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\(3^x.3^{x+1}.3^{x+2}=36\)
\(\Rightarrow\)\(3^{x+x+1+x+2}=36\)
\(\Rightarrow\)\(3^{3x+3}=36\)
\(\Rightarrow\) \(3^{3x}\cdot3^3=36\)
\(\Rightarrow\) \(3^{3x}.27=36\)
\(\Rightarrow\) \(3^{3x}=\frac{4}{3}\)
\(\Rightarrow\)Không có giá trị x thỏa mãn
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36 x 12 + 36 x 45 + 36 x 33
= 36 x (12 + 45 + 33)
= 36 x 90
= 3240
b) 78 x 31 + 78 x 24 + 78 x 17 + 22 x 72
=78 x (31 + 24 + 17) + 22 x 72
= 78 x 72 + 22 x 72
= (78 + 22) x 72
= 100 x 72
= 7200
c) ( 34 - 17 )2 - 23 x 32
\(=17^2-8.9\)
\(=289-72\)
\(=217\)
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a) Ta có: \(\left[\frac{\left(8x-12\right)}{4}\right]\cdot3^3=3^6\)
\(\Leftrightarrow\frac{8x-12}{4}=3^3\)
\(\Leftrightarrow2x-3=3^3\)
\(\Leftrightarrow2x-3=27\)
\(\Leftrightarrow2x=30\)
hay x=15
Vậy: x=15
b) Ta có: \(3^{2+1}+3^x=36\)
\(\Leftrightarrow3^3+3^x=36\)
\(\Leftrightarrow3^x=9\)
hay x=2
Vậy: x=2
c) Ta có: \(2^{x+1}=16\)
\(\Leftrightarrow2^{x+1}=2^4\)
\(\Leftrightarrow x+1=4\)
hay x=3
Vậy: x=3
1.a)
[(8x - 12) : 4] . 33 = 36
(8x - 12) : 4 = 36 : 33 = 33 = 27
8x - 12 = 27 . 4 = 108
8x = 108 + 12 = 120
x = 120 : 8 = 15
b)
32+1 + 3x = 36
27 + 3x = 36
3x = 36 - 27 = 9 = 32
=> x = 2
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3[2x+1]=5+10
3[2x+1]=15
2x+1=15:3
2x+1=5
2x+1=22+1
=>x=2
5x+1=36-10
5x+1=26
5x+1=52+1
=>x=2
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3x + 3x + 1=36
3x . 1 + 3x . 3 =36
3x . (1+3)=36
3x . 4 = 36
3x = 36 : 4
3x = 9
33 = 9
=> x = 3
vậy x = 3
3x + 3x + 1 = 36
\(\Rightarrow\) 3x + 3x + 1 = 32 + 32 + 1
\(\Rightarrow\) x = 2
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1) (x+2)2=36
(x+2)2=62
=>x+2=6 hoặc x+2=-6
x=6-2 hoặc x=-6-2
x=4 hoặc x=-8
2)\(\left(\frac{3}{4}\right)^x=\frac{2^8}{3^4}\)
\(\left(\frac{3}{4}\right)^x:\frac{2^8}{3^4}=0\)
\(\left(\frac{3}{4}\right)^x.\frac{3^4}{2^8}=0\)
không có giá trị x nào thỏa mãn vì \(\left(\frac{3}{4}\right)^x>0;\frac{3^4}{2^8}>0\)
5(x-2)(x+3)=1
5(x-5)(x+3)=50
=>(x-2)(x+3)=0
=>x-2=0 hoặc x+3=0
x=2 hoặc x=-3
(x-2)8=(x-2)6
vì 8 là mũ chẵn nên (x-2)8=(x-2)6 khi:
x-2=0 hoặc x-2=1 hoặc x-2=-1
x=2 hoặc x=1/2 hoặc x=1
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3x.3x+1=36
=>3x.1.3x.31=36
=>3x.(1+31)=36
=>3x.4=36
=>3x=36:4
=>3x=9
=>3x=32
=>x=2
Vậy x =2
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a.\(36-\left(2x-1\right)^x=3^0.3^1.3^2\)
=> \(36-\left(2x-1\right)^x=1.3.9\)
=>\(\left(2x-1\right)^x=36-27\)
=>\(\left(2x-1\right)^x=9\)
=>\(\left(2x-1\right)^x=9^1=3^2\)
+) Xét trường hợp 1:
\(\left(2x-1\right)^x=9^1\)
=> 2x-1=9 thì x=1
=> 2x=9+1 thì x=1
=> 2x=10 thì x=1
=> x=5 thì x=1
mà \(5\ne1\)=> loại.
+ ) Xét trường hợp 2:
\(\left(2x-1\right)^x=3^2\)
=> 2x-1=3 thì x=2
=> 2x=3+1 thì x=2
=> 2x=4 thì x=2
=> x=2 thì x=2 (2=2) => chọn.
Vậy x=2.
b. \(3^{x+2}+3^x=10\)
=> \(3^x.\left(3^2+1\right)=10\)
=>\(3^x.10=10\)
=>3x=1
=>3x=30
Vậy x=0.