Operations on real numbers and the laws of exponents

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.

The direct answer is option B: 3. Start with \(4^{x+1}-4^x=192\). Using \(4^{x+1}=4\cdot4^x\), the left side becomes \(4\cdot4^x-4^x=(4-1)4^x=3\cdot4^x\). Thus 3·4^x=192. Dividing by 3 gives 4^x=64. Since 64=4^3, we get x=3. Option A, 2, would give 4^2=16 and the original difference 4^3−4^2=64−16=48, not 192. Option B, 3, gives 4^4−4^3=256−64=192, so it works exactly. Option C, 4, gives 4^5−4^4=1024−256=768, too large. Option D, 5, gives 4^6−4^5=4096−1024=3072, also wrong. The key is to factor out the smaller common power rather than trying to calculate an unknown power directly. Remember: \(a^{x+1}=a\cdot a^x\).

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).

The direct answer is option B, x=2. Factor out the common factor 3^x: 3^x+3^{x+1}+3^{x+2}=3^x(1+3+3^2)=3^x(1+3+9)=13\cdot3^x. The equation becomes 13\cdot3^x=117. Divide by 13: 3^x=9. Since 9=3^2, x=2. Option B is correct. Option A, x=1, would give 3+9+27=39, not 117. Option C, x=3, would give 9+27+81=117? Check carefully: for x=3 the sum is 27+81+243=351, so it is not correct. Option D, x=4, gives an even larger value, 1053, and is also wrong. Another quick check for x=2 is 9+27+81=117. The key exponent rule is 3^{x+1}=3^x\cdot3 and 3^{x+2}=3^x\cdot9. The supplied answer is consistent.

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 product of denominators is (10-9=1), and the numerator is ((\sqrt{10}+3)-(\sqrt{10}-3)=6). In exams, find the product of conjugate denominators first.

Using the laws of exponents, \(x^{12}=(x^4)^3=5^3=125\) and \(x^8=(x^4)^2=5^2=25\). Therefore, \(x^{12}-x^8=125-25=100\), so option C is correct. Option D is only the value of \(x^{12}\), not of the complete expression. Exam tip: When the required exponents are multiples of a given exponent, rewrite them as powers of that given expression.

Since \(\left(\frac{16}{81}\right)^{\frac{1}{4}}=\frac{2}{3}\), \(\left(\frac{16}{81}\right)^{-\frac{3}{4}}=\left(\frac{2}{3}\right)^{-3}=\frac{27}{8}\). In exams, take the fourth root first.

Here ((3\sqrt{5})^{2}=45), ((2\sqrt{7})^{2}=28), and the middle term is (12\sqrt{35}). Therefore, the expansion is (73-12\sqrt{35}).

We have \((4x^{-1})^2=16x^{-2}\) and \((3x^3)^2=9x^6\). Thus, the numerator becomes \(16x^{-2}\cdot9x^6=144x^4\). Therefore, \(\frac{144x^4}{12x^4}=12\), since \(x^4\) cancels for \(x\ne0\). Options B and C retain unnecessary powers of \(x\), while option D is only the numerator. Exam tip: when multiplying like bases, add their exponents; when dividing, subtract them.

Here (\sqrt{108}=6\sqrt{3}), (\sqrt{75}=5\sqrt{3}), and (\sqrt{12}=2\sqrt{3}). The numerator is (9\sqrt{3}), so the value is (9).

The conjugate of \(5+2\sqrt{6}\) is \(5-2\sqrt{6}\). Since \((5+2\sqrt{6})(5-2\sqrt{6})=25-24=1\), we get \(\frac{1}{5+2\sqrt{6}}=5-2\sqrt{6}\). Therefore, \(y+\frac{1}{y}=(5+2\sqrt{6})+(5-2\sqrt{6})=10\). Option A considers only the integer part of \(y\), while option C may result from incorrectly combining the radical terms. Exam tip: use the conjugate to rationalize the denominator in such expressions.

Inside, \(\frac{x^{-2}y^{4}}{z^{-3}}=x^{-2}y^{4}z^{3}\), so its reciprocal is \(x^{2}y^{-4}z^{-3}\). Multiplying by \(\frac{y^{2}}{x^{3}z^{2}}\) gives \(\frac{1}{xy^{2}z^{5}}\).