পশ্চিমবঙ্গ মধ্যশিক্ষা পর্ষদের (WBBSE) অষ্টম শ্রেণির গণিত পাঠ্যবইয়ের পঞ্চদশ অধ্যায় হলো ‘বীজগাণিতিক সংখ্যামালার সরলীকরণ’। এই পোস্টে আমরা ‘কষে দেখি – 15’-এর সমস্ত প্রশ্নের সহজ ও নির্ভুল সমাধান নিয়ে আলোচনা করেছি। আশা করি, এই নোটসগুলো তোমাদের গণিত শিখতে এবং পরীক্ষার প্রস্তুতিতে দারুণভাবে সহায়তা করবে।
1. প্রদত্ত সম্পর্কগুলি দেখি এবং কোনটি সত্য, কোনটি মিথ্যা তা লিখি।
(i) \(\frac{a+b}{c} = \frac{a}{c} + \frac{b}{c}\)
সমাধান –
বামপক্ষ = \(\frac{a+b}{c}\)
ডানপক্ষ = \(\frac{a}{c} + \frac{b}{c} = \frac{a+b}{c}\)
\(\therefore\) বামপক্ষ = ডানপক্ষ
\(\therefore\) প্রদত্ত সম্পর্কটি সত্য।
(ii) \(\frac{a}{x+y} = \frac{a}{x} + \frac{b}{y}\)
সমাধান –
বামপক্ষ = \(\frac{a}{x+y}\)
ডানপক্ষ = \(\frac{a}{x} + \frac{b}{y} = \frac{ay+bx}{xy}\)
\(\therefore\) বামপক্ষ \(\neq\) ডানপক্ষ
\(\therefore\) প্রদত্ত সম্পর্কটি মিথ্যা।
(iii) \(\frac{x-y}{a-b} = \frac{y-x}{b-a}\)
সমাধান –
বামপক্ষ = \(\frac{x-y}{a-b}\) = \(\frac{-(y-x)}{-(b-a)}\) = \(\frac{y-x}{b-a}\) = ডানপক্ষ
\(\therefore\) প্রদত্ত সম্পর্কটি সত্য।
(iv) \(\frac{1}{x} + \frac{1}{y} = \frac{1}{x+y}\)
সমাধান –
বামপক্ষ = \(\frac{1}{x} + \frac{1}{y} = \frac{y+x}{xy}\)
ডানপক্ষ = \(\frac{1}{x+y}\)
\(\therefore\) বামপক্ষ \(\neq\) ডানপক্ষ
\(\therefore\) সম্পর্কটি মিথ্যা।
2. নিচের বীজগাণিতিক সংখ্যামালাগুলি লঘিষ্ঠ আকারে পরিণত করি।
(i) \(\frac{63a^{3}b^{4}}{77b^{5}}\)
সমাধান –
= \( \frac{9a^{3}}{11b^{5-4}}\)
= \( \frac{9a^{3}}{11b} \quad \text{[উত্তর]}\)
(ii) \(\frac{18a^{4}b^{5}c^{2}}{21a^{7}b^{2}}\)
সমাধান –
\(\frac{18a^{4}b^{5}c^{2}}{21a^{7}b^{2}}\)= \( \frac{6b^{5-2}c^{2}}{7a^{7-4}}\)
= \( \frac{6b^{3}c^{2}}{7a^{3}} \quad \text{[উত্তর]}\)
(iii) \(\frac{x^{2}-3x+2}{x^{2}-1}\)
সমাধান –
\(\frac{x^{2}-3x+2}{x^{2}-1}\)= \( \frac{x^{2}-2x-x+2}{(x+1)(x-1)}\)
= \( \frac{x(x-2)-1(x-2)}{(x+1)(x-1)}\)
= \( \frac{(x-2)(x-1)}{(x+1)(x-1)}\)
= \( \frac{x-2}{x+1} \quad \text{[উত্তর]}\)
(iv) \(\frac{a+1}{a-2} \times \frac{a^{2}-a-2}{a^{2}+a}\)
সমাধান –
\(\frac{a+1}{a-2} \times \frac{a^{2}-a-2}{a^{2}+a}\)= \( \frac{a+1}{a-2} \times \frac{a^{2}-(2-1)a-2}{a(a+1)}\)
= \( \frac{a+1}{a-2} \times \frac{a^{2}-2a+a-2}{a(a+1)}\)
