Math  /  Algebra

Question16 Simplify the following expressions. 1912×(13)0.51^{\circ} 9^{-\frac{1}{2}} \times\left(\frac{1}{3}\right)^{-0.5} 67n+14n7n14n(nN)6^{\circ} \frac{7^{n}+14^{n}}{7^{n}-14^{n}}(n \in \mathbb{N}) 28×103×4×1040.00162^{\circ} \frac{8 \times 10^{3} \times 4 \times 10^{-4}}{0.0016} 7A=52n+1n53n+1n+1÷5n+3n+15n1n7^{\circ} A=\frac{\sqrt[n]{5^{2 n+1}}}{\sqrt[n+1]{5^{3 n+1}}} \div \frac{\sqrt[n+1]{5^{n+3}}}{\sqrt[n]{5^{n-1}}} 33n+46×3n+17×3n+1(nN)3^{\circ} \frac{3^{n+4}-6 \times 3^{n+1}}{7 \times 3^{n+1}}(n \in \mathbb{N}) 8B=2n+3+5×2n+13×2n+1\mathbf{8}^{\circ} B=\frac{2^{n+3}+5 \times 2^{n+1}}{3 \times 2^{n+1}} 44n+24n2n+12n(nN)4^{\circ} \frac{4^{n+2}-4^{n}}{2^{n+1}-2^{n}}(n \in \mathbb{N}) 9C=4x2+8x2+82x249^{\circ} C=\frac{\sqrt{4 x^{2}+8 x \sqrt{2}+8}}{2 x^{2}-4} for x>2x>\sqrt{2} 58n+2n20n+5n(nN)5^{\circ} \frac{8^{n}+2^{n}}{20^{n}+5^{n}}(n \in \mathbb{N}) 10D=64(23)66+32(2+510^{\circ} D=\sqrt[6]{64(\sqrt{2}-3)^{6}}+\sqrt[5]{-32(\sqrt{2}+}

Studdy Solution

STEP 1

What is this asking? We need to simplify a bunch of expressions with exponents and radicals, making them look as neat and tidy as possible! Watch out! Exponent rules can be tricky!
Remember, a negative exponent means "flip it," and fractional exponents are like hidden radicals.
Also, don't forget to consider any restrictions on the variables, like x>2x > \sqrt{2} in problem 9.

STEP 2

1. Simplify Expression 1
2. Simplify Expression 2
3. Simplify Expression 3
4. Simplify Expression 4
5. Simplify Expression 5
6. Simplify Expression 6
7. Simplify Expression 7
8. Simplify Expression 8
9. Simplify Expression 9
10. Simplify Expression 10

STEP 3

We have 912×(13)0.59^{-\frac{1}{2}} \times \left(\frac{1}{3}\right)^{-0.5}.
A **negative exponent** means we take the reciprocal!
So, 9129^{-\frac{1}{2}} becomes 1912\frac{1}{9^{\frac{1}{2}}}.
And (13)0.5\left(\frac{1}{3}\right)^{-0.5} flips to become 30.53^{0.5} or 3123^{\frac{1}{2}}.

STEP 4

Remember, fractional exponents are just **radical** in disguise! 9129^{\frac{1}{2}} is the same as 9\sqrt{9}, which is **3**.
Also, 3123^{\frac{1}{2}} is 3\sqrt{3}.

STEP 5

Putting it all together, we get 133=33\frac{1}{3} \cdot \sqrt{3} = \frac{\sqrt{3}}{3}.
Boom!

STEP 6

We've got 8×103×4×1040.0016\frac{8 \times 10^3 \times 4 \times 10^{-4}}{0.0016}.
Let's **tackle the numerator** first. 8×4=328 \times 4 = 32.
And 103×104=103+(4)=10110^3 \times 10^{-4} = 10^{3 + (-4)} = 10^{-1}.
So the numerator becomes 32×101=3.232 \times 10^{-1} = 3.2.

STEP 7

Now, the **denominator**: 0.0016=16×104=1.6×1030.0016 = 16 \times 10^{-4} = 1.6 \times 10^{-3}.

STEP 8

Dividing, we have 3.21.6×103=3.21.6×103=2×103=2000\frac{3.2}{1.6 \times 10^{-3}} = \frac{3.2}{1.6} \times 10^3 = 2 \times 10^3 = 2000.
Nailed it!

