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\(5^{x-1}=125\)
\(5^{x-1}=5^3\)
\(\Rightarrow x-1=3\)nên x = 4
4(x - 3) = 72 - 110
=> 4x - 12 = 49 - 1
=> 4x = 48 - 12
=> 4x = 24
=> x = 6
5x + x = 39 - 311 : 39
=> 6x = 39 - 32
=> 6x = 39 - 9
=> 6x = 30
=> x = 5
\(4\left(x-3\right)=7^2-1^{10}\)
\(4x-4\cdot3=49-1\)
\(4x\cdot12=48\)
\(4x=48:12\)
\(4x=4\)
\(x=4:4\)
\(x=1\)
\(5x+x=39-3^{11}:3^9\)
\(6x=39-3^{11-9}\)
\(6x=39-3^2\)
\(6x=39-9\)
\(6x=30\)
\(x=30:6\)
\(x=5\)
c/ 2x - 1 = \(5^{98}:5^{96}\)
2x - 1 = \(5^2\) = 25
2x = 25 + 1 = 26
x = 26 : 2
x = 13
d/ 7x + 3 = \(3^5.2^3.9\)
7x + 3 = \(3^5.3^2.8=3^7.8=2187.8\)
7x + 3 = \(17496\)
7x = 17496 - 3 = 17493
x = 17493 : 7
x = 2499
e/\(2^{2x+6}=1\)
\(2^{2x+6}=2^0\)
2x + 6 = 0
2x = 0 - 6 = - 6
x = - 6 : 2
x = - 3
j/ \(2^x=8\)
\(2^x=2^3\)
x = 3
g/ \(2^x:2^3=16\)
\(2^{x-3}=2^4\)
x - 3 = 4
x = 4 + 3
x = 7
h/ \(2^x+2^{x+1}+2^{x+2}=56\)
\(2^x\left(1+2+2^2\right)\) = 56
\(2^x.7=56\)
\(2^x=56:7\)
\(2^x=8\)
\(2^x=2^3\)
x = 3
Bài a, b thiên phong giải r, mk chỉ làm những bài còn lại thôi. Chúc bạn học tốt!!!
ne ban minh biet cau tra loi nhung lam the nao bam ngoac vuong
a) (x : 3 - 4) . 5 = 15
x : 3 - 4 = 15 : 5
x : 3 - 4 = 3
x : 3 = 3 + 4
x : 3 = 7
x = 7 . 3
x = 21.
b) 128 - 3 . (x + 4) = 23
3 . (x + 4) = 128 - 23
3 . (x + 4) = 105
x + 4 = 105 : 3
x + 4 = 35
x = 35 - 4
x = 31.
c) (3x - 24) . 73 = 2 . 74
3x - 24 = 2 . 74 : 73
3x - 24 = 2 . 7
3x - 16 = 14
3x = 14 + 16
3x = 30
x = 30 : 3
x = 10.
d) 5x + x = 39 - 311 : 39
5x + x = 39 - 32
5x + x = 39 - 9
6x = 30
x = 30 : 6
x = 5.
a) \(\left(\dfrac{x}{3}-4\right).5=15\)
\(\Leftrightarrow\dfrac{x}{3}-4=\dfrac{15}{5}\)
\(\Leftrightarrow\dfrac{x}{3}-4=3\)
\(\Leftrightarrow\dfrac{x}{3}=3+4\)
\(\Leftrightarrow\dfrac{x}{3}=7\)
\(\Leftrightarrow x=7.3\)
\(\Leftrightarrow x=21\)
Vậy \(x=21\)
b) \(128-3\left(x+4\right)=23\)
\(\Leftrightarrow3\left(x+4\right)=128-23\)
\(\Leftrightarrow3\left(x+4\right)=105\)
\(\Leftrightarrow x+4=\dfrac{105}{3}\)
\(\Leftrightarrow x+4=35\)
\(\Leftrightarrow x=35-4\)
\(\Leftrightarrow x=31\)
Vậy \(x=31\)
c) \(\left(3x-2^4\right).7^3=2.7^4\)
\(\Leftrightarrow3x-2^4=\dfrac{2.7^4}{7^3}\)
\(\Leftrightarrow3x-2^4=2.7\)
\(\Leftrightarrow3x-2^4=14\)
\(\Leftrightarrow3x-16=14\)
\(\Leftrightarrow3x=14+16\)
\(\Leftrightarrow3x=30\)
\(\Leftrightarrow x=\dfrac{30}{3}\)
\(\Leftrightarrow x=10\)
Vậy \(x=10\)
d) \(5x+x=39-3^{11}:3^9\)
\(\Leftrightarrow6x=39-3^{11-9}\)
\(\Leftrightarrow6x=39-3^2\)
\(\Leftrightarrow6x=39-9\)
\(\Leftrightarrow6x=30\)
\(\Leftrightarrow x=\dfrac{30}{6}\)
\(\Leftrightarrow x=5\)
Vậy \(x=5\)
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.
3ˣ⁻¹ + 3ˣ + 3ˣ⁺¹ = 39
3ˣ⁻¹.(1 + 3 + 3²) = 39
3ˣ⁻¹.13 = 39
3ˣ⁻¹ = 39 : 13
3ˣ⁻¹ = 3
x - 1 = 1
x = 1 + 1
x = 2
\(3^{x-1}+3^x+3^{x+1}=39\)
\(=>3^x:3+3^x+3^x.3=39\)
\(=>3^x.\dfrac{1}{3}+3^x+3^x.3=39\)
\(=>3^x.\left(\dfrac{1}{3}+1+3\right)=39\)
\(=>3^x.\dfrac{13}{3}=39\)
\(=>3^x=39:\dfrac{13}{3}=39.\dfrac{3}{13}\)
\(=>3^x=9=3^2\)
\(=>x=2\)