= \( \frac{a+1}{a-2} \times \frac{a(a-2)+1(a-2)}{a(a+1)}\)
= \( \frac{a+1}{a-2} \times \frac{(a-2)(a+1)}{a(a+1)}\)
= \( \frac{a+1}{a-2} \times \frac{a-2}{a}\)
= \( \frac{a+1}{a} \quad \text{[উত্তর]}\)
(v) \(\frac{p^{3}+q^{3}}{p^{2}-q^{2}} \div \frac{p+q}{p-q}\)
সমাধান –
\(\frac{p^{3}+q^{3}}{p^{2}-q^{2}} \div \frac{p+q}{p-q}\)= \( \frac{(p+q)(p^{2}-pq+q^{2})}{(p+q)(p-q)} \times \frac{p-q}{p+q}\)
= \( \frac{p^{2}-pq+q^{2}}{p+q} \quad \text{[উত্তর]}\)
(vi) \(\frac{x^{2}-x-6}{x^{2}+4x-5} \times \frac{x^{2}+6x+5}{x^{2}-4x+3}\)
সমাধান –
\(\frac{x^{2}-x-6}{x^{2}+4x-5} \times \frac{x^{2}+6x+5}{x^{2}-4x+3}\)= \( \frac{x^{2}-(3-2)x-6}{x^{2}+(5-1)x-5} \times \frac{x^{2}+(5+1)x+5}{x^{2}-(3+1)x+3}\)
= \( \frac{x^{2}-3x+2x-6}{x^{2}+5x-x-5} \times \frac{x^{2}+5x+x+5}{x^{2}-3x-x+3}\)
= \( \frac{x(x-3)+2(x-3)}{x(x+5)-1(x+5)} \times \frac{x(x+5)+1(x+5)}{x(x-3)-1(x-3)}\)
= \( \frac{(x-3)(x+2)}{(x+5)(x-1)} \times \frac{(x+5)(x+1)}{(x-3)(x-1)}\)
= \( \frac{(x+2)(x+1)}{(x-1)^{2}}\)
= \( \frac{x^{2}+2x+x+2}{x^{2}-2x+1}\)
= \( \frac{x^{2}+3x+2}{x^{2}-2x+1} \quad \text{[উত্তর]}\)
(vii) \(\frac{a^{2}-ab+b^{2}}{a^{2}+ab} \div \frac{a^{3}+b^{3}}{a^{2}-b^{2}}\)
সমাধান –
\(\frac{a^{2}-ab+b^{2}}{a^{2}+ab} \div \frac{a^{3}+b^{3}}{a^{2}-b^{2}}\)= \( \frac{a^{2}-ab+b^{2}}{a(a+b)} \div \frac{(a+b)(a^{2}-ab+b^{2})}{(a+b)(a-b)}\)
= \( \frac{a^{2}-ab+b^{2}}{a(a+b)} \times \frac{(a+b)(a-b)}{(a+b)(a^{2}-ab+b^{2})}\)
= \( \frac{a-b}{a^{2}+ab} \quad \text{[উত্তর]}\)
3. নীচের বীজগাণিতিক সংখ্যামালাগুলিকে সরলতম আকারে প্রকাশ করি
(i) \(\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}\)
সমাধান –
\(\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}\)= \( \frac{c+a+b}{abc}\) [উত্তর]
(ii) \(\frac{a-c-b}{a} + \frac{a+c+b}{a}\)
সমাধান –
\(\frac{a-c-b}{a} + \frac{a+c+b}{a}\)= \( \frac{a-c-b+a+c+b}{a}\)
= \( \frac{2a}{a}\)
= \( 2\) [উত্তর]
(iii) \(\frac{x^{2}+a^{2}}{ab} + \frac{x-a}{ax} – \frac{x^{3}}{b}\)
সমাধান –
\(\frac{x^{2}+a^{2}}{ab} + \frac{x-a}{ax} – \frac{x^{3}}{b}\)= \( \frac{x(x^{2}+a^{2})+b(x-a)-ax^{4}}{abx}\)
