Operations on real numbers and the laws of exponents

The square of a binomial is given by \((a+b)^2=a^2+2ab+b^2\). Option A is this expansion, option C only reverses the order of the terms, and in option D, \(ab+ab=2ab\). Option B lacks the middle term \(2ab\), so it is not equal to \((a+b)^2\). Exam tip: For \((a+b)^2\), remember the middle term as \(2ab\).

Using the distributive law, \(2(x+3)=2\times x+2\times 3=2x+6\). Therefore, option B is correct. In option A, the 3 has not been multiplied by 2. In such questions, multiply the factor outside the bracket by every term inside it.

Coefficients of like terms are added, so (3a+5a=8a). Only terms with the same variable and same exponent combine directly.

The terms 7x and 2x are like terms, so subtract their coefficients: 7x-2x=(7-2)x=5x. The variable x and its exponent remain unchanged. Exam tip: combine only like terms by operating on their coefficients.

Multiply the numerical coefficients and the variable parts separately: \(4\times 3=12\) and \(x\cdot x^2=x^{1+2}=x^3\). Hence, the product is \(12x^3\). Option B is incorrect because exponents of the same base are added during multiplication, not replaced by the larger exponent. Exam tip: for \(a^m\cdot a^n\), use \(a^{m+n}\).

To simplify a fraction containing powers of the same nonzero variable, simplify the numerical coefficients and subtract the denominator exponent from the numerator exponent. This uses the law \\(x^m/x^n=x^{m-n}\\), which is valid here because \\(x\\ne0\\). Thus the expression keeps the factor 3 and the remaining power is \\(x^2\\). The simplified result is therefore \\(3x^2\\), corresponding to option A.

First divide the coefficients: \\(15/5=3\\). Next divide the powers: \\(x^4/x^2=x^{4-2}=x^2\\). Multiplying these simplified parts gives \\(3x^2\\). The condition \\(x\\ne0\\) ensures that the original denominator \\(5x^2\\) is not zero. A common error is to add exponents while dividing; exponents are subtracted in division.

Using the distributive law, \(5(2x+1)=5\times 2x+5\times 1=10x+5\). Therefore, option B is correct. In option A, the constant term has not been multiplied by 5. In an exam, remember to multiply the factor outside the brackets by every term inside them.

The terms are being added, not multiplied, so their exponents cannot be added directly. \(2^3+2^4=8+16=24\), so the correct answer is 24. Choosing 128 (=\(2^7\)) reflects the common mistake of applying the exponent rule for multiplication to addition. Exam tip: evaluate each power first and then add the results.

(a) is multiplied by both (b) and (c), so this is the distributive law. This law is most useful while opening brackets.

The equation \(a+b=b+a\) says that the two addends may exchange their positions without changing the sum. This property concerns the order of terms, not the way a sum is multiplied or expanded. For example, \(3+5=5+3\), and both sides equal 8. The letters may represent any suitable real numbers, so the rule is general.

This property is called the commutative law of addition. The word “commutative” means that the order can be changed. It is different from the distributive law, such as \(a(b+c)=ab+ac\), and from an exponent law. Therefore the statement matches option A, the commutative law. The equality is not claiming that addition and multiplication are interchangeable; it only changes the order of the addends.

In this equation, the order of the numbers remains unchanged, but the grouping changes from \\((a+b)+c\\) to \\(a+(b+c)\\). The sum remains the same, so it represents the associative law of addition. The distributive law involves multiplication over addition, such as \\(a(b+c)=ab+ac\\). Exam tip: associative law changes grouping, whereas commutative law changes the order of terms.

Aligning the decimal places gives 0.50 + 0.25 = 0.75. The answer 1.25 is incorrect because the decimal numbers have been combined improperly. In an exam, always place the decimal points in the same column before adding.

Using common denominator (6), (\frac{2}{3}=\frac{4}{6}). Hence (\frac{4}{6}+\frac{1}{6}=\frac{5}{6}).

