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a: \(2^3-5^3:5^2+12\cdot2^2\)
\(=8-5+12\cdot4\)
=3+48
=51
b: \(142-\left\lbrack50-\left(2^3\cdot10-2^3\cdot5\right)\right\rbrack\)
\(=142-50+\left(2^3\cdot10-2^3\cdot5\right)\)
\(=92+2^3\left(10-5\right)\)
\(=92+8\cdot5\)
=92+40
=132
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
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
a,5.(x-4) mũ 2-7=13
5.(x-4) mũ 2 =13+7
5.(x-4) mũ 2 =20
(x-4) mũ 2 = 20:5
(x-4)mũ 2= 4
(x-4) mũ 2=2 mũ 2
x-4=2
x=6
Vậy x = 6
b, phần này hình như thiếu gì đó
c,2 mũ x+ 3 - 3.2 mũ x+ 1=32
2 mũ x . 2 mũ 3 - 3 . 2 mũ x . 2 = 32
2 mũ x . 8 -( 3.2).2 mũ x = 32
2 mũ x . 8 -6 . 2 mũ x =32
2 mũ x .( 8- 6) = 32
2 mũ x = 32 : 2
2 mũ x = 16
2 mũ x=2 mũ 4
x = 4
vậy x = 4
k cho mình nha !!!
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. 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
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) \(205-\left[1200-\left(4^2-2.3\right)^3\right]\div40\)
\(=\)\(205-\left[1200-\left(16-6\right)^3\right]\div40\)
\(=\)\(205-\left[1200-10^3\right]\div40\)
\(=\)\(205-\left[1200-1000\right]\div40\)
\(=\)\(205-200\div40\)
\(=\)\(205-5\)
\(=\)\(200\)
a)x=9
b)x=6
c)x=10
d)x=7
a) x^3 = 27
x^3 = 3^3
b) ( 2x-1 )^3 =8
( 2x-1 )^3 = 2^3
2x-1 = 2
2x = 2+1
2x = 3
x = 3 : 2
x = 1,5
c) ( x-2 )^2 = 16
( x-2 )^2 = 4^2
x-2 = 4
x = 4+2
x = 6
d) ( 2x-3 )^2 = 9
( 2x-3 )^2 = 3^2
2x-3 = 3
2x = 3+3
2x = 6
x = 6 : 2
x = 3
a) 3
b)1,5
c)6
d)3
a: \(x^3=27\)
=>\(x^3=3^3\)
=>x=3(nhận)
b: \(\left(2x-1\right)^3=8\)
=>2x-1=2
=>2x=3
=>\(x=\frac32\) (loại)
c: \(\left(x-2\right)^2=16\)
=>\(\left[\begin{array}{l}x-2=4\\ x-2=-4\end{array}\right.\Rightarrow\left[\begin{array}{l}x=6\left(nhận\right)\\ x=-2\left(loại\right)\end{array}\right.\)
d: \(\left(2x-3\right)^2=9\)
=>\(\left[\begin{array}{l}2x-3=3\\ 2x-3=-3\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=6\\ 2x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\left(nhận\right)\\ x=0\left(nhận\right)\end{array}\right.\)