2x+1 +4x+3=264
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Bài 3:
\(\Leftrightarrow x^3+64-x^3+25x=264\)
hay x=8
\(1,C=6x^2+23x-55-6x^2-23x-21=-76\\ 2,=\left(2x^4-x^2+2x^3-x-6x^2+6-3\right):\left(2x^2-1\right)\\ =\left[\left(2x^2-1\right)\left(x^2+x-6\right)-3\right]:\left(2x^2-1\right)\\ =x^2+x-6\left(dư.-3\right)\\ 3,\Leftrightarrow x^3+64-x^3+25x=264\\ \Leftrightarrow25x=200\Leftrightarrow x=8\)
\(2^{2x+1}+4^{x+3}=264\)
\(=>2^{2x+1}+2^{2x+6}=264\)
\(=>2^{2x+1}.\left(1+2^5\right)=264\)
\(=>2^{2x+1}.\left(1+32\right)=264\)
\(=>2^{2x+1}.33=264\)
\(=>2^{2x+1}=264:33\)
\(=>2^{2x+1}=8\)
\(=>2^{2x+1}=2^3\)
\(=>2x+1=3\)
\(=>2x=3-1\)
\(=>2x=2\)
\(=>x=2:2\)
\(=>x=1\)
\(2^{2x+1}+4^{x+3}=264\\ \Leftrightarrow2^{2x+1}+2^{2x+6}=264\\ \Leftrightarrow2^{2x+1}\left(1+2^5\right)=264\\ \Leftrightarrow2^{2x+1}\cdot33=264\\ \Leftrightarrow2^{2x+1}=8=2^3\\ \Leftrightarrow2x+1=3\Leftrightarrow x=1\)
a) \(\left|-4x+1\frac{1}{3}\right|=x+2\frac{1}{7}\)
TH1: \(-4x+1\frac{1}{3}=x+2\frac{1}{7}\)
\(-4x-x=2\frac{1}{7}-1\frac{1}{3}\)
\(-5x=\frac{17}{21}\)
=> ...
TH2: \(-4x+1\frac{1}{3}=-x-2\frac{1}{7}\)
...
rùi bn tự lm típ nha!
b) 22x-1+4x+2 = 264
=> 22x: 2 + (22)x+2=264
22x.1/2 + 22x+4=264
22x.1/2 + 22x.24 = 264
22x.(1/2 + 24) = 264
22x. 33/2 = 264
22x = 16
22x = 24
=> 2x = 4
x = 2
a) Ta có: \(9\cdot5^x=6\cdot5^6+3\cdot5^6\)
\(\Leftrightarrow9\cdot5^x=9\cdot5^6\)
\(\Leftrightarrow5^x=5^6\)
hay x=6
b) Ta có: \(2^{2x+1}+4^{x+3}=264\)
\(\Leftrightarrow4^x\cdot2+4^x\cdot64=264\)
\(\Leftrightarrow4^x=4\)
hay x=1
264+12x=1800
12x=1536
x=128
(x+73):19=321
x+73=6099
x=6026
2x+3x+4x=18
9x=18
x=2
10<2x<20
=>5<x<10
x=6,7,8,9
\(264+12x=1800\)
\(\Leftrightarrow12x=1536\)
\(\Leftrightarrow x=128\)
==============
\(\left(x+73\right):19=321\)
\(\Leftrightarrow x+73=6099\)
\(\Leftrightarrow x=6026\)
===============
\(2x+3x+4x=18\)
\(\Leftrightarrow x.\left(2+3+4\right)=18\)
\(\Leftrightarrow x.9=18\)
\(\Leftrightarrow x=2\)
================
\(10< 2x< 20\)
..................................
\(2x=12\Leftrightarrow x=6\)
\(2x=14\Leftrightarrow x=7\)
\(2x=16\Leftrightarrow x=8\)
\(2x=18\Leftrightarrow x=9\)
\(x\in\left\{6;7;8;9\right\}\)
12x = 1800 - 264
12x = 1536
x = 1536 : 12
x = 128
( x + 73 ) : 19 = 321
( x + 73 ) = 321 x 19
x + 73 = 6099
x = 6099 - 73
x = 6026
2x + 3x + 4x = 18
x . ( 2 + 3 + 4 ) = 18
x . 9 = 18
x = 18 : 9
x = 2
10 < 2x < 20
x = 6 ; 7 ; 8 ; 9
2^x+1 + 4^x+3 = 264
c) Ta có:
2^2x+1+4^x+3=264
⇒2^2x+1+(2^2)^x+3=264
⇒2^2x+1+2^2x+6=2264
⇒2^2x+1+2^2x+1+5=264
⇒2^2x+1+2^2x+1⋅25=264
⇒2^2x+1⋅(1+25)=264
⇒2^2x+1⋅26=264
⇒2^2x+1=132/13
⇒2^2x+1∉N
⇒2^x+1∉N
⇒x∉N
⇒⇒ Không có giá trị x thỏa mãn.
Vậy không có giá trị x thỏa mãn.
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