K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
PN
0
MS
2
4 tháng 9 2017
ban viet lai de bai duoc ko minh ko hieu neu viet so mu thi an shift roi an so 7 la mu vi du 3^2
TB
0
NN
2
6 tháng 5 2017
Ta thấy :
\(\dfrac{1}{4^2}>\dfrac{1}{4.5}\)
\(\dfrac{1}{5^2}>\dfrac{1}{5.6}\)
..............
\(\dfrac{1}{99^2}>\dfrac{1}{99.100}\)
\(\Rightarrow\) \(K>\dfrac{1}{4.5}+\dfrac{1}{5.6}+.....+\dfrac{1}{99.100}\)
Ta có công thức \(\dfrac{a}{b.c}=\dfrac{a}{c-b}.\left(\dfrac{1}{b}-\dfrac{1}{c}\right)\)
Dựa vào công thức ta có :
\(\dfrac{1}{4.5}=\dfrac{1}{4}-\dfrac{1}{5}\)
\(\dfrac{1}{5.6}=\dfrac{1}{5}-\dfrac{1}{6}\)
.......................
\(\dfrac{1}{99.100}=\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow\) \(K>\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+......+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Leftrightarrow\) \(K>\dfrac{1}{4}-\dfrac{1}{100}\)
\(\Rightarrow K>\dfrac{6}{25}>\dfrac{1}{5}\Rightarrow dpcm\) (1)
Ta có :
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
\(\dfrac{1}{5^2}< \dfrac{1}{4.5}\)
................
\(\dfrac{1}{99^2}< \dfrac{1}{98.99}\)
Dựa vào công thức \(\dfrac{a}{b.c}=\dfrac{a}{c-b}.\left(\dfrac{1}{b}-\dfrac{1}{c}\right)\) ta có :
\(K< \dfrac{1}{3.4}+\dfrac{1}{4.5}+......+\dfrac{1}{98.99}\)
\(\Rightarrow\) \(K< \dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+.......+\dfrac{1}{98}-\dfrac{1}{99}\)
\(\Rightarrow\) \(K< \dfrac{1}{3}-\dfrac{1}{99}\)
Vậy \(K< \dfrac{32}{99}< \dfrac{1}{3}\Rightarrow dpcm\) (2)
(1) ; (2) \(\Rightarrow\) \(\dfrac{1}{5}< K< \dfrac{1}{3}\)
Ai thấy đúng thì ủng hộ nha !!!
6 tháng 5 2017
Công thức tổng quát: \(\dfrac{1}{n\left(n+1\right)}< \dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)n}\)
=>\(\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}< K< \dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{98.99}\)
=>\(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}< K< \dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{98}-\dfrac{1}{99}\)
=>\(\dfrac{1}{4}-\dfrac{1}{100}< K< \dfrac{1}{3}-\dfrac{1}{100}\)
=>\(\dfrac{1}{4}< K< \dfrac{1}{3}\)
=>\(\dfrac{1}{5}< K< \dfrac{1}{3}\left(do\dfrac{1}{4}>\dfrac{1}{5}\right)\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
3 tháng 7
1; 5.22 + (\(x\) + 3) = 52
5.4 + (\(x\) + 3) = 25
20 + (\(x\) + 3) = 25
\(x\) + 3 = 25 - 20
\(x+3\) = 5
\(x\) = 5 - 3
\(x\) = 2
Vậy \(x=2\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
3 tháng 7
2; 23 + (\(x\) - 32) = 53 - 43
8 + (\(x\) - 9) = 125 - 64
8 + (\(x\) - 9) = 61
\(x\) - 9 = 61 - 8
\(x\) - 9 = 53
\(x\) = 53 + 9
\(x\) = 62
Vậy \(x\) = 62
KV
27 tháng 10 2023
a) 5.2² + (x + 3) = 5²
5.4 + x + 3 = 25
20 + x + 3 = 25
x + 23 = 25
x = 25 - 23
x = 2
b) 2³ + (x - 3²) = 5³ - 4³
8 + (x - 9) = 125 - 64
8 + x - 9 = 61
x - 1 = 61
x = 61 + 1
x = 62
c) 4.(x - 5) - 2³ = 2⁴.3
4x - 20 - 8 = 16.3
4x - 28 = 48
4x = 48 + 28
4x = 76
x = 76 : 4
x = 19
d) 5.(x + 7) - 10 = 2³.5
5x + 35 - 10 = 8.5
5x + 25 = 40
5x = 40 - 25
5x = 15
x = 15 : 5
x = 3
e) 7² - 7.(13 - x) = 14
49 - 91 + 7x = 14
