Operations on real numbers and the laws of exponents

Since (25^{\frac{3}{2}}=125) and (125^{\frac{2}{3}}=25), the value is (5). In exams, understand the root first in fractional powers.

Because (x^3-27=(x-3)(x^2+3x+9)), the quotient is (x^2+3x+9). In exams, identify the difference of cubes.

(\sqrt{75}=5\sqrt{3}), (\sqrt{12}=2\sqrt{3}), and (\sqrt{48}=4\sqrt{3}), so the answer is (7\sqrt{3}). In exams, combine only terms with the same radical part.

Multiplying by (\sqrt{7}-\sqrt{5}) makes the denominator (7-5=2) and gives (\sqrt{7}-\sqrt{5}). In exams, use the conjugate.

Since (4^{x+1}=2^{2x+2}) and (128=2^7), we get (2x+2=7) and (x=\dfrac{5}{2}). In exams, write both sides with the same base.

Rewrite \(16\) as \(2^4\): \(16^p=(2^4)^p=(2^p)^4=5^4=625\). Therefore, the correct answer is 625. The distractor 125 equals \(5^3\), but the required exponent is 4. Exam tip: Express the new base as a power of the given base and apply \((a^m)^n=a^{mn}\).

Because (x^2-y^2=(x-y)(x+y)), ((x+y)) cancels and (x-y) remains. In exams, identify difference of squares quickly.

On expansion, ((2m-n)^2=4m^2-4mn+n^2) and ((m+n)^2=m^2+2mn+n^2), so the difference is (3m^2-6mn). In exams, check the signs carefully.

The product of coefficients (-4) and (3) is (-12), and (a^{2-1}b^{-3+5}=ab^2). In exams, handle coefficients and exponents separately.

Since (0.0001=10^{-4}), ((10^{-4})^{\frac{3}{2}}=10^{-6}). In exams, convert decimals into powers of (10).

Because ((5x^2)^0=1), (x^0=1), and (2^{-1}=\dfrac{1}{2}), the value is (4). In exams, apply the zero exponent rule only to a non-zero base.

\(\left(\dfrac{27}{8}\right)^{\frac{1}{3}}=\dfrac{3}{2}\), so \(\left(\dfrac{27}{8}\right)^{-\frac{2}{3}}=\left(\dfrac{3}{2}\right)^{-2}=\dfrac{4}{9}\). In exams, take the reciprocal for a negative exponent.

(\sqrt{12}=2\sqrt{3}) and (\sqrt{27}=3\sqrt{3}), so the inside value is (-\sqrt{3}) and the product is (-6). In exams, simplify the surds first.

This matches ((a-b)(a^2+ab+b^2)=a^3-b^3), so the answer is (x^3-8). In exams, identifying the identity makes expansion faster.

Applying the outside exponent (\dfrac{1}{2}) gives (a^2b^{-1}=\dfrac{a^2}{b}). In exams, apply the fractional power to every factor.

Taking (10^4) common in the numerator gives (\dfrac{10^4(10-1)}{9\times 10^3}=10). In exams, taking a common factor makes calculation easier.

Since (6^4=(2\times 3)^4=2^4\times 3^4), the value is (3^2=9). In exams, write a composite base in prime factors.

(\dfrac{\sqrt{48}}{\sqrt{3}}=\sqrt{16}=4) and (\dfrac{\sqrt{75}}{\sqrt{3}}=\sqrt{25}=5), so the sum is (9). In exams, simplify the division inside the root.

Since (81=3^4), we get (2x-1=4) and (x=\dfrac{5}{2}). In exams, equate exponents when the bases are the same.

Dividing both terms by (x^{-3}) gives (1+x). In exams, divide each term separately by the denominator.

Inside, \(\dfrac{a^2}{b^{-3}}=a^2b^3\), and applying the power (-2) gives \(\dfrac{1}{a^4b^6}\). In exams, simplify the inside part first.

Since (\sqrt{8}=2\sqrt{2}), ((\sqrt{2}+\sqrt{8})^2=(3\sqrt{2})^2=18). In exams, simplify the surd before squaring.

This is of the form ((A+B)^2-(A-B)^2=4AB), where (A=3x) and (B=2), so the answer is (24x). In exams, identities save time.

The direct answer is A, \(4x^2-2x+1\). Recognize the numerator as a sum of cubes: \(8x^3+1=(2x)^3+1^3\). Use \(a^3+b^3=(a+b)(a^2-ab+b^2)\). With a=2x and b=1, the factorization is \((2x+1)(4x^2-2x+1)\). Dividing by \(2x+1\) leaves the quotient \(4x^2-2x+1\), with the usual restriction that the divisor is not zero. Option A is correct. Option B has the wrong sign in the middle term; multiplying it back would not reproduce the numerator. Option C is missing a linear term and is not the sum-of-cubes quotient. Option D has the wrong leading coefficient and therefore cannot produce \(8x^3\). Memory cue: for a sum of cubes, the signs in the second factor are plus, minus, plus.

(N=\dfrac{0.45}{10^{-3}}=0.45\times 10^3=450). In exams, dividing by (10^{-3}) is like multiplying by (10^3).