15.(x-2)3=64
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a) \(2^x=16\)
\(2^x=2^4\)
\(\Rightarrow x=4\)
vay \(x=4\)
b) \(4^x=64\)
\(4^x=4^3\)
\(\Rightarrow x=3\)
vay \(x=3\)
c) \(15^x=225\)
\(15^x=15^2\)
\(\Rightarrow x=2\)
vay \(x=2\)
d) \(2^x-15=17\)
\(2^x=32\)
\(2^x=2^5\)
\(\Rightarrow x=5\)
vay \(x=5\)
e) \(2^x-64=2^6\)
\(2^x=2^6+64\)
\(2^x=128\)
\(2^x=2^7\)
\(\Rightarrow x=7\)
vay \(x=7\)
f) \(\left(7x-11\right)^3=2^5.5^2+200\)
\(\left(7x-11\right)^3=800+200\)
\(\left(7x-11\right)^3=1000\)
\(\left(7x-11\right)^3=10^3\)
\(\Rightarrow7x-11=10\)
\(\Rightarrow7x=21\)
\(\Rightarrow x=3\)
vay \(x=3\)
`a,`\(2^x -15= 2^4+1\)
`-> 2^x-15=17`
`-> 2^x=17+15`
`-> 2^x=32`
`-> 2^x=2^5`
`-> x=5`
`b,` Có phải đề là \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\) ?
`=>`\(\dfrac{x+1}{65}+1+\dfrac{x+2}{64}+1=\dfrac{x+3}{63}+1+\dfrac{x+4}{62}+1\)
`=>`\(\dfrac{x+1+65}{65}+\dfrac{x+2+64}{64}-\dfrac{x+3+63}{63}-\dfrac{x+4+62}{62}=0\)
`=>`\(\dfrac{x+66}{65}+\dfrac{x+66}{64}-\dfrac{x+66}{63}-\dfrac{x+66}{62}=0\)
`=>`\(\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{64}-\dfrac{1}{63}-\dfrac{1}{62}\right)=0\)
Mà `1/65+1/64-1/63-1/62 \ne 0`
`-> x+66=0`
`-> x=-66`
a: =>2^x=2^4+16=32
=>x=5
b: Sửa đề: \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\)
=>\(\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+2}{64}+1\right)=\left(\dfrac{x+3}{63}+1\right)+\left(\dfrac{x+4}{62}+1\right)\)
=>x+66=0
=>x=-66
( 7.x - 15 ) : 3 = 2
7.x - 15 = 6
7.x = 6 + 15
7.x = 21
x = 3
không biết vì chưa học
88 - 3 ( 7 + x ) = 64
3 ( 7 + x ) = 88 - 64
3 ( 7 + x ) = 24
7 + x = 8
x = 1
(7.x -15):3=2
7.x-15=2.3
7.x-15=6
7.x=6+15
7.x=21
x=21:7
x=3
12.(x+37)=504
x+37=504:12
x+37=42
x=42-37
x=5
88-3(7+x)=64
3(7+x)=88-64
3(7+x)=24
7+x=24:3
7+x=8
x=8-7
x=1
dấu . là nhân nhé
Bài 3:
a: Ta có: \(23\left(42-x\right)=23\)
\(\Leftrightarrow42-x=1\)
hay x=41
b: Ta có: 15(x-3)=30
nên x-3=2
hay x=5
Bài 1:
a: 32+89+68=100+89=189
b: 64+112+236=300+112=412
c: \(1350+360+650+40=2000+400=2400\)
a) Ta có: \(\left(x-1\right)^{x+2}-\left(x-1\right)^{x+4}=0\)
\(\Leftrightarrow\left(x-1\right)^x\cdot\left(x-1\right)^2-\left(x-1\right)^x\cdot\left(x-1\right)^4=0\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\cdot\left[1-\left(x-1\right)^2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
b) Ta có: \(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot\dfrac{5}{15}\cdot...\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}=2x\)
\(\Leftrightarrow2x=\dfrac{1}{64}\)
hay \(x=\dfrac{1}{128}\)
2x = 64
⇒ 2x = 26
⇒ x = 6.
7x = 710 : 75
⇒ 7x = 75
⇒ x = 5.
3x = 315 : 243
⇒ 3x = 315 : 35
⇒ 3x = 310
⇒ x = 10.
\(2^x=64=2^6\Rightarrow x=6\\ ---\\ 7^x=7^{10}:7^5=7^5\Rightarrow x=5\\ ----\\ 3^x=3^{15}:243=3^{15}:3^5=3^{10}\Rightarrow x=10\)