STEP 9

We have 3n+46×3n+17×3n+1\frac{3^{n+4} - 6 \times 3^{n+1}}{7 \times 3^{n+1}}.
Notice that both terms in the numerator have a factor of 3n+13^{n+1}.
Let's **factor that out**: 3n+1(336)3^{n+1}(3^3 - 6).

STEP 10

Now, 33=273^3 = 27, so the numerator becomes 3n+1(276)=3n+1(21)=3n+1×3×73^{n+1}(27 - 6) = 3^{n+1}(21) = 3^{n+1} \times 3 \times 7.

STEP 11

Dividing by the denominator, we get 3n+1×3×77×3n+1\frac{3^{n+1} \times 3 \times 7}{7 \times 3^{n+1}}.
The 3n+13^{n+1} and the 77 divide to one, leaving us with just **3**!

STEP 12

We're looking at 4n+24n2n+12n\frac{4^{n+2} - 4^n}{2^{n+1} - 2^n}.
Let's **rewrite** 44 as 222^2, so 4n+2=(22)n+2=22n+44^{n+2} = (2^2)^{n+2} = 2^{2n+4} and 4n=(22)n=22n4^n = (2^2)^n = 2^{2n}.

STEP 13

Now, we can **factor** the numerator: 22n(241)=22n(161)=22n(15)2^{2n}(2^4 - 1) = 2^{2n}(16 - 1) = 2^{2n}(15).
And the denominator factors as 2n(21)=2n(1)=2n2^n(2 - 1) = 2^n(1) = 2^n.

STEP 14

Dividing, we get 22n(15)2n=2n(15)=152n\frac{2^{2n}(15)}{2^n} = 2^n(15) = 15 \cdot 2^n.
Perfect!

STEP 15

We have 8n+2n20n+5n\frac{8^n + 2^n}{20^n + 5^n}.
Let's rewrite 88 as 232^3, so 8n=(23)n=23n8^n = (2^3)^n = 2^{3n}.
Also, 20n=(45)n=4n5n=(22)n5n=22n5n20^n = (4 \cdot 5)^n = 4^n \cdot 5^n = (2^2)^n \cdot 5^n = 2^{2n} \cdot 5^n.

STEP 16

Now our expression is 23n+2n22n5n+5n\frac{2^{3n} + 2^n}{2^{2n} \cdot 5^n + 5^n}.
We can **factor** 2n2^n from the numerator and 5n5^n from the denominator: 2n(22n+1)5n(22n+1)\frac{2^n(2^{2n} + 1)}{5^n(2^{2n} + 1)}.

STEP 17

The (22n+1)(2^{2n} + 1) terms divide to one, leaving us with 2n5n=(25)n\frac{2^n}{5^n} = \left(\frac{2}{5}\right)^n.
Awesome!

STEP 18

We have 7n+14n7n14n\frac{7^n + 14^n}{7^n - 14^n}.
Notice that 14n=(27)n=2n7n14^n = (2 \cdot 7)^n = 2^n \cdot 7^n.

STEP 19

Substituting, we get 7n+2n7n7n2n7n\frac{7^n + 2^n \cdot 7^n}{7^n - 2^n \cdot 7^n}.
Now we can **factor** out 7n7^n: 7n(1+2n)7n(12n)\frac{7^n(1 + 2^n)}{7^n(1 - 2^n)}.

STEP 20

The 7n7^n terms divide to one, leaving us with 1+2n12n\frac{1 + 2^n}{1 - 2^n}.
Fantastic!

STEP 21

We have A=52n+1n53n+1n+1÷5n+3n+15n1nA = \frac{\sqrt[n]{5^{2n+1}}}{\sqrt[n+1]{5^{3n+1}}} \div \frac{\sqrt[n+1]{5^{n+3}}}{\sqrt[n]{5^{n-1}}}.
Let's rewrite the radicals as fractional exponents: A=52n+1n53n+1n+1÷5n+3n+15n1nA = \frac{5^{\frac{2n+1}{n}}}{5^{\frac{3n+1}{n+1}}} \div \frac{5^{\frac{n+3}{n+1}}}{5^{\frac{n-1}{n}}}.

STEP 22

Remember, dividing by a fraction is the same as multiplying by its reciprocal: A=52n+1n53n+1n+15n1n5n+3n+1A = \frac{5^{\frac{2n+1}{n}}}{5^{\frac{3n+1}{n+1}}} \cdot \frac{5^{\frac{n-1}{n}}}{5^{\frac{n+3}{n+1}}}.

