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i) \(2345-1000\div\left[19-2\left(21-18\right)^2\right]\)
\(=\)\(2345-1000\div\left[19-2.3^2\right]\)
\(=\)\(2345-1000\div\left[19-2.9\right]\)
\(=\)\(2345-1000\div\left[19-18\right]\)
\(=\)\(2345-1000\div1\)
\(=\)\(2345-1000\)
\(=\)\(1345\)
j) \(128-\left[68+8\left(37-35\right)^2\right]\div4\)
\(=\)\(128-\left[68+8.2^2\right]\div4\)
\(=\)\(128-\left[68+8.4\right]\div4\)
\(=\)\(128-\left[68+32\right]\div4\)
\(=\)\(128-100\div4\)
\(=\)\(128-25\)
\(=\)\(3\)
k) \(568-\left\{5\left[143-\left(4-1\right)^2\right]+10\right\}\div10\)
\(=\)\(568-\left\{5\left[143-3^2\right]+10\right\}\div10\)
\(=\)\(568-\left\{5\left[143-9\right]+10\right\}\div10\)
\(=\)\(568-\left\{5.134+10\right\}\div10\)
\(=\)\(568-\left\{670+10\right\}\div10\)
\(=\)\(568-680\div10\)
\(=\)\(568-68\)
\(=\)\(500\)
a) \(107-\left\{38+\left[7.3^2-24\div6+\left(9-7\right)^3\right]\right\}\div15\)
\(=\)\(107-\left\{38+\left[7.3^2-24\div6+2^3\right]\right\}\div15\)
\(=\)\(107-\left\{38+\left[7.9-4+8\right]\right\}\div15\)
\(=\)\(107-\left\{38+\left[63-4+8\right]\right\}\div15\)
\(=\)\(107-\left\{38+67\right\}\div15\)
\(=\)\(107-105\div15\)
\(=\)\(107-7\)
\(=\)\(7\)
b) \(307-\left[\left(180-160\right)\div2^2+9\right]\div2\)
\(=\)\(307-\left[20\div4+9\right]\div2\)
\(=\)\(307-\left[5+9\right]\div2\)
\(=\)\(307-14\div2\)
\(=\)\(307-7\)
\(=\)\(300\)
c) \(205-\left[1200-\left(4^2-2.3\right)^3\right]\div40\)
\(=\)\(205-\left[1200-\left(16-6\right)^3\right]\div40\)
\(=\)\(205-\left[1200-10^3\right]\div40\)
\(=\)\(205-\left[1200-1000\right]\div40\)
\(=\)\(205-200\div40\)
\(=\)\(205-5\)
\(=\)\(200\)
Tìm x :
a) 35 + 3. |x| = 50
b)72 - [ 41 - (2x - 5) - 23.5
c) 70 - 5 (x - 3) = 45
giup mk voi mk dang vội
Tìm x :
a) 35 + 3. |x| = 50
3.|x|=50-35
3.|x|=15
|x|=15:3
|x|=5
x=5 hoặc x=-5
b)72 - [41 - (2x - 5) ]- 23.5
41-(2x-5)=72-40
41-(2x-5)=32
2x-5=41-32
2x-5=9
2x=9+5
2x=14
x=14:2
x=7
c) 70 - 5 (x - 3) = 45
5(x-3)=70-45
5(x-3)=25
x-3=25:5
x-3=5
x-5+3
x=8
S = 1 + 3 + 32 + 33 + ... + 38 + 39
S = ( 1 + 3 ) + ( 32 + 33 ) + ... + ( 38 + 39 )
S = 4 + ( 1 . 32 + 3 .32 ) + .. + ( 1. 38 + 3 . 38 )
S = 4 + 4 .32 + .. + 4 . 38
S = 4 ( 1 + 32 + ... + 38 ) \(⋮\)4
Vậy S \(⋮\)4 ( đpcm )
Học tốt
#Dương
S = 1 + 3 + 32 + 33 + 34+35+ 36 + 37 + 38+39
S=( 1 + 3)+(32 + 33)+(34+35)+(36 + 37)+(38+39)
s=4+32.(3+1)+32.(3+1)+34.(3+1)+36.(3+1)+38.(3+1)
S=4.(1+32+34+36+38)
CHIA HẾT CHO 4
a> =>x+1=3
=>x=2
vậy x=2
b> =>32x-1=33
=>2x-1=3
=>2x=4
=>x=2
vậy x=2
C> =>22+x=27
=>2+x=7
=>x=5
vậy x=5
D> =>33x-2=34
=>3x-2=4
=>3x=6
=>x=2
vậy x=2
