Respuesta :
Answer:
1. 18 (sqrt21 - sqrt2a)
2. 3
3. x^m/n
4. (5x^4√10)/(√2x)
5. x = -2 or -6
6. x = 30
Step-by-step explanation:
Number 1
3^3sqrt21 - 6^3sqrt2a
3 * 6 * sqrt21 - sqrt2a
18 (sqrt21 - sqrt2a)
Number 2
3^1/2 * 3^1/2 =
3^1/2+1/2 =
3^1 =
3
Number 3
^nsqrtx^m =
x^m/n
Number 4
(√250x^16)/(√2x) =
(√25 * 10 * x^16)/√(2x )=
(5x^4√10)/(√2x)
Number 5
√2x + 13 - 5 = x
√2x + 13 = x + 5
square both side to take away the sqrt sign
(√2x + 13)^2 = (x + 5)^2
expand the equation on the RHS
2x + 13 = x(x+5) + 5(x+5)
2x+13 = x^2 + 10x +25
substract 13 from both sides
2x = x^s + 10x +12
subtract 2x from both sides
0 = x^2 +8x + 12
Factorize equation
x^2 + 6x +2x + 12 = 0
x(x+6) + 2(x+6) = 0
(x+2)(x+6) = 0
x = -2 or -6
Number 6
3 ^5sqrt(x+2)^3 + 3 = 27
subtract 3 from both sides
3 ^5sqrt(x+2)^3 = 27 - 3
3 ^5sqrt(x+2)^3 = 24
divide through by 3
^5sqrt(x+2)^3 = 8
square both sides by 5 to take away the 5th root sign
(x+2)^3 = (8)^5
(x+2)^3 = 32,768
take the cube root of both sides to take away the ^3
x+2 = ^3sqrt 32,768
x+2 = 32
x = 32 - 2
x = 30
Answer:
1.
The expression is [tex]3^{3}\sqrt{21} -6^{3}\sqrt{2a}[/tex]
We need to solve each power: [tex]27\sqrt{21}-216\sqrt{2a}[/tex].
The Greatest common factor between 27 and 216 is 27, so we extract that
[tex]27(\sqrt{21}- 8\sqrt{2a})[/tex], which is the simplest form.
2.
The expression is [tex]3^{\frac{1}{2} } \times 3^{\frac{1}{2} }[/tex]
Notice that bases are equal, that means we need to sum exponents only to find the simplest form
[tex]3^{\frac{1}{2} +\frac{1}{2} }=3^{1}=3[/tex]
3.
The expression is [tex]\sqrt[n]{x^{m} }[/tex]
Here we transform the root into a fractional exponent.
[tex]\sqrt[n]{x^{m} }=x^{\frac{m}{n} }[/tex]
4.
The expression is
[tex]\frac{\sqrt{250x^{16} } }{\sqrt{2x} }[/tex]
Here we need to express it as the root of a fraction
[tex]\sqrt{\frac{250x^{16} }{2x} }[/tex]
Then, we divide
[tex]\sqrt{\frac{250x^{16} }{2x} }=\sqrt{125x^{10-1} } =\sqrt{125x^{9} }[/tex]
5.
The equation is [tex]\sqrt{2x+13}-5=x[/tex]
First, we move the term 5 to other side, then we elevate the equality to the square power to eliminate the square root. Consequently, we have to solve the square power of the binomial x+5:
[tex](\sqrt{2x+13} )^{2} =(x+5)^{2} \\2x+13=x^{2} +10x+25[/tex]
Then, we move all terms to one side
[tex]x^{2} +10x+25-2x-13=0\\x^{2} +8x+12=0[/tex]
Now, we have to find to numbers which product is 12 and which sum is 8, those numbers are 6 and 2:
[tex]\x^{2} +8x+12=(x+6)(x+2)[/tex]
The solutions are -6 and -2.
6.
The expression is
[tex]3\sqrt[5]{(x+2)^{3} } +3=27[/tex]
First, we subtract the equation by 3, then we divide by 3:
[tex]3\sqrt[5]{(x+2)^{3} } 3-3 =27-3\\3\sqrt[5]{(x+2)^{3} } =24\\\frac{3\sqrt[5]{(x+2)^{3} } }{3}=\frac{24}{3}\\ \sqrt[5]{(x+2)^{3} } =8[/tex]
Then, we elevate each side to the fifth power to eliminate the root
[tex](\sqrt[5]{(x+2)^{3} } )^{5} =8^{5} \\(x+2)^{3} =32768[/tex]
Now, we apply a cubic root to each side
[tex]\sqrt[3]{(x+2)^{3}} =\sqrt[3]{32768} \\x+2=32\\x+2-2=32-2\\ \therefore x=30[/tex]