STEP 23

Now we can use our exponent rules!
When multiplying, we add the exponents, and when dividing, we subtract: A=52n+1n3n+1n+1+n1nn+3n+1A = 5^{\frac{2n+1}{n} - \frac{3n+1}{n+1} + \frac{n-1}{n} - \frac{n+3}{n+1}}.

STEP 24

Simplifying the exponent, we get 2n+1+n1n3n+1+n+3n+1=3nn4n+4n+1=34(n+1)n+1=34=1\frac{2n+1+n-1}{n} - \frac{3n+1+n+3}{n+1} = \frac{3n}{n} - \frac{4n+4}{n+1} = 3 - \frac{4(n+1)}{n+1} = 3 - 4 = -1.

STEP 25

So, A=51=15A = 5^{-1} = \frac{1}{5}.
Excellent!

STEP 26

We have B=2n+3+5×2n+13×2n+1B = \frac{2^{n+3} + 5 \times 2^{n+1}}{3 \times 2^{n+1}}.
Let's factor out 2n+12^{n+1} from the numerator: B=2n+1(22+5)3×2n+1B = \frac{2^{n+1}(2^2 + 5)}{3 \times 2^{n+1}}.

STEP 27

Simplifying, we get B=2n+1(4+5)3×2n+1=2n+1(9)3×2n+1=93=3B = \frac{2^{n+1}(4+5)}{3 \times 2^{n+1}} = \frac{2^{n+1}(9)}{3 \times 2^{n+1}} = \frac{9}{3} = 3.
Wonderful!

STEP 28

We have C=4x2+8x2+82x24C = \frac{\sqrt{4x^2 + 8x\sqrt{2} + 8}}{2x^2 - 4} for x>2x > \sqrt{2}.
Notice that 4x2+8x2+8=(2x+22)24x^2 + 8x\sqrt{2} + 8 = (2x + 2\sqrt{2})^2.

STEP 29

So, 4x2+8x2+8=(2x+22)2=2x+22=2x+22\sqrt{4x^2 + 8x\sqrt{2} + 8} = \sqrt{(2x + 2\sqrt{2})^2} = |2x + 2\sqrt{2}| = 2x + 2\sqrt{2} since x>2x > \sqrt{2}.

STEP 30

Now, C=2x+222x24=2(x+2)2(x22)=x+2(x2)(x+2)=1x2C = \frac{2x + 2\sqrt{2}}{2x^2 - 4} = \frac{2(x + \sqrt{2})}{2(x^2 - 2)} = \frac{x + \sqrt{2}}{(x-\sqrt{2})(x+\sqrt{2})} = \frac{1}{x-\sqrt{2}}.
Superb!

STEP 31

We have D=64(23)66+32(2+3)55D = \sqrt[6]{64(\sqrt{2}-3)^6} + \sqrt[5]{-32(\sqrt{2}+3)^5}.
Remember, ann=a\sqrt[n]{a^n} = |a|.

STEP 32

So, 64(23)66=26(23)66=2(23)=2(23)=2(32)\sqrt[6]{64(\sqrt{2}-3)^6} = \sqrt[6]{2^6(\sqrt{2}-3)^6} = |2(\sqrt{2}-3)| = 2|(\sqrt{2}-3)| = 2(3-\sqrt{2}) since 3>23 > \sqrt{2}.

STEP 33

And 32(2+3)55=(2)5(2+3)55=2(2+3)\sqrt[5]{-32(\sqrt{2}+3)^5} = \sqrt[5]{(-2)^5(\sqrt{2}+3)^5} = -2(\sqrt{2}+3).

STEP 34

Therefore, D=2(32)2(2+3)=622226=42D = 2(3-\sqrt{2}) - 2(\sqrt{2}+3) = 6 - 2\sqrt{2} - 2\sqrt{2} - 6 = -4\sqrt{2}.
Amazing!

STEP 35

1. 33\frac{\sqrt{3}}{3}
2. 2000
3. 3
4. 152n15 \cdot 2^n
5. (25)n\left(\frac{2}{5}\right)^n
6. 1+2n12n\frac{1 + 2^n}{1 - 2^n}
7. 15\frac{1}{5}
8. 3
9. 1x2\frac{1}{x-\sqrt{2}}
10. 42-4\sqrt{2}

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