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suy ra 1000=25[30+(2^x-6)]
suy ra 4=30+2^x-6
suy ra 2^x=-20
suy ra ko có x thỏa mãn giả thiết
1. [|x| + 3 ] . 15 - 5 = 70
2. 86 : { 2. [ 2x - 1 ]2 - 7 } + 42 = 2. 32
3. 20 - { 42 + [ x - 6 }] = 90
4. 24- |x + 8| = 3. [25 - 52 ]
5. 1000 : [ 30 + { 2x - 6 }] = 32 + 42
6. [ x + 1 ] chia hết [ x + 2 ] và x thuộc N
Baì làm
1) . [|x| + 3 ] . 15 - 5 = 70
.[|x| + 3 ] . 15 = 70 + 5
[|x| + 3 ] . 15 = 75
.[|x| + 3 ] = 75 : 15
.[|x| + 3 ] = 5
|x| = 5 - 3
|x| = 2
=> \(\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
2). 86 : { 2. [ 2x - 1 ]2 - 7 } + 42 = 2. 32
86 : { 2. [ 2x - 1 ]2 - 7 } + 42 = 2. 9
86 : { 2. [ 2x - 1 ]2 - 7 } + 42 = 18
86 : { 2. [ 2x - 1 ]2 - 7 } = 18 - 16
86 : { 2. [ 2x - 1 ]2 - 7 } = 2
{ 2. [ 2x - 1 ]2 - 7 } = 86 : 2
{ 2. [ 2x - 1 ]2 - 7 } = 43
2. [ 2x - 1 ]2 = 43 + 7
2. [ 2x - 1 ]2 = 50
. [ 2x - 1 ]2 = 50 : 2
[ 2x - 1 ]2 = 25
. [ 2x - 1 ]2 = 52
2x -1 = 5
2x = 5+1
2x = 6
x = 6: 2
x = 3
3) 20 - { 42 + [ x - 6 }] = 90
20 - { 42 + [ x - 6 }] = 90: 3
20 - { 42 + [ x - 6 }] = 30
{ 42 + [ x - 6 }] = 30 + 20
{ 42 + [ x - 6 }] = 50
{ 16 + [ x - 6 }] = 50
[ x - 6 ] = 50 -16
[ x - 6 ] = 34
x = 34 + 6
x =40
4) 24- |x + 8| = 3. [25 - 52 ]
4. 24- |x + 8| = 3.(32-25)
4. 24- |x + 8| = 3. 7
4. 24- |x + 8| = 21
4. 24- |x + 8| = 21
96-|x + 8| = 21
|x + 8| = 96 - 21
|x + 8| = 75
=> \(\orbr{\begin{cases}x+8=75\\x+8=-75\end{cases}}=>\orbr{\begin{cases}x=75-8\\x=-75-8\end{cases}}=>\orbr{\begin{cases}x=67\\x=-83\end{cases}}\)
5) 1000 : [ 30 + { 2x - 6 }] = 32 + 42
1000 : [ 30 + { 2x - 6 }] = 9 + 16
1000 : [ 30 + { 2x - 6 }] = 25
[ 30 + { 2x - 6 }] = 1000 : 25
[ 30 + { 2x - 6 }] = 40
{ 2x - 6 } = 40 + 30
{ 2x - 6 } = 70
2x = 70 - 6
2x = 64
2x = 26
x = 6
6. [ x + 1 ] chia hết [ x + 2 ] và x thuộc N
Ta có : x + 1 = x + 2 - 1
Vì x + 1 \(⋮\)x + 2
=> x + 2 - 1\(⋮\) x + 2
vì x + 2 -1\(⋮\) x + 2 => 1 \(⋮\)x+2
=> x + 2 \(\in\)Ư(1)
=>\(x+2=1=>x=1-2=>x=-1\)
=> mà x \(\in\)N lại có x = -1
=> x \(\in\) \(\varnothing\)
Bạn ơi đề sai rồi
1000 ko chia hết cho 12
a) 6x / 2 - 84 / 2 - 72 =201
3x - 42 -72 = 201
3x - 114 =201
3x = 315
x = 105
b)3x -3^4 =6^5 / 6^3
3x - 3^4 = 36
3x - 3 * 3^3 = 36
3*(x-27) = 36
x - 27 = 36 / 3
x - 27 = 12
x = 39
\(a,\left(7x-11\right)^3=2^5.5^2+200.\)
\(\left(7x+11\right)^3=32.25+200.\)
\(\left(7x+11\right)^3=800+200.\)
\(\left(7x-11\right)^3=1000.\)
\(\left(7x-11\right)^3=10^3.\)
\(\Rightarrow7x-11=10.\)
\(\Rightarrow x=\left(10+11\right):3=7\in Z.\)
Vậy.....
