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T
9 tháng 4 2022
Câu 14)
\(a,\\ =-\dfrac{3}{8}+\dfrac{8}{17}+\dfrac{-5}{8}-\dfrac{3}{5}+\dfrac{9}{17}\\ =\left(\dfrac{-3}{8}+\dfrac{-5}{8}\right)+\left(\dfrac{8}{17}+\dfrac{9}{17}\right)-\dfrac{3}{5}\\ =\left(-1\right)+1-\dfrac{3}{5}=0-\dfrac{3}{5}=\dfrac{-3}{5}\\ b,\\ =\dfrac{7}{15}.\dfrac{-15}{14}+\left(\dfrac{27}{16}-\dfrac{1}{8}\right):\dfrac{5}{8}\)
\(=\dfrac{-1}{2}+\dfrac{25}{16}.\dfrac{8}{5}=\dfrac{-1}{2}+\dfrac{5}{2}=2\\ c,\\ =\dfrac{2}{2}-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{4}+.....+\dfrac{2}{99}-\dfrac{2}{100}\\ =1-\dfrac{1}{50}=\dfrac{49}{50}\)
Câu 15
\(a,2x+\dfrac{-1}{4}=\dfrac{3}{2}\\ 2x=\dfrac{3}{2}-\dfrac{-1}{4}=\dfrac{7}{4}\\ x=\dfrac{7}{4}:2=\dfrac{7}{8}\\ b,\dfrac{15}{x}=\dfrac{-3}{4}\\ x=\dfrac{15.4}{-3}=-20\)
NT
12 tháng 2 2017
Bài này có mẹo á ; giải ra dễ lắm !!!
\(\left(100-1^2\right)\left(100-2^2\right)....\left(100-10^2\right)......\left(100-20^2\right)\\ =\left(100-1\right).\left(100-4\right)....0....\left(100-400\right)=0\\ \)
Chúc bạn học tốt !!!
D
2
AR
17 tháng 8 2017
A = 10,11 + 11,12 + 12,13 + . . .+ 98,99 + 99,10
Ta có :
10,11 = 10 + 0,11
11,12 = 11 + 0,12
12,13 = 12 + 0,13
. . . . . . . . . . . . . .
97,98 = 97 + 0,98
98,99 = 98 + 0,99
99,10 = 99 + 0,10
Đặt B = 10 + 11 + 12 + 13 + . .. +98 + 99
và C = 0,11 + 0,12 + 0,13 + . . . .+ 0,98 + 0,99 + 0,10
- - > 100C = 11 + 12 + 13 + . . .+ 98 + 99 + 10
Ta chỉ việc tính B là suy ra C !
B = 10 + 11 + 12 + 13 + . .. +98 + 99
B = (10+99)+(11+98)+(12+97)+. . . +(44+65) + (45 + 64)
Vì từ 10 đến 99 có tất cả 90 số . Ta sẽ có 90/2 = 45 cặp
Mỗi cặp có tổng là 10 + 99 = 11 + 98 = . .= 45 +64 = 109
Vậy ta có B = 45.109 = 4905
Với A = 4905 . Ta thấy 100C = 10 + 11 + 12 +. . + 98 + 99 =B
- - > 100C = 4905 . Hay C = 4905/100 = 49,05
Vậy A = B + C = 4905 + 49,05 = 4954,05
AT
12 tháng 6 2017
i) \(5\dfrac{8}{17}:x+\left(-\dfrac{4}{17}\right):x+3\dfrac{1}{7}:17\dfrac{1}{3}=\dfrac{4}{11}\)
\(\Rightarrow\dfrac{93}{17}:x-\dfrac{4}{17}:x+\dfrac{33}{182}=\dfrac{4}{11}\)
\(\Rightarrow\left(\dfrac{93}{17}-\dfrac{4}{17}\right):x=\dfrac{4}{11}-\dfrac{33}{182}\)
\(\Rightarrow\dfrac{89}{17}:x=\dfrac{365}{2002}\)
\(\Rightarrow x=\dfrac{89}{17}:\dfrac{365}{2002}=\dfrac{178178}{6205}\)
j) \(\dfrac{17}{2}-\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\)
\(\Rightarrow\left|2x-\dfrac{3}{4}\right|=\dfrac{17}{2}-\left(-\dfrac{7}{4}\right)=\dfrac{41}{4}\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{41}{4}\\2x-\dfrac{3}{4}=-\dfrac{41}{4}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}2x=11\Rightarrow x=\dfrac{11}{2}\\2x=-\dfrac{19}{2}\Rightarrow x=-\dfrac{19}{4}\end{matrix}\right.\)
k) \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)
