a)5 mũ x =125
b)x mũ 3 =64
c)(x-1) mũ 2=25
d)5 mũ x +5 mũ x+2=130
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Bài 1:
2\(x\) = 4
2\(^x\) = 22
\(x=2\)
Vậy \(x=2\)
Bài 2:
2\(^x\) = 8
2\(^x\) = 23
\(x=3\)
Vậy \(x=3\)
a)<=>
A,=(x+y)(x-y)=x^2-y^2
x=(-1/2)^5:(1/2)^4=-1/2
x^2=1/4
y=8^2/(-2)^5=-2
y^2=4
A=1/4-4=-15/4
a) \(\dfrac{10^{12}+5^{11}.2^9-5^{13}.2^8}{4.5^5.10^6}\)
\(=\dfrac{2^{12}.5^{12}+5^{11}.2^9-5^{13}.2^8}{2^2.5^5.2^6.5^6}\)
\(=\dfrac{2^{12}.5^{12}+5^{11}.2^9-5^{13}.2^8}{2^8.5^{11}}\)
\(=\dfrac{\left(2^8.5^{11}\right)\left(2^4.5+2-5^2\right)}{2^8.5^{11}}\)
\(=2^4.5+2-5^2\)
\(=57\)
b) \(\dfrac{\left[5\left(x-y\right)^4-3\left(x-y\right)^3+4\left(x-y\right)^2\right]}{\left(y-x\right)^2}\)
\(=\dfrac{\left(x-y\right)^2\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(y-x\right)^2}\)
\(=\dfrac{\left(x^2+y^2-2xy\right)\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(y^2+x^2-2xy\right)}\)
\(=5\left(x-y\right)^2-3\left(x-y\right)+4\)
c) \(\dfrac{\left(x+y\right)^5-2\left(x+y\right)^4+3\left(x+y\right)^3}{-5\left(x+y\right)^3}\)
\(=\dfrac{\left(x+y\right)^3\left[5\left(x+y\right)^2-2\left(x+y\right)+3\right]}{-5\left(x+y\right)^3}\)
\(=\dfrac{5\left(x+y\right)^2-2\left(x+y\right)+3}{-5}\)
a) x=50-37
x= 13
b) 2.x=11+3
2.x=14
x=14:2=7
c) 2+x=6.5
2.x=30
x=30:2=15
e) \(^{^{^{ }}x^9}\)=25
\(\Rightarrow\)x=5
a) 5x = 125
5x = 53
=> x = 3
b) x3 = 64
x3 = 43
=> x = 4
c) ( x - 1 ) 2 = 25
( x - 1 ) 2 = 52
=> x - 1 = 5
=> x = 5 + 1
=> x = 6
d) 5x + 5x+2 = 130
5x . 1 + 5x . 52 = 130
5x . ( 1 + 52 ) = 130
5x . 26 = 130
5x = 130 : 26
5x = 5
=> x = 1
\(a.\)\(5^x=125\)
\(5^x=5^3\Rightarrow x=3\)
\(b.\)\(x^3=64\)
\(x^3=4^3\Rightarrow x=4\)
\(c.\)\(\left(x-1\right)^2=25\)
\(x^2=5^2\)mà \(\left(x-1\right)\)nên \(x=5+1=6\)
P/s: Ý d thiếu nhé bạn