Operations on real numbers and the laws of exponents

Use the identity \(a^2+b^2=(a-b)^2+2ab\). Thus, \(a^2+b^2=8^2+2(12)=64+24=88\). The value 76 results from adding only \(ab\) instead of \(2ab\), so it is incorrect. In exams, carefully identify the sign and the coefficient of \(2ab\) in the relevant identity.

Use the identity \\(a-b)^2=a^2-2ab+b^2. Since \\(97=100-3, \\(97)^2=(100-3)^2=10000-600+9=9409, so option A is correct. The nearby numerical distractors result from small place-value or middle-term errors; in an exam, check the term \\(-2ab carefully.

\(104^2=(100+4)^2=100^2+2(100)(4)+4^2=10000+800+16=10816\). Therefore, the correct answer is 10816. The nearby distractor 10840 is incorrect because the sum of the middle term 800 and the last term 16 is 816, not 840. Exam tip: Always include the middle term \(2ab\) when using \((a+b)^2=a^2+2ab+b^2\).

Here, \(72=70+2\) and \(68=70-2\). Therefore, \((70+2)(70-2)=70^2-2^2=4900-4=4896\). The identity used is \((a+b)(a-b)=a^2-b^2\). Exam tip: When two numbers are equally spaced around a central number, use the difference-of-squares identity.

Factor the expression first: \(a^2b+ab^2=ab(a+b)\). Substituting \(a=2\) and \(b=5\) gives \(2\times5\times(2+5)=10\times7=70\). Therefore, the correct answer is 70. The distractor 60 can result from incorrectly evaluating one of the common factors or the sum. In an exam, look for the common factor \(ab\) to simplify such expressions quickly and accurately.

The greatest common factor of \(4x^5\) and \(12x^4\) is \(4x^4\). Taking it outside gives \(4x^5+12x^4=4x^4(x+3)\), since \(x^5\div x^4=x\) and \(12x^4\div 4x^4=3\). Therefore, option A is correct. Exam tip: divide every term by the common factor to determine the expression inside the parentheses.

When powers have the same non-zero base, their exponents are subtracted, so \(a^m\div a^n=a^{m-n}\). For example, \(a^5\div a^2=a^3\). Exam tip: addition of exponents is used in multiplication, not division.

Substituting \(x=4\) gives \(x^3-3x^2+2x=4^3-3(4^2)+2(4)=64-48+8=24\). Therefore, 24 is correct. The value 22 results from an arithmetic error. In the exam, calculate powers first, then multiplication, followed by addition or subtraction.

\((-4)^4=256\) because an even power gives a positive result, while \((-4)^3=-64\) because an odd power retains the negative sign. Therefore, \(x^4+x^3=256+(-64)=192\). Exam tip: Carefully check the sign when raising a negative number to an even or odd power.

When dividing powers with the same base, subtract the exponents: \(x^{8-5}y^{-4-(-6)}=x^3y^2\). Therefore, option A is correct. In option B, the exponent of \(y\) is calculated incorrectly, while options C and D add the exponents instead of subtracting them. Exam tip: use \(a^m\div a^n=a^{m-n}\) when dividing like bases.

First, \(\frac{a^{-2}}{b^{-4}}=a^{-2}\times b^4=\frac{b^4}{a^2}\). Taking the power \(-1\) inverts the fraction: \(\left(\frac{b^4}{a^2}\right)^{-1}=\frac{a^2}{b^4}\). Therefore, option B is correct. Exam tip: the power \(-1\) of a non-zero number or fraction gives its reciprocal.

By the zero-exponent rule, the zeroth power of any non-zero number is 1, so \(\left(5^2\right)^0=1\). Using the negative-exponent rule, \(2^{-3}=\frac{1}{2^3}=\frac{1}{8}\). Therefore, \(1+\frac{1}{8}=\frac{9}{8}\). In an exam, first convert a negative exponent into a reciprocal to avoid sign and calculation errors.

