac{1}{5}\))\(^2\)+\(\dfrac{16}{25}\)=1
e,(\(\dfrac{2}{5}\)-3x)\(^2\)-\(\dfrac{1}{5}\)=\(\dfrac{4}{25}\)