\(b,3^x+25=26.2^2+2.3^0.\)
\(3^x+25=26.4+2.\)
\(3^x+25=104+2.\)
\(3^x+25=106.\)
\(3^x=106-25.\)
\(3^x=81.\)
\(3^x=3^4\Rightarrow x=4\in Z.\)
Vậy.....
\(c,2^x+3.2=64.\)(có vấn đề).
\(d,5^{x+1}+5^x=750.\)
\(5^x.5^1+5^x+1=750.\)
\(5^x\left(5^1+1\right)=750.\)
\(5^x\left(5+1\right)=750.\)
\(5^x.6=750.\)
\(5^x=750:6.\)
\(5^x=125.\)
\(5^x=5^3\Rightarrow x=3\in Z.\)
Vậy.....
\(e,x^{15}=x.\)
\(\Rightarrow x\left(x^{14}-1\right)=0\Rightarrow\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right..\)
\(f,\left(x-5\right)^4=\left(x-5\right)^6.\)
\(\Leftrightarrow\left(x-5\right)^4-\left(x-5^6\right)=0.\)
\(\Leftrightarrow\left(x-5\right)^4\left[1-\left(x-5\right)^2\right]=0.\)
\(\Leftrightarrow\left(x-5\right)^4\left(1-x+5\right)\left(1+x-5\right)=0.\)
\(\Leftrightarrow\left(x-5\right)^4\left(6-x\right)\left(x-4\right)=0.\)
\(\Leftrightarrow\left(x-5\right)^4=0\Rightarrow x-5=0\Rightarrow x=5\in Z.\)
\(6-x=0\Rightarrow x=6\in Z.\)
\(x-4=0\Rightarrow x=4\in Z.\)
Vậy.....
a) 100 - 7.(x-5) = 68
7.(x - 5) = 100 - 68 = 32
x - 5 = \(\frac{32}{7}\)
x = \(\frac{32}{7}+5=\frac{67}{7}\)
b) 2(x-6)+2227:17=151
2( x -6) + 131 = 151
2 (x - 6) = 20
x - 6 = 20 : 2 = 10
x = 10 + 6 = 16
Bài 2:
a: =>20-16-x+6=90
=>10-x=90
=>x=-80
b: =>|x+8|=24-3*(32-25)=24-3*7=3
=>x+8=3 hoặc x+8=-3
=>x=-11 hoặc x=-5
c: \(\Leftrightarrow30+2^x-6=1000:25=40\)
=>2^x=16
=>x=4
c: =>x+2+9 chia hết cho x+2
mà x>=0
nên \(x+2\in\left\{3;9\right\}\)
hay \(x\in\left\{1;7\right\}\)
a) 57 . 53 : 510 = 510 : 510 = 1
b) 24 : 23 . 27 = 2 . 27 = 28
c) 93 : 33 . 272 = ( 32 ) 3 : 33 . ( 33 ) 2 = 36 : 33 . 36 = 33 . 36 = 39
d) 25 . 13 - 25 . 5 = 25 . ( 13 - 5 ) = 25 . 8 = 25 . 23 = 28
e) \(\frac{4^6.8^3}{4^8}=\frac{\left(2^2\right)^6.\left(2^3\right)^3}{\left(2^2\right)^8}=\frac{2^{12}.2^9}{2^{16}}=\frac{2^{21}}{2^{16}}=2^5\)
\(\left(19x+2.5^2\right):14=\left(13-8\right)^2-4^2\)
<=> \(\left(19x+50\right):14=25-16\)
<=> \(\left(19x+50\right):14=9\)
<=> \(19x+50=126\)
<=> \(19x=76\)
<=> \(x=4\)
câu B làm j có \(x\)để tìm
\(390-\left(x-7\right)=169:13\)
<=> \(390-x+7=13\)
<=> \(390-x=6\)
<=> \(x=384\)
\(x-6:2-\left(48:24\right):2:6-3=0\)
<=> \(x-3-2:2:6=3\)
<=> \(x-3-\frac{1}{6}=3\)
<=> \(x=\frac{37}{6}\)
\(x+5.2-\left(32+16.3:6-15\right)=0\)
<=> \(x+10-25=0\)
<=> \(x=15\)
1000 : [30 + (2x - 6)] = 32 + 42
= 1000 : [ 30 + (2x-6)] = 9 + 16 = 25
= 30 + (2x - 6) = 1000 : 25 = 40
= 2x - 6 = 40 - 30 = 10
2x = 10 + 6 = 16
⇒ 2x = 24
⇒ x = 4
\(\Leftrightarrow2^x+24=40\)
hay x=4