When powers with the same non-zero base are divided, their exponents are subtracted; hence this is the quotient law. The product law adds exponents instead. Exam tip: division of like bases means subtract exponents.

Since \(16=2^4\), we have \(16^{-1}=(2^4)^{-1}=2^{-4}\). Therefore, \(2^7\cdot16^{-1}=2^7\cdot2^{-4}=2^{7-4}=2^3=8\). Hence, the correct answer is 8. Exam tip: When multiplying powers with the same base, add their exponents.

Apply the identity \((a-b)^2=a^2-2ab+b^2\). Here, \(a=4x\) and \(b=5\). Thus, \((4x-5)^2=(4x)^2-2(4x)(5)+5^2=16x^2-40x+25\). Option B incorrectly calculates the middle term as \(-20x\). Exam tip: in the square of a binomial, the middle term is twice the product of the two terms, with the appropriate sign.

Use the identity \\((a+b)^2=a^2+2ab+b^2\\). Here, \\(a=5x\\) and \\(b=2\\). Thus, \\(a^2=25x^2\\), \\(2ab=2\\cdot5x\\cdot2=20x\\), and \\(b^2=4\\). Therefore, the expansion is \\(25x^2+20x+4\\). Option B incorrectly gives the middle term as \\(10x\\) by missing the factor 2. In an exam, check the middle term using \\(2ab\\).

This product matches the identity (a+b)(a-b)=a^2-b^2. Here, \(a=7x\) and \(b=6\), so \((7x+6)(7x-6)=(7x)^2-6^2=49x^2-36\). Option B has the wrong sign, while option D incorrectly includes a middle term. Exam tip: the middle terms cancel when multiplying conjugate binomials.

Since \(243=3^5\), the equation becomes \(3^{x+2}=3^5\). With equal bases, the exponents are equal, so \(x+2=5\) and \(x=3\). Option C would give an exponent of 6, so it is incorrect. Exam tip: Rewrite both sides with the same base before comparing exponents.

Since \(32=2^5\), equality of the bases gives \(x-1=5\). Therefore, \(x=6\). Option B results from incorrectly treating the exponent \(x-1\) as \(x=5\). Exam tip: Rewrite both sides with the same base before comparing exponents.

\((2^4)^2=2^8\), \(8^{-1}=2^{-3}\), and \(4=2^2\). Hence, \(\frac{2^8\cdot2^{-3}}{2^2}=2^{8-3-2}=2^3=8\), so option A is correct. Exam tip: for powers with the same base, add exponents when multiplying and subtract them when dividing. Choosing option C (16) results from making an error while subtracting the exponents.

Here, \(100x^2=(10x)^2\) and \(121y^2=(11y)^2\). Thus, the expression is \((10x)^2-(11y)^2\). Applying the difference of squares identity \(a^2-b^2=(a-b)(a+b)\) gives \((10x-11y)(10x+11y)\). Options B and C are squares of a single binomial, not a difference of two squares. Exam tip: first check whether both terms are perfect squares before factorizing.

Using the power rule, \((2x^2y^{-2})^3=2^3x^6y^{-6}=8x^6y^{-6}\). Therefore, \(\frac{8x^6y^{-6}}{8x^3y^{-7}}=x^{6-3}y^{-6-(-7)}=x^3y\). Hence, option A is correct. Exam tip: when dividing like bases, subtract the exponents carefully; here \(-6-(-7)=1\), not \(-13\).

Use the identity ((x+y)^2=x^2+y^2+2xy). Substituting the given values gives (10^2=58+2xy), or (100=58+2xy). Hence (2xy=42) and (xy=21). The value 42 is the value of (2xy), not of (xy). Exam tip: In such questions, square the sum and subtract (x^2+y^2) to find (2xy).

By exponent laws, ((3^2)^3=3^6) and (3^6 \div 3^4=3^2). In exams, apply power of a power first and then the division law.

Here (8=2^3) and (4^2=(2^2)^2=2^4), so (\dfrac{2^5 \times 2^3}{2^4}=2^4=16). In exams, converting numbers to the same base is useful.