Multiply and simplify: (\frac{3}{5}\cdot\frac{10}{9}=\frac{30}{45}=\frac{2}{3}). Cancelling first saves time.

According to the order of operations, multiplication is performed before subtraction. Thus, \\(3\\cdot2=6\\), and then \\(12-6=6\\). Option B results from incorrectly calculating \\(12-3\\) first. Exam tip: perform multiplication and division before addition and subtraction.

Evaluate the exponent first: \(3^2=9\). Then multiply to get \(9\cdot2=18\), and finally add \(2+18=20\). Therefore, option A is correct. Option D results from omitting the initial 2 in the final addition. Exam tip: follow the order exponents, multiplication/division, and then addition/subtraction.

4x^2 and 3x^2 are like terms because they have the same variable with the same exponent. Add only their coefficients: 4+3=7, while x^2 remains unchanged. Therefore, the simplified form is 7x^2. Option B is incorrect because exponents are not added when like terms are combined. Exam tip: For like terms, add or subtract the coefficients and keep the variable part unchanged.

The two terms are like terms because both contain x^3. Subtract their coefficients while keeping the common variable part unchanged: 6x^3 − 2x^3 = (6 − 2)x^3 = 4x^3. The exponent is not subtracted or added in this operation. Therefore option A is correct; 8x^3 would result from addition, while the other exponents use an invalid rule.

The expression \((2x+3y)\) has two terms, \(2x\) and \(3y\), separated by the plus sign. Parentheses do not change the number of terms. In an exam, count the parts separated by plus or minus signs.

The governing concept is the definition of like terms. Two algebraic terms are like terms when their variable parts are identical, including every variable and its exponent; only the numerical coefficients may differ. The term x² has x as its variable with exponent 2. In 3x², the variable part is also x², while the coefficient changes from the implied 1 to 3. Therefore 3x² is a like term and option A is correct. The term 3x has exponent 1, so it cannot be combined with x². The term x³ has the same letter but a different exponent. The term 2y² has the same exponent but a different variable. Thus likeness depends on the complete variable-and-exponent pattern, not merely on a shared letter, exponent, or numerical coefficient.

Multiplying the coefficients gives \(3\times2=6\). For the same base, add the exponents: \(x^2\cdot x^1=x^{2+1}=x^3\). Therefore, the product is \(6x^3\). Exam tip: multiply the numerical coefficients separately and add the exponents of identical variables.

Divide the numerical coefficients and subtract the exponents of the same base: \(18\div 6=3\) and \(a^5\div a^2=a^{5-2}=a^3\). Therefore, \(18a^5\div 6a^2=3a^3\), so option A is correct. Exam tip: when dividing powers with the same non-zero base, subtract the exponents; do not add them.

When powers with the same base are multiplied, their exponents are added: \(2^3\cdot2^0\cdot2^2=2^{3+0+2}=2^5\). Therefore, \(2^5\) is correct. The distractor \(2^6\) results from incorrectly treating the exponent 0 as 1. Exam tip: for every non-zero number, \(a^0=1\).

For powers with the same nonzero base, multiplication adds exponents and division subtracts exponents: aᵐ·aⁿ=aᵐ⁺ⁿ and aᵐ÷aⁿ=aᵐ⁻ⁿ. First simplify the numerator: 5²·5³=5⁵. Then divide by 5⁴: 5⁵÷5⁴=5⁵⁻⁴=5¹. Hence option A is correct, and the numerical value is 5. Option B results from stopping after multiplying and forgetting the division. Option C incorrectly adds all three exponents, even though the final operation is division. Option D subtracts in the reverse order, 4−5, rather than numerator exponent minus denominator exponent. Since the common base 5 is nonzero, the exponent laws apply without any restriction problem.

Here \(\left(\frac{2}{3}\right)^0=1\) and \(\left(\frac{1}{2}\right)^2=\frac{1}{4}\). Therefore the sum is \(\frac{5}{4}\).