7x - 42 = 14
7x = 14 + 42
7x = 56
x = 56 : 7
x = 8
H9
HT.Phong (9A5)
CTVHS
27 tháng 10 2023
a) \(5\cdot2^2+\left(x+3\right)=5^2\)
\(\Rightarrow x+3=5^2-5\cdot2^2\)
\(\Rightarrow x+3=25-5\cdot4\)
\(\Rightarrow x+3=5\)
\(\Rightarrow x=5-3\)
\(\Rightarrow x=2\)
b) \(2^3+\left(x-3^2\right)=5^3-4^3\)
\(\Rightarrow8+\left(x-9\right)=125-64\)
\(\Rightarrow8+x-9=61\)
\(\Rightarrow x-1=61\)
\(\Rightarrow x=61+1\)
\(\Rightarrow x=62\)
c) \(4\left(x-5\right)-2^3=2^4\cdot3\)
\(\Rightarrow4\left(x-5\right)=2^4\cdot3+2^3\)
\(\Rightarrow4\cdot\left(x-5\right)=16\cdot3+8\)
\(\Rightarrow4\cdot\left(x-5\right)=56\)
\(\Rightarrow x-5=56:4\)
\(\Rightarrow x-5=14\)
\(\Rightarrow x=19\)
d) \(5\left(x+7\right)-10=2^3\cdot5\)
\(\Rightarrow5\left(x+7\right)=8\cdot5+10\)
\(\Rightarrow5\left(x+7\right)=40+10\)
\(\Rightarrow5\left(x+7\right)=50\)
\(\Rightarrow x+7=10\)
\(\Rightarrow x=10-7\)
\(\Rightarrow x=3\)
e) \(7^2-7\left(13-x\right)=14\)
\(\Rightarrow7\left(13-x\right)=7^2-14\)
\(\Rightarrow7\left(13-x\right)=49-14\)
\(\Rightarrow7\left(13-x\right)=35\)
\(\Rightarrow13-x=5\)
\(\Rightarrow x=13-5\)
\(\Rightarrow x=8\)
f) \(5x-5^2=10\)
\(\Rightarrow5x=10+5^2\)
\(\Rightarrow5x=10+25\)
\(\Rightarrow5x=35\)
\(\Rightarrow x=\dfrac{35}{5}\)
\(\Rightarrow x=7\)
g) \(9x-2\cdot3^2=3^4\)
\(\Rightarrow9x=3^4+2\cdot3^2\)
\(\Rightarrow9x=81+2\cdot9\)
\(\Rightarrow9x=99\)
\(\Rightarrow x=\dfrac{99}{9}\)
\(\Rightarrow x=11\)
h) \(10x+2^2\cdot5=10^2\)
\(\Rightarrow10x=10^2-2^2\cdot5\)
\(\Rightarrow10x=100-4\cdot5\)
\(\Rightarrow10x=80\)
\(\Rightarrow x=\dfrac{80}{10}\)
\(\Rightarrow x=8\)
i) \(125-5\left(4+x\right)=15\)
\(\Rightarrow5\left(4+x\right)=125-5\)
\(\Rightarrow5\left(4+x\right)=120\)
\(\Rightarrow4+x=\dfrac{120}{5}\)
\(\Rightarrow4+x=24\)
\(\Rightarrow x=24-4\)
\(\Rightarrow x=20\)
j) \(2^6+\left(5+x\right)=3^4\)
\(\Rightarrow5+x=3^4-2^6\)
\(\Rightarrow5+x=81-64\)
\(\Rightarrow5+x=17\)
\(\Rightarrow x=17-5\)
\(\Rightarrow x=12\)
7 tháng 7 2023
a: =16-2+91=14+91=105
b: =9*5+8*10-27=45+53=98
c: =32+65-3*8=8+65=73
d; \(=5^3-10^2=125-100=25\)
e: \(=4^2-3^2+1=8\)
f: =9*16-16*8-8+16*4
=16(9-8+4)-8
=16*5-8
=72
H9
HT.Phong (9A5)
CTVHS
7 tháng 7 2023
a) \(2^4-50:25+13\cdot7\)
\(=2^4-2+91\)
\(=16-2+91\)
\(=14+91\)
\(=105\)
b) \(3^2\cdot5+2^3\cdot10-3^4:3\)
\(=9\cdot5+8\cdot10-3^3\)
\(=45+80-27\)
\(=98\)
c) \(2^5+5\cdot13-3\cdot2^3\)
\(=32+65-3\cdot8\)
\(=32+65-24\)
\(=73\)
d) \(5^{13}:5^{10}-5^2\cdot2^2\)
\(=5^{13-10}-\left(5\cdot2\right)^2\)
\(=5^3-10^2\)
\(=125-100\)
\(=25\)
e) \(4^5:4^3-3^9:3^7+5^0\)
\(=4^{5-3}-3^{9-7}+1\)
\(=4^2-3^2+1\)
\(=16-9+1\)
\(=8\)
f) \(3^2\cdot2^4-2^3\cdot4^2-2^3\cdot5^0+4^2\cdot2^2\)
\(=3^2\cdot4^2-2^3\cdot4^2-2^3\cdot1+4^2\cdot2^2\)
\(=4^2\cdot\left(3^2-2^3+2^2\right)-2^3\)
\(=4^2\cdot\left(9-8+4\right)-8\)
\(=16\cdot5-8\)
\(=72\)
29 tháng 7 2023
1: 2*2*2*2*2*2=2^6
2: x*x*x*x=x^4
3: =7^4*6^4=42^4
4: =5^2*3^2*4^2=60^2
5: =2*3^2*5^5
6: =10^3*10^3=10^6
\(10+2\times x=4^5:4\)
\(10+2\times x=256\)
\(2\times x=256-10\)
\(2\times x=246\)
\(x=246:2\)
\(x=123\)