Operations on real numbers and the laws of exponents

Since \(\left(\frac{81}{256}\right)^{\frac{1}{4}}=\frac{3}{4}\), \(\left(\frac{81}{256}\right)^{-\frac{3}{4}}=\left(\frac{3}{4}\right)^{-3}=\frac{64}{27}\). In exams, take the fourth root first.

We use \(\left(x+\frac{1}{x}\right)^{2}=x^{2}+\frac{1}{x^{2}}+2\). Thus \(49=x^{2}+\frac{1}{x^{2}}+2\), so the value is (47).

Here (169^{-1}=13^{-2}) and (2197^{-1}=13^{-3}), so (\frac{13^{4}\cdot13^{-2}}{13^{-3}}=13^{5}). In exams, division by a negative power adds the exponent.

We have (\sqrt{242}=11\sqrt{2}), (\sqrt{128}=8\sqrt{2}), (\sqrt{98}=7\sqrt{2}), and (\sqrt{72}=6\sqrt{2}). The total is (4\sqrt{2}).

Let (A=x^{-2}) and (B=y^{-2}). Then (\frac{A^{2}-B^{2}}{A-B}=A+B), so the answer is (x^{-2}+y^{-2}=\frac{x^{2}+y^{2}}{x^{2}y^{2}}).

Because ((3+\sqrt{10})^{2}=9+10+6\sqrt{10}=19+6\sqrt{10}), (\sqrt{A}=3+\sqrt{10}). In exams, identify perfect-square surd forms.

Inside, \(\frac{8x^{-3}y^{2}}{2x^{5}y^{-4}}=4x^{-8}y^{6}\), and its square is \(16x^{-16}y^{12}\). Multiplying by \(\frac{x^{16}}{16y^{12}}\) gives (1).

The direct answer is option B, 2. Factor the common smallest power \(4^x\). Since \(4^{x+1}=4\cdot4^x\) and \(4^{x+2}=16\cdot4^x\), the equation becomes \(4^x(1+4+16)=336\). Hence \(21\cdot4^x=336\), so \(4^x=16=4^2\), giving \(x=2\). Option A, 1, gives \(4+16+64=84\), not 336. Option B gives \(16+64+256=336\), so it works. Option C, 3, gives four times this total, 1344, and option D, 4, gives an even larger value, 5376. The important skill is using the exponent law \(a^{x+r}=a^r a^x\), then solving the simple power equation. Memory cue: after factoring, add the multipliers \(1+4+16=21\).

Here \(25^{\frac{3}{2}}=(5^{2})^{\frac{3}{2}}=5^{3}\) and \(125^{-\frac{2}{3}}=(5^{3})^{-\frac{2}{3}}=5^{-2}\). The product is (5).

Using the square of a binomial, \(r^2=(\sqrt{21}+\sqrt{14})^2=21+14+2\sqrt{294}\). Since \(\sqrt{294}=\sqrt{49\times6}=7\sqrt{6}\), we get \(r^2=35+14\sqrt{6}\). Therefore, \(r^2-14\sqrt{6}=35\), so option C is correct. Exam tip: In such problems, simplify the cross-term \(2\sqrt{ab}\) carefully before subtracting.

We use (x^{10}-1024=(x^{5})^{2}-32^{2}=(x^{5}-32)(x^{5}+32)). Cancelling the common factor leaves (x^{5}+32).

The product of denominators is (26-25=1), and the numerator is ((\sqrt{26}+5)+(\sqrt{26}-5)=2\sqrt{26}). In exams, add conjugate fractions together.

Using the laws of exponents, \(x^{15}=(x^5)^3=3^3=27\) and \(x^{10}=(x^5)^2=3^2=9\). Hence, \(x^{15}+x^{10}=27+9=36\), so option B is correct. Exam tip: rewrite each exponent as a multiple of 5 before substituting the given value of \(x^5\).

Since \(\left(\frac{25}{49}\right)^{\frac{1}{2}}=\frac{5}{7}\), \(\left(\frac{25}{49}\right)^{-\frac{3}{2}}=\left(\frac{5}{7}\right)^{-3}=\frac{343}{125}\). In exams, take the square root first.

Here ((4\sqrt{3})^{2}=48), ((3\sqrt{5})^{2}=45), and the middle term is (24\sqrt{15}). Therefore, the expansion is (93-24\sqrt{15}).

We get (a=4), and (9^{b}=3^{2b}=3^{6}) gives (b=3). Thus (a^{b}-b^{a}=4^{3}-3^{4}=64-81=-17), which is not among the options.

Using the laws of exponents, \((5x^{-2})^2=25x^{-4}\) and \((2x^4)^2=4x^8\). Thus, the numerator becomes \(25x^{-4}\cdot4x^8=100x^4\). Therefore, \(\frac{100x^4}{20x^4}=5\), since \(x\ne0\). Hence, the correct answer is 5. Exam tip: square each bracket first, combine powers with the same base, and then cancel common factors.

Here (\sqrt{300}=10\sqrt{3}), (\sqrt{192}=8\sqrt{3}), and (\sqrt{108}=6\sqrt{3}). The numerator is (12\sqrt{3}), so the value is (12).

Since \((7+4\sqrt{3})(7-4\sqrt{3})=49-48=1\), we get \(\frac{1}{7+4\sqrt{3}}=7-4\sqrt{3}\). Therefore, \(y+\frac{1}{y}=(7+4\sqrt{3})+(7-4\sqrt{3})=14\). Exam tip: For expressions of the form \(a+b\sqrt{c}\), use the conjugate \(a-b\sqrt{c}\) to rationalise the denominator.

Inside, \(\frac{x^{-4}y^{5}}{z^{-2}}=x^{-4}y^{5}z^{2}\), so its reciprocal is \(x^{4}y^{-5}z^{-2}\). Multiplying by \(\frac{y^{3}}{x^{2}z^{4}}\) gives \(\frac{x^{2}}{y^{2}z^{6}}\).

Here \(125^{\frac{2}{3}}=(5)^{2}=25\) and \(25^{-\frac{3}{2}}=(5)^{-3}=\frac{1}{125}\). The product is \(\frac{1}{5}\).

When powers with the same base are multiplied, their exponents are added; hence \(a^m\cdot a^n=a^{m+n}\). In division, exponents are subtracted, not multiplied. Exam tip: check whether the bases are identical first.

We have (\sqrt[3]{343}=7), (\sqrt[3]{a^{15}}=a^{5}), and (\sqrt[3]{b^{12}}=b^{4}). In exams, divide exponents by (3) under a cube root.

Since (x^{12}-4096=(x^{6})^{2}-64^{2}=(x^{6}-64)(x^{6}+64)), cancelling the common factor gives (x^{6}+64).

The direct answer is option A, 2\sqrt{63}. Let p=8-\sqrt{63}. Rationalize its reciprocal by multiplying numerator and denominator by the conjugate 8+\sqrt{63}: 1/p=(8+\sqrt{63})/(8^2-(\sqrt{63})^2)=(8+\sqrt{63})/(64-63)=8+\sqrt{63}. Therefore 1/p-p=(8+\sqrt{63})-(8-\sqrt{63})=2\sqrt{63}. Option A is correct. Option B, 16, incorrectly adds or otherwise combines the constants while losing the radical terms. Option C, \sqrt{63}, misses the second radical; subtraction produces two copies of it. Option D, 8\sqrt{63}, introduces an extra factor of 4 and does not follow from the algebra. It is also useful to notice that p is nonzero because 8^2-63=1, so the reciprocal exists. The conjugate method works because (u-v)(u+v)=u^2-v^2. The supplied answer and explanation are mathematically consistent.