\(\Rightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{26}{25}-\dfrac{17}{25}=\dfrac{9}{25}=\left(\dfrac{3}{5}\right)^2\)\(=\left(-\dfrac{3}{5}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\Rightarrow x=\dfrac{2}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\Rightarrow x=-\dfrac{4}{5}\end{matrix}\right.\)
l) \(-1\dfrac{5}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)
\(\Rightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-32}{27}-\left(-\dfrac{24}{27}\right)=-\dfrac{8}{27}=\left(-\dfrac{2}{3}\right)^3\)
\(\Rightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)
\(\Rightarrow3x=-\dfrac{2}{3}+\dfrac{7}{9}=\dfrac{1}{9}\)
\(\Rightarrow x=\dfrac{1}{27}\)
S
12 tháng 6 2017
j, \(\dfrac{17}{2}-\left|2x-\dfrac{3}{4}\right|=\dfrac{-7}{4}\)
\(\Rightarrow-\left|2x-\dfrac{3}{4}\right|=\dfrac{-7}{4}-\dfrac{17}{2}\)
\(\Rightarrow-\left|2x-\dfrac{3}{4}\right|=\dfrac{-41}{4}\)
\(\Rightarrow\left|2x-\dfrac{3}{4}\right|=\dfrac{41}{4}\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{41}{4}\\2x-\dfrac{3}{4}=\dfrac{-41}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=\dfrac{-19}{4}\end{matrix}\right.\)
k, \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)
\(\Rightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)
\(\Rightarrow x+\dfrac{1}{5}=\pm\dfrac{3}{5}\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=\dfrac{-3}{5}\end{matrix}\right.\Rightarrow}\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=\dfrac{-4}{5}\end{matrix}\right.\)
l, \(-1\dfrac{5}{27}-\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-24}{27}\)
\(\Rightarrow-\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-19}{27}\)
\(\Rightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{19}{27}\)
\(\Rightarrow3x-\dfrac{7}{9}=\dfrac{\sqrt[3]{19}}{3}\)
\(\Rightarrow3x=\dfrac{\sqrt[3]{19}}{3}+\dfrac{7}{19}\)
\(\Rightarrow...\)
giải giùm tớ nha
DH
14 tháng 5 2017
Từ đề bài ta có:
\(T=\dfrac{1+2}{2}.\dfrac{1+3}{3}.\dfrac{1+4}{4}...\dfrac{1+98}{98}.\dfrac{1+99}{99}\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{99}{98}.\dfrac{100}{99}\)
\(=\dfrac{100}{2}\)
\(=50\).
JH
15 tháng 5 2017
\(T=\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{98}+1\right)\left(\dfrac{1}{99}+1\right)\)
\(T=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}....\dfrac{99}{98}.\dfrac{100}{99}\)
\(T=\dfrac{3.4.5......99}{3.4.5......99}.\dfrac{100}{2}\)
\(T=50\)
10 tháng 8 2018
Tuy có vẻ hơi muộn nhưng thôi 
Nếu A là số tự nhiên ⇒ \(\dfrac{1}{10}\left(7^{2004}-3^{92^{94}}\right)\in N\)
\(\Rightarrow7^{2004}-3^{92^{94}}⋮10\)
Thật vậy, ta có :
72004 với lũy thừa là 2004 ⋮ 4
⇒ 72004 = ( .......... 9 )
392^94 với lũy thừa là 9294 mà 92 ⋮ 4 ⇒ 9294 ⋮ 4
⇒ 392^94 = ( .......... 9 )
⇒ 72004 - 392^94 = ( .......... 9 ) - ( ............ 9) = ( ........... 0 ) ⋮ 10
⇒ \(\dfrac{1}{10}\left(7^{2004}-3^{92^{94}}\right)\in N\)
A=1/10.(72004-392^94) là số tự nhiên.
