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SubjectsMathematics
बीजीय व्यंजक
Algebraic Expressions introduces Class 9 Mathematics students to the language used for representing numbers and relationships with variables. As part of the chapter Introduction to Polynomials, students learn to identify terms, coefficients, constants, and variables; distinguish like and unlike terms; and simplify expressions by combining like terms. They also practise substituting values to evaluate expressions and applying addition, subtraction, multiplication, and division carefully, building a foundation for understanding polynomials and solving algebraic problems.
TOPIC PRACTICE
Up to 8 questions from this page. Select your focus, then start.
Because \(x\neq0\), \(\frac{x}{x}=1\). Therefore, \(x^2+\frac{x}{x}=x^2+1\). Option \(x^2+x\) is incorrect because \(\frac{x}{x}\) equals 1, not \(x\). Exam tip: before cancelling terms in a fraction, ensure that the denominator is non-zero.
View question detailsAs \(x\neq0\), \(\frac{2x^2}{x}=2x^{2-1}=2x\). Therefore, the expression becomes \(x^3+2x\). The option \(x^2+2x\) incorrectly changes \(x^3\) to \(x^2\). Exam tip: subtract exponents only within the term where powers of the same base are being divided.
View question detailsFactorising gives \(x^2-1=(x-1)(x+1)\). Since \(x\neq-1\), \(\frac{(x-1)(x+1)}{x+1}=x-1\). Hence the expression is \(x^2+(x-1)=x^2+x-1\). The close distractor \(x^2+x+1\) has the wrong sign for the constant term. Exam tip: factorise before cancelling and retain values that make the original denominator zero.
View question detailsUsing the difference-of-squares identity, \(x^2-4=(x-2)(x+2)\). Since \(x\neq2\), the factor \(x-2\) can be cancelled, giving \(\frac{x^2-4}{x-2}=x+2\). Therefore, the expression is \(x^3+(x+2)=x^3+x+2\), so option A is correct. Option B has an incorrect sign on the constant term. Exam tip: even after cancellation, retain the original restriction \(x\neq2\).
View question details\(x^2-9\) is a difference of squares: \(x^2-9=(x-3)(x+3)\). Since \(x\neq3\), the factor \(x-3\) cancels, giving \(\frac{x^2-9}{x-3}=x+3\). Thus the expression becomes \(x^3+x+3\), so option B is correct. Option A incorrectly gives \(x-3\) after simplification. Exam tip: factorise the numerator before cancelling factors, and retain the condition that the denominator is non-zero.
View question details\(x^2-4\) is a difference of squares: \(x^2-4=(x-2)(x+2)\). Thus, for \(x\neq2\), \(\frac{(x-2)(x+2)}{x-2}=x+2\). The option \(x-2\) is only a factor of the numerator, not the simplified expression. Exam tip: factorise the numerator before cancelling, and note values that make the denominator zero.
View question detailsDivide both terms of the numerator by \(x^2\): \(\frac{x^4}{x^2}-\frac{3x^2}{x^2}=x^{4-2}-3x^{2-2}=x^2-3\). Therefore, \(x^2-3\) is the correct answer. Option B incorrectly divides the second term, since \(\frac{3x^2}{x^2}=3\), not \(\frac{3}{x^2}\). The simplification is valid only for \(x\neq0\), as the original expression is undefined at \(x=0\). Exam tip: when dividing powers with the same base, subtract the exponents.
View question detailsFactor the numerator: \(x^2-1=(x-1)(x+1)\). Since \(x\neq1\), \(\frac{x^2-1}{x-1}=x+1\). Hence the given expression becomes \(x^2+(x+1)=x^2+x+1\), so option A is correct. Option B incorrectly has a constant term of \(-1\); the simplified expression has \(+1\). Exam tip: Factor the numerator fully before cancelling a common factor, and retain the condition that the denominator must not be zero.
View question detailsQUIZ COMPLETE
