Operations on real numbers and the laws of exponents

Multiplying both sides by (\sqrt{a}+\sqrt{b}), we get (1=(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})=a-b). In exams, apply the conjugate product directly.

Inside, \(\frac{2x^{-3}}{x^{2}}=2x^{-5}\), so \(\left(2x^{-5}\right)^{-2}x^{-4}=\frac{x^{10}}{4}x^{-4}=\frac{x^{6}}{4}\). In exams, subtract the inner exponents first.

Using the laws of exponents, \(\frac{a^{2p+1}\cdot a^{p-3}}{a^{p+4}}=a^{(2p+1)+(p-3)-(p+4)}=a^{2p-6}\). Since \(a>0\) and \(a\neq1\), equal powers with the same base have equal exponents; hence \(2p-6=6\). Therefore, \(2p=12\) and \(p=6\), so option C is correct. Exam tip: add exponents when multiplying powers with the same base and subtract them when dividing.

Writing all terms with base (3), the total exponent is (8-3+8-10=3). Therefore, the value is (3^{3}), so choose the option (3^{3}).

Here (u^{2}-v^{2}=(u-v)(u+v)=2\sqrt{5}\cdot2\sqrt{13}=4\sqrt{65}) and (uv=8). Hence the value is (\frac{\sqrt{65}}{2}).

The product of the denominators is (16-15=1), and the numerator becomes (8). In exams, adding conjugate denominators together is a fast method.

Here (4^{x+1}-4^{x}=4\cdot4^{x}-4^{x}=3\cdot4^{x}=192), so (4^{x}=64). Since (64=4^{3}), (x=3).

Using the quotient rule for exponents, \(\frac{m^{-4}n^3}{m^2n^{-5}}=m^{-4-2}n^{3-(-5)}=m^{-6}n^8\). Raising this result to the power \(-1\) takes its reciprocal: \((m^{-6}n^8)^{-1}=m^6n^{-8}\). Therefore, option B is correct. Exam tip: a negative exponent represents a reciprocal, so \(n^{-8}=\frac{1}{n^8}\).

Since \(\left(\frac{64}{125}\right)^{\frac{1}{3}}=\frac{4}{5}\), \(\left(\frac{64}{125}\right)^{-\frac{2}{3}}=\left(\frac{4}{5}\right)^{-2}=\frac{25}{16}\). In exams, take the cube root first.

We use \(\left(x-\frac{1}{x}\right)^{2}=x^{2}+\frac{1}{x^{2}}-2\). Thus \(36=x^{2}+\frac{1}{x^{2}}-2\), so the value is (38).

Here (121^{-2}=11^{-4}) and (1331^{-1}=11^{-3}), so (\frac{11^{5}\cdot11^{-4}}{11^{-3}}=11^{4}). In exams, division by a negative power adds the exponent.

We have (\sqrt{162}=9\sqrt{2}), (\sqrt{98}=7\sqrt{2}), (\sqrt{50}=5\sqrt{2}), and (\sqrt{18}=3\sqrt{2}). The total is (4\sqrt{2}).

The key step is to express every factor with the same base, 2. Since \\(8=2^3\\) and \\(64=2^6\\), the equation can be compared using powers of 2. The exponent on the left becomes \\(4x-6\\). Equating exponents gives \\(x=3\\), so option B is correct.

Rewrite the equation as \\(2^x(2^3)^{x-2}=2^6\\). Using the power rule, \\((2^3)^{x-2}=2^{3x-6}\\), and multiplying like bases gives \\(2^{x+3x-6}=2^{4x-6}\\). Therefore \\(4x-6=6\\), so \\(4x=12\\) and \\(x=3\\). Substitution checks it: \\(2^3\\cdot8^1=8\\cdot8=64\\).

The numerator is (\frac{y^{3}-x^{3}}{x^{3}y^{3}}), and the denominator is (\frac{y-x}{xy}). Division gives (\frac{x^{2}+xy+y^{2}}{x^{2}y^{2}}).

Because ((3+\sqrt{5})^{2}=9+5+6\sqrt{5}=14+6\sqrt{5}), (\sqrt{A}=3+\sqrt{5}). In exams, identify perfect-square surd forms.

Inside, \(\frac{6x^{-2}y^{3}}{3x^{4}y^{-1}}=2x^{-6}y^{4}\), and its square is \(4x^{-12}y^{8}\). Multiplying by \(\frac{x^{12}}{4y^{8}}\) gives (1).

Factoring (3^{x}), we get (3^{x}(1+3+9)=117). Thus (13\cdot3^{x}=117), (3^{x}=9), and (x=2).

Here \(49^{\frac{3}{2}}=(7^{2})^{\frac{3}{2}}=7^{3}\) and \(343^{-\frac{2}{3}}=(7^{3})^{-\frac{2}{3}}=7^{-2}\). The product is \(7^{1}=7\).

Using the square of a binomial, \(r^2=(\sqrt{15}+\sqrt{6})^2=15+6+2\sqrt{90}\). Since \(\sqrt{90}=3\sqrt{10}\), we get \(r^2=21+6\sqrt{10}\). Therefore, \(r^2-6\sqrt{10}=21\). Option 27 results from incorrectly retaining the radical term instead of cancelling it. Exam tip: always include the middle term \(2ab\) when expanding \((a+b)^2\).

The governing concept is the difference-of-squares identity, A^2−B^2=(A−B)(A+B). Rewrite the numerator as x^6−64=(x^3)^2−8^2. It therefore factors as (x^3−8)(x^3+8). Since the condition x^3≠8 guarantees that the denominator is non-zero, the common factor x^3−8 may be cancelled, leaving x^3+8. Thus option B is correct. Option A is only the cancelled denominator factor, option C does not follow from a valid factorization, and option D incorrectly adds the two square terms instead of applying the difference identity. The restriction remains important even after simplification because the original expression was undefined when x^3=8.