11 tháng 7 2017
Nếu là z+x thì mik biết làm nè:
Đặt x-y=2011(1)
y-z=-2012(2)
z+x=2013(3)
Cộng (1);(2);(3) lại với nhau ta được :
2x=2012=>x=1006
Từ (1) => y=-1005
Từ (3) => z=1007
Bài 3:
a) \(\dfrac{3}{4}.\dfrac{16}{9}-\dfrac{7}{5}:\dfrac{-21}{20}=\dfrac{4}{3}-\dfrac{-4}{3}=\dfrac{8}{3}\)
b) \(2\dfrac{1}{3}-\dfrac{1}{3}.\left[\dfrac{-3}{2}+\left(\dfrac{2}{3}+0,4.5\right)\right]\)
\(=\dfrac{7}{3}-\dfrac{1}{3}.\left[\dfrac{-3}{2}+\left(\dfrac{2}{3}+\dfrac{2}{5}.5\right)\right]\)
\(=\dfrac{7}{3}-\dfrac{1}{3}.\left[\dfrac{-3}{2}+\left(\dfrac{2}{3}+2\right)\right]\)
\(=\dfrac{7}{3}-\dfrac{1}{3}.\left[\dfrac{-3}{2}+\dfrac{8}{3}\right]\)
\(=\dfrac{7}{3}-\dfrac{1}{3}.\dfrac{7}{6}\)
\(=\dfrac{7}{3}-\dfrac{7}{18}\)
\(=\dfrac{35}{18}\)
c) \(\left(20+9\dfrac{1}{4}\right):2\dfrac{1}{4}=\left(20+\dfrac{37}{4}\right):\dfrac{9}{4}=\dfrac{117}{4}:\dfrac{9}{4}=13\)
d) \(\left(6-2\dfrac{4}{5}\right).3\dfrac{1}{8}-1\dfrac{3}{5}:\dfrac{1}{4}\)
\(=\left(6-\dfrac{14}{5}\right).\dfrac{25}{8}-\dfrac{8}{5}:\dfrac{1}{4}\)
\(=\dfrac{16}{5}.\dfrac{25}{8}-\dfrac{32}{5}\)
\(=10-\dfrac{32}{5}\)
\(=\dfrac{18}{5}\)
e) \(\dfrac{32}{15}:\left(-1\dfrac{1}{5}+1\dfrac{1}{3}\right)=\dfrac{32}{15}:\left(\dfrac{-6}{5}+\dfrac{4}{3}\right)=\dfrac{32}{15}:\dfrac{2}{15}=16\)
g) \(0,2.\dfrac{15}{36}-\left(\dfrac{2}{5}+\dfrac{2}{3}\right):1\dfrac{1}{5}\)
\(=\dfrac{1}{5}.\dfrac{5}{12}-\dfrac{16}{15}:\dfrac{6}{5}\)
\(=\dfrac{1}{12}-\dfrac{8}{9}\)
\(=\dfrac{-29}{36}\)
h) \(1\dfrac{13}{15}.0,75-\left(\dfrac{8}{15}+0,25\right).\dfrac{24}{27}\)
\(=\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{8}{15}+\dfrac{1}{4}\right).\dfrac{8}{9}\)
\(=\dfrac{7}{5}-\dfrac{47}{60}.\dfrac{8}{9}\)
\(=\dfrac{7}{5}-\dfrac{94}{135}\)
\(=\dfrac{19}{27}\)
g) \(5:\left(4\dfrac{3}{4}-1\dfrac{25}{28}\right)-1\dfrac{3}{8}:\left(\dfrac{3}{8}+\dfrac{9}{20}\right)\)
\(=5:\left(\dfrac{19}{4}-\dfrac{53}{28}\right)-\dfrac{11}{8}:\dfrac{33}{40}\)
\(=5:\dfrac{20}{7}-\dfrac{5}{3}\)
\(=\dfrac{7}{4}-\dfrac{5}{3}\)
\(=\dfrac{1}{12}\)
\(a\)) \(\dfrac{3}{4}.\dfrac{16}{9}-\dfrac{7}{5}:\dfrac{-21}{20}\)
\(=\dfrac{4}{3}-\dfrac{7}{5}.\dfrac{-20}{21}\)
\(=\dfrac{4}{3}-\dfrac{-4}{3}\)
\(=\dfrac{8}{3}\)
\(=2\dfrac{2}{3}\)