a) 6x+1 = 63 <=> x + 1 = 3 <=> x = 2
b) 32x-1 = 27 <=> 32x-1 = 33 <=> 2x-1 = 3 <=> 2x = 4 <=> x = 2
c)4 . 2x = 128 <=> 2x = 128 : 4 <=> 2x = 32 <=> 2x = 25 <=> x = 5
d) 33x-2 = 81 <=> 33x-2 = 34 <=> 3x - 2 = 4 <=> 3x = 6 <=> x = 2
a. 52 + (x+3) = 52
=> x + 3 = 52 - 52
=> x + 3 = 0
=> x = -3
b. 23 + (x-32) = 53 - 43
=> 8 + (x-9) = 125 - 64
=> x - 9 = 125 - 64 - 8
=> x - 9 = 53
=> x = 53 + 9
=> x = 62
A= 1+2+22+23+.......+298+299
A= (1+2)+(22+23)+.......+(298+299 )
A=3+22.(1+2)+...+298.(1+2)
A= 3+22.3+...+298.3
A=3.(22+...+298)
Vid 3 chia hết cho 3 nên A chia hết cho 3
Đơn giản như đang giỡn
HT
a) \(2^x\cdot4=128\)
\(\Rightarrow2^x\cdot2^2=2^7\)
\(\Rightarrow2^{x+2}=2^7\)
\(\Rightarrow x+2=7\)
\(\Rightarrow x=5\)
b) \(\left(2x+1\right)^3=125\)
\(\Rightarrow\left(2x+1\right)^3=5^3\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=4:2\)
\(\Rightarrow x=2\)
c) \(2x-2^6=6\)
\(\Rightarrow2x-64=6\)
\(\Rightarrow2x=70\)
\(\Rightarrow x=70:2\)
\(\Rightarrow x=35\)
d) \(64\cdot4^x=45\)
\(\Rightarrow4^3\cdot4^x=45\)
\(\Rightarrow4^{x+3}=45\)
Xem lại đề
e) \(27\cdot3^x=243\)
\(\Rightarrow3^3\cdot3^x=3^5\)
\(\Rightarrow3^{x+3}=3^5\)
\(\Rightarrow x+3=5\)
\(\Rightarrow x=2\)
g) \(49\cdot7^x=2401\)
\(\Rightarrow7^2\cdot7^x=7^4\)
\(\Rightarrow7^{x+2}=7^4\)
\(\Rightarrow x+2=4\)
\(\Rightarrow x=2\)
h) \(3^x=81\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
k) \(3^4\cdot3^x=3^7\)
\(\Rightarrow3^{x+4}=3^7\)
\(\Rightarrow x+4=7\)
\(\Rightarrow x=3\)
n) \(3^x+25=26\cdot2^2+2\cdot3^0\)
\(\Rightarrow3^x+25=104+2\)
\(\Rightarrow3^x+25=106\)
\(\Rightarrow3^x=81\)
\(\Rightarrow3^x=3^4\)
\(x=4\)
`@` `\text {Ans}`
`\downarrow`
`a)`
`2^x*4 = 128`
`=> 2^x = 128 \div 4`
`=> 2^x = 2^7 \div 2^2`
`=> 2^x = 2^5`
`=> x = 5`
Vậy, `x = 5.`
`b)`
\(\left(2x+1\right)^3=125\)
`=> (2x + 1)^3 = 5^3`
`=> 2x + 1 = 5`
`=> 2x = 5-1`
`=> 2x = 4`
`=> x = 4 \div 2`
`=> x = 2`
Vậy, `x = 2`
`c)`
\(2x-2^6=6\)
`=> 2x = 6+2^6`
`=> 2x = 70`
`=> x = 70 \div 2`
`=> x = 35`
Vậy, `x = 35`
`d)`
\(64\cdot4^x=45\) Bạn xem lại đề
`e)`
`27*3^x = 243`
`=> 3^3 * 3^x = 3^5`
`=> 3^(3 + x) = 3^5`
`=> 3 + x = 5`
`=> x = 5 - 3`
`=> x = 2`
Vậy, `x = 2`
`g)`
`49* 7^x = 2401`
`=> 7^2 * 7^x = 7^4`
`=> 7^(2 + x) = 7^4`
`=> 2 + x = 4`
`=> x = 4 - 2`
`=> x = 2`
Vậy, `x = 2`
`h)`
`3^x = 81`
`=> 3^x = 3^4`
`=> x = 4`
Vậy, `x = 4`
`k)`
`3^4 * 3^x = 3^7`
`=> 3^(4 + x) = 3^7`
`=> 4 + x = 7`
`=> x = 7 - 4`
`=> x = 3`
Vậy, `x = 3`
`n)`
`3^x + 25 = 26*2^2 + 2*3^0`
`=> 3^x + 25 = 104 + 2`
`=> 3^x + 25 = 106`
`=> 3^x = 106 - 25`
`=> 3^x = 81`
`=> 3^x = 3^4`
`=> x = 4`
Vậy, `x = 4.`
\(#48Cd\)