(17x-25):2mũ3+65=9mũ2
17x – 25): 8 + 65 = 9²
`(17x -25) : 8 + 65 =9^2`
`=> (17x -25) : 8 =81 -65`
`=> (17x -25) : 8 =16`
`=>17x -25=16 xx 8`
`=>17x -25=128`
`=>17x=128 + 25`
`=>17x=153`
`=>x=153:17`
`=>x=9`
tìm x
(17x-25):8+65=92
(17x-25):8+65=92
(17x-25):8 =92-65
(17x-25):8 =27
(17x-25) =\(27\times8\)
(17x-25) =216
17x =216+25
17x =241
x =241:7
x =\(\frac{241}{17}\)
T mk nhé bạn ^...^ ^_^
(17x-25):8+65=92
(17x-25):8=92-65
(17x-25):8=27
(17x-25)=27.8
17x-25=216
17x=216+25
17x=241
x=241:17
x=
( 17x - 25 ) : 8 + 65 = 92
( 17x - 25 ) : 8 = 92 - 65
( 17x - 25 ) : 8 = 27
17x - 25 = 27 x 8
17x - 25 = 216
17x = 216 + 25
17x = 241
x = 241 : 17
x = \(\frac{241}{17}\)
2.(3x^2-8)=64:23
(17x-25):8+65=9^5:9^3
b: \(\left(17x-25\right):8+65=9^5:9^3\)
\(\Leftrightarrow\left(17x-25\right):8=9^2-65=81-65=16\)
\(\Leftrightarrow17x-25=128\)
hay x=9
b: (17x−25):8+65=95:93(17x−25):8+65=95:93
⇔(17x−25):8=92−65=81−65=16⇔(17x−25):8=92−65=81−65=16
⇔17x−25=128⇔17x−25=128
hay x=9
(17x-25):8+65=92
giúp mìh nhé
( 17x - 25 ) : 8 + 65 = 9^2
( 17x - 25 ) : 8 + 65 = 81
( 17x - 25 ) : 8 = 81 - 65
( 17x - 25 ) : 8 = 16
( 17x - 25 ) = 16 x 8
( 17x - 25 ) = 128
17x = 128 + 25
17x = 153
x = 153 : 17
x = 9
\(\left(17x-25\right):8+65=9^2\)
\(\Leftrightarrow\left(17x-25\right):8=81-65\)
\(\Leftrightarrow\left(17x-25\right):8=16\)
\(\Leftrightarrow17x-25=16.8\Leftrightarrow17x=128+25\Leftrightarrow17x=153\Leftrightarrow x=153:17=9\)
(k cho mình với nhá!)
(17x-25):8+65=81
(17x-25):8 =81-65
(17x-25):8 =16
17x-25 =16x8
17x-25 =128
17x =128+25
17x =153
x =153:17
x =9
1.tim x
a) ( 17x-25):8+65=81
b) 720 :[41-(2x-5)]=8.5
c) 231-(x-6)=1339:13
\(a.\left(17x-25\right)\div8+65=81\)
\(\left(17x-25\right):8=81-65\)
\(\left(17x-25\right):8=16\)
\(17x-25=16\times8\)
\(17x-25=128\)
\(17x=128+25\)
\(17x=153\)
\(x=153\div17\)
\(x=9.\)
\(b.720\div\left[41-\left(2x-5\right)\right]=8\times5\)
\(720\div\left[41-\left(2x-5\right)\right]=40\)
\(\left[41-\left(2x-5\right)\right]=720\div40\)
\(\left[41-\left(2x-5\right)\right]=18\)
\(2x-5=41-18\)
\(2x-5=23\)
\(2x=23+5\)
\(2x=28\)
\(x=28\div2\)
\(x=14.\)
\(c.231-\left(x-6\right)=1339\div13\)
\(231-\left(x-6\right)=103\)
\(x-6=231-103\)
\(x-6=128\)
\(x=128+6\)
\(x=134.\)
a) => ( 17x-25 ) : 8 = 16 => ( 17x -25 ) = 128 => 17x = 103 = >x = 103/7
a) 4x + 7x = 88
b) (17x - 25) : 2^3 + 65 = 9^2
c)3^7x : 3^5x = 81
ai trả lời đúng thì mình sẽ tick nhé^^
a>4x+7x=88
=>x(4+7)=88
x.11=88
x=88/11
x=8
May cau sau de lam !!!!
Tìm x ∈ N, biết: ( 17x - 25 ) : 8 + 65 = 92 ( giải giùm mình nhé )
\(\left(17x-25\right):8+65=9^2\)
=> \(\frac{17x}{8}-\frac{25}{8}+65=81\)
=> \(\frac{17x}{8}=81-65+\frac{25}{8}\)
=> \(\frac{17x}{8}=\frac{153}{8}\)
=> \(17x=\left(153.8\right):8\)
=> \(17x=153\)
=> \(x=153:17=9\)
(17x - 25 ) : 8 + 65 = 92
(17x - 25 ) : 8 + 65 = 81
(17x - 25) : 8 = 81 - 65 = 16
17x - 25 = 16 x 8 = 128
17x = 128+25=153
x= 153:17 =9
5 nhân (2 mũ 5-10):2mũ3
A.15. B.20. C.25. Đ.10
5.(2⁵ - 10) : 2³
= 5.22 : 8
= 110 : 8
= 13,75
Cả 4 đáp án đều sai
\(5.\left(2^5-10\right):2^3\\ =5.\left(32-10\right):8\\ =5.22:8\\ =110:8\\ =13,75\)
Vậy không có đáp án nào đúng
bài 1 thục hiện phép tính
35.16+35.28-44.15
240-2(3.52 - 20:22)
bài 2 tìm x
(17x - 25):8+65=95: 22
( 8-3x)4-1=15
218-5(x-8)=25:22
(5-3x)4 - 1 = 15
bài 3 cô giáo muốn chia 60 cuốn vở , 72 bút bi , 36 bút chì là nhiều nhất bao nhiêu phần thưởng ? 1 số phần thưởng như nhau . hỏi cô giáo có thể chia dc bao nhiêu phần thưởng
1:
a: \(35\cdot16+35\cdot28-44\cdot15\)
\(=35\left(16+28\right)-44\cdot15\)
\(=44\left(35-15\right)\)
\(=44\cdot20=880\)
b: \(240-2\left(3\cdot5^2-20:2^2\right)\)
\(=240-2\left(3\cdot25-20:4\right)\)
\(=240-150+10=10+90=100\)
2:
b: \(\left(8-3x\right)^4-1=15\)
=>\(\left(3x-8\right)^4=16\)
=>\(\left[{}\begin{matrix}3x-8=2\\3x-8=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=10\\3x=6\end{matrix}\right.\)
=>x=10/3 hoặc x=2
c: \(218-5\left(x-8\right)=2^5:2^2\)
=>\(218-5\left(x-8\right)=2^3=8\)
=>5(x-8)=210
=>x-8=42
=>x=50
d: \(\left(5-3x\right)^4-1=15\)
=>\(\left(3x-5\right)^4-1=15\)
=>\(\left(3x-5\right)^4=16\)
=>\(\left[{}\begin{matrix}3x-5=-4\\3x-5=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\\3x=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=3\end{matrix}\right.\)