Operations on real numbers and the laws of exponents

We get \(\left(\frac{64x^{-6}}{27y^{9}}\right)^{\frac{1}{3}}=\frac{4x^{-2}}{3y^{3}}\). The power \(-\frac{1}{3}\) gives the reciprocal \(\frac{3x^{2}y^{3}}{4}\).

We use \(x^{2}-\frac{1}{x^{2}}=\left(x-\frac{1}{x}\right)\left(x+\frac{1}{x}\right)\). Thus \(40=5\left(x+\frac{1}{x}\right)\), so the value is (8).

Combining like terms gives \(4b^{-2}+6b^{-2}=10b^{-2}\). Therefore, \(\frac{10b^{-2}}{5b^{-3}}=2b^{-2-(-3)}=2b\), so option A is correct. Exam tip: when dividing powers with the same nonzero base, subtract the exponents: \(b^m\div b^n=b^{m-n}\).

From (\sqrt{x}=4\sqrt{3}), (x=48), and (x^{\frac{3}{2}}=x\sqrt{x}=48\cdot4\sqrt{3}=192\sqrt{3}). In exams, write (x^{\frac{3}{2}}) as (x\sqrt{x}).

Here \(\left(\frac{4}{7}\right)^{-2}=\frac{49}{16}\) and \(\left(\frac{7}{4}\right)^{-2}=\frac{16}{49}\). The sum is \(\frac{2401+256}{784}=\frac{2657}{784}\).

The left side is \(5^{2x}\cdot5^{x-2}=5^{3x-2}\), and \(3125=5^{5}\). Hence \(3x-2=5\), so \(x=\frac{7}{3}\).

Here (\sqrt{192}=8\sqrt{3}), (2\sqrt{48}=8\sqrt{3}), and (3\sqrt{12}=6\sqrt{3}). The numerator is (6\sqrt{3}), so the value is (6).

The expression is \(8x^{-3}y^{6}\cdot\frac{1}{4}x^{-2}y=2x^{-5}y^{7}\). Hence \(c+r+s=2-5+7=4\).

Here (x^{2}=7+2\sqrt{10}), so (x^{3}=17\sqrt{2}+11\sqrt{5}) and (x^{3}-7x=10\sqrt{2}+4\sqrt{5}). In exams, first find (x^{2}) and then multiply by (x).

Here \(\frac{4x^{-2}}{x^{3}}=4x^{-5}\), so its reciprocal is \(\frac{x^{5}}{4}\), and multiplying by \(x^{-4}\) gives \(\frac{x}{4}\). In exams, simplify the bracket first.

By the laws of exponents, exponents are added when like bases are multiplied and subtracted when they are divided. Thus, the exponent on the left side is \((3p-2)+(p+5)-(2p-1)=2p+4\). Therefore, \(2p+4=10\), giving \(2p=6\) and hence \(p=3\). Option C can result from mishandling the subtraction of the denominator’s exponent. In an exam, remember to change the sign of the exponent in the denominator while simplifying.

Since (25^{-2}=5^{-4}) and (125=5^{3}), the total exponent is (9-4+3-4=4). In exams, convert all terms to the same base.

Here (u^{2}-v^{2}=(u-v)(u+v)=2\sqrt{8}\cdot2\sqrt{17}=8\sqrt{34}), and (uv=9). Hence the value is (\frac{8\sqrt{34}}{9}).

The product of the denominators is (25-24=1), and the numerator becomes (2\sqrt{24}). In exams, first find the product of conjugate denominators.

Here (6^{x+1}-6^{x}=5\cdot6^{x}=900), so (6^{x}=180), which is not a listed integral power. Therefore none of the listed options is correct.

Using the quotient rule for exponents, \(\frac{p^{-5}q^4}{p^{-1}q^{-2}}=p^{-5-(-1)}q^{4-(-2)}=p^{-4}q^6\). Applying \((a^m)^n=a^{mn}\), we get \((p^{-4}q^6)^{-2}=p^8q^{-12}\), so option A is correct. Option B results from mishandling the signs while multiplying the exponents by the outer \(-2\). Exam tip: subtract exponents when dividing like bases, then multiply by the outside exponent.

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}).