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\(318-5\left(x-64\right)=103\)
\(\Rightarrow5\left(x-64\right)=318-103\)
\(\Rightarrow5\left(x-64\right)=215\)
\(\Rightarrow x-64=43\)
\(\Rightarrow x=43+64\)
\(\Rightarrow x=107\)
_____________
\(4^x\cdot5+216=296\)
\(\Rightarrow4^x\cdot5=296-216\)
\(\Rightarrow4^x\cdot5-80\)
\(\Rightarrow4^x=16\)
\(\Rightarrow4^x=4^2\)
\(\Rightarrow x=2\)
___________
\(376-6^x:3=364\)
\(\Rightarrow6^x:3=376-364\)
\(\Rightarrow6^x:3=12\)
\(\Rightarrow6^x=36\)
\(\Rightarrow6^x=6^2\)
\(\Rightarrow x=2\)
___________
\(\left(4x-1\right)^2=121\)
\(\Rightarrow\left(4x-1\right)^2=11^2\)
\(\Rightarrow\left[{}\begin{matrix}4x-1=11\\4x-1=-11\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4x=12\\4x=-10\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
a.2x - 138 =72
\(\Leftrightarrow\)2x = 210 \(\Leftrightarrow\)x= 105
b.32 - 3x =66 : 65
\(\Leftrightarrow\)9 - 3x =6 \(\Leftrightarrow\)3x =3 \(\Leftrightarrow\)x=1
c.( 64 -4x).3 =24
\(\Leftrightarrow\)64-4x =8 \(\Leftrightarrow\)4x = 56 \(\Leftrightarrow\)x=14
d.3x.6 - 39 =123
\(\Leftrightarrow\) 3x . 6 = 162 \(\Leftrightarrow\)3x =27 \(\Leftrightarrow\)x =3
\(2^{x+3}.4^2=64\Leftrightarrow2^{x+3}.2^4=64\Leftrightarrow2^{x+7}=2^6\Leftrightarrow x+7=6\Leftrightarrow x=-1\)
a,\(\left(x-15\right):50+22=24\)
\(< =>\frac{\left(x-15\right)}{50}=2< =>x-15=100\)
\(< =>x=100+15=115\)
b,\(42-\left(2x+32\right)+12:2=6\)
\(< =>42-2x-32=0\)
\(< =>10-2x=0< =>x=\frac{10}{2}=5\)
Làm nốt :
c) \(134-2\left\{156-6\cdot\left[54-2\cdot\left(9+6\right)\right]\right\}\cdot x=86\)
=> 134 - 2{156 - 6 . [54 - 2 . 15]} . x = 86
=> 134 - 2{156 - 6 . [54 - 30]} . x = 86
=> 134 - 2{156 - 6. 24} . x = 86
=> 134 - 2{156 - 144} . x = 86
=> 134 - 2.12 . x = 86
=> 134 - 24 . x = 86
=> 24.x = 48
=> x = 2
Bài 2 : a) 120 : [21 - (4x - 4)] = 23.3
=> 120 : [21 - (4x - 4)] = 8.3
=> 120 : [21 - (4x - 4)] = 24
=> 21 - (4x - 4) = 5
=> 4x - 4 = 16
=> 4x = 20
=> x = 5
b) 3.[205 - (x - 9)] - 486 = 0
=> 3.[205 - (x - 9)] = 486
=> 205 - (x - 9) = 162
=> x - 9 = 205 - 162 = 43
=> x = 43 + 9 = 52
c) 204 - 2{200 - 5.[64 - 2.(11 + 6)]} . x = 4
=> 204 - 2{200 - 5.[64 - 2.17]} . x = 4
=> 204 - 2{200 - 5 .[64 - 34]}.x = 4
=> 204 - 2{200 - 5.30} . x = 4
=> 204 - 2{200 - 150}.x = 4
=> 204 - 2.50 . x = 4
=> 2.50.x = 200
=> 100.x = 200
=> x = 2
(428-4x) .27=1728
<=> 428-4x = 1728 : 27
<=> 428-4x = 64
<=> 4x = 428 - 64
<=> 4x = 364
<=> x= 364 :4
<=> x= 91
Vậy x= 91
x.x2.x3=64
<=> x1+2+3 = 64
<=> x6 = 64
<=> x6= (+ - 2 )6
=> x = cộng trừ 2
Vậy x= cộng trừ 2
a, \(\left(1+3x\right)^3=64\Leftrightarrow\left(1+3x\right)^3=4^3\)
\(\Leftrightarrow1+3x=4\Leftrightarrow3x=3\Rightarrow x=1\in Z\)
Vậy x = 1
b, \(\left(11-4x\right)^3=343\Leftrightarrow\left(11-4x\right)^3=7^3\)
\(\Leftrightarrow11-4x=7\Rightarrow4x=11-7\Rightarrow4x=4\Rightarrow x=1\in Z\)
Vậy x = 1
1.
( 1 + 3x )3 = 64
( 1 + 3x )3 = 43
( 1 + 3x ) = 4
3x = 4 - 1
3x = 3
x = 3 : 3
x = 1
2.
( 11 - 4x )3 = 343
( 11 - 4x )3 = 73
( 11 - 4x ) = 7
4x = 11 - 7
4x = 4
x = 4 : 4
x = 1
`64.4^x=168`
`<=>4^x=168/64=21/8`
Vì `x\inNN` nên không tồn tại `x`
Kết luận:.....
\(6\cdot6\cdot6\cdot6\cdot3\cdot2\)
\(=6\cdot6\cdot6\cdot6\cdot\left(3\cdot2\right)\)
\(=6\cdot6\cdot6\cdot6\cdot6\)
\(=6^5\)
⇒ Chọn D
`(4x-2)^6 = 64`
`(4x-2)^6 = ` \(\left(\pm2\right)^6\)
`@TH1:`
`4x-2=2`
`4x=2+2`
`4x=4`
`x=4:4`
`x=1`
`@TH2:1`
`4x-2=-2`
`4x=-2+2`
`4x=0`
`x=0:4`
`x=0`
Vậy `x={0;1}`
( 4x - 2 )6 = 64
( 4x - 2 )6 = 26
4x - 2 = 2
4x = 2 + 2
4x = 4
x = 1