Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\text{a) Ta co }\) \(4^{x+3}-3.4^{x+1}=13.4^{11}\)
\(\Rightarrow\) \(4^{x+1}\left(16-3\right)=13.4^{11}\)
\(\Rightarrow4^{x+1}.13=13.4^{11}\)
\(\Rightarrow4^{x+1}=4^{11}\)
\(\Rightarrow x+1=11\)
\(\Rightarrow\text{x=10}\)
a)
\(4^{x+3}-3.4^{x+1}=13.4^{11}\)
<=> \(4^{x+1}\left(16-3\right)=13.4^{11}\)
<=> \(4^{x+1}.13=13.4^{11}\)
<=> \(4^{x+1}=4^{11}\)
<=> \(x+1=11\)
<=> x=10
2.3x + 3x - 1 = 7 . (32 + 2 . 62)
=> 2.3x + 3x - 1 = 567
=> 7 . 3x - 1 = 567
=> 3x - 1 = 567 : 7 = 81
=> x - 1 = 4
=> x = 5
a)2*3x+3x-1=7(32+2*62)
2*3x+3x-1=7(9+72)=7*81
2*3x+3x/3=567
2*3x+3x*1/3=567
(2+1/3)*3x=567
7/3*3x=567
3x=567:7/3
3x=243=35
=>x=5
b) mk ko hiểu đề mấy, cái chỗ 7x+2 là nhân vs 2 ak
a, 3 : ( 1 - 3/2x ) = 4 : ( 2 - x )
<=> \(\frac{3}{1-\frac{3}{2}x}=\frac{4}{2-x}\)
<=> 3 ( 2 - x ) = 4 ( 1 - 3/2x )
<=> 6 - 3x = 4 - 6x
<=> -3x + 6x = 4 - 6
<=> 3x = -2
<=> x = -2/3
b, 2.3x + 3x-1 = 7( 32 + 2.62 )
b, 2.3x + 3x-1 = 7( 32 + 2.62 )
<=> 2.3x + 3x-1 = 7.81
<=> 3x-1(2.3 + 1) = 7.81
<=> 3x-1.7 = 7.81
<=> 3x-1=81
<=> 3x-1 = 34
=> x - 1 = 4 => x = 5
a)Viết dưới dạng phân số rồi sử dụng tích chéo ý
b)\(\frac{-1}{7}.2^3-2x:1\frac{4}{3}=-2^{x-1}\)
\(\Rightarrow\frac{-8}{7}-2x:\frac{7}{3}=-2^{x-1}\)
\(\Rightarrow\frac{-8}{7}-\frac{6x}{7}=-2^{x-1}\)
\(\Rightarrow\frac{-8-6x}{7}=\frac{2^{x-1}}{-1}\)
\(\Rightarrow-1\left(-8-6x\right)=7.2^{x-1}\)
\(\Rightarrow6x+8=7.2^{x-1}\)
.........
a) 2x.(1 + 23) = 144
2x . 9 = 144
2x = 16
=> x = 4
b) (2x - 1)10 = (2x - 1)100
(2x - 1)100 - (2x - 1)10 = 0
(2x - 1)10.[ (2x - 1)90 - 1] = 0
=> (2x - 1)10 = 0 hoặc (2x - 1)90 - 1 = 0
=> 2x = 1 hoặc (2x - 1)90 = 1
=> x = \(\frac{1}{2}\) hoặc \(2x-1=\orbr{\begin{cases}1\\-1\end{cases}}\)
=> \(2x=\orbr{\begin{cases}2\\0\end{cases}}\)
=> x = {\(\frac{1}{2};1;0\)}
=> \(11.6^{x-1}+2.6^{x-1+2}=11.6^{11}+2.6^{11+2}\)
=>\(11.6^{x-1}+2.6^{x-1}.6^2=11.6^{11}+2.6^{11}.6^2\)
=>\(6^{x-1}\left(11+2.6^2\right)=6^{11}\left(11+2.6^2\right)\)
=>\(6^{x-1}=6^{11}\)
=>\(x-1=11\)
=>\(x=12\)
Tíck giùm mik nha !