= \( \frac{x^{3}+a^{2}x+bx-ab-ax^{4}}{abx}\) [উত্তর]
(iv) \(\frac{2a^{2}b}{3b^{2}c} \times \frac{c^{4}}{3a^{3}} \div \frac{4bc^{3}}{9a^{2}}\)
সমাধান –
\(\frac{2a^{2}b}{3b^{2}c} \times \frac{c^{4}}{3a^{3}} \div \frac{4bc^{3}}{9a^{2}}\)= \( \frac{2a^{2}b}{3b^{2}c} \times \frac{c^{4}}{3a^{3}} \times \frac{9a^{2}}{4bc^{3}}\)
= \( \frac{a}{2b^{2}}\) [উত্তর]
(v) \(\frac{1}{x^{2}-3x+2} + \frac{1}{x^{2}-5x+6} + \frac{1}{x^{2}-4x+3}\)
সমাধান –
\(\frac{1}{x^{2}-3x+2} + \frac{1}{x^{2}-5x+6} + \frac{1}{x^{2}-4x+3}\)= \( \frac{1}{x^{2}-2x-x+2} + \frac{1}{x^{2}-3x-2x+6} + \frac{1}{x^{2}-3x-x+3}\)
= \( \frac{1}{x(x-2)-1(x-2)} + \frac{1}{x(x-3)-2(x-3)} + \frac{1}{x(x-3)-1(x-3)}\)
= \( \frac{1}{(x-2)(x-1)} + \frac{1}{(x-3)(x-2)} + \frac{1}{(x-3)(x-1)}\)
= \( \frac{x-3+x-1+x-2}{(x-1)(x-2)(x-3)}\)
= \( \frac{3x-6}{(x-1)(x-2)(x-3)}\)
= \( \frac{3(x-2)}{(x-1)(x-2)(x-3)}\)
= \( \frac{3}{(x-1)(x-3)}\)
= \( \frac{3}{x^{2}-x-3x+3}\)
= \( \frac{3}{x^{2}-4x+3}\) [উত্তর]
(vi) \(\frac{1}{x-1} + \frac{1}{x+1} + \frac{2x}{x^{2}+1} + \frac{4x^{3}}{x^{4}+1}\)
সমাধান –
\(\frac{1}{x-1} + \frac{1}{x+1} + \frac{2x}{x^{2}+1} + \frac{4x^{3}}{x^{4}+1}\)= \( \frac{x+1+x-1}{(x-1)(x+1)} + \frac{2x}{x^{2}+1} + \frac{4x^{3}}{x^{4}+1}\)
= \( \frac{2x}{x^{2}-1} + \frac{2x}{x^{2}+1} + \frac{4x^{3}}{x^{4}+1}\)
= \( \frac{2x(x^{2}+1)+2x(x^{2}-1)}{(x^{2}-1)(x^{2}+1)} + \frac{4x^{3}}{x^{4}+1}\)
= \( \frac{2x^{3}+2x+2x^{3}-2x}{(x^{2})^{2}-1^{2}} + \frac{4x^{3}}{x^{4}+1}\)
= \( \frac{4x^{3}}{x^{4}-1} + \frac{4x^{3}}{x^{4}+1}\)
= \( \frac{4x^{3}(x^{4}+1)+4x^{3}(x^{4}-1)}{(x^{4}-1)(x^{4}+1)}\)
= \( \frac{4x^{7}+4x^{3}+4x^{7}-4x^{3}}{(x^{4})^{2}-1^{2}}\)
= \( \frac{8x^{7}}{x^{8}-1}\) [উত্তর]
(vii) \(\frac{b^{2}-5b}{3b-4a} \times \frac{9b^{2}-16a^{2}}{b^{2}-25} \div \frac{3b^{2}+4ab}{ab+5a}\)
সমাধান –
\(\frac{b^{2}-5b}{3b-4a} \times \frac{9b^{2}-16a^{2}}{b^{2}-25} \div \frac{3b^{2}+4ab}{ab+5a}\)= \( \frac{b(b-5)}{(3b-4a)} \times \frac{(3b)^{2}-(4a)^{2}}{(b)^{2}-(5)^{2}} \div \frac{b(3b+4a)}{a(b+5)}\)
= \( \frac{b(b-5)}{(3b-4a)} \times \frac{(3b+4a)(3b-4a)}{(b+5)(b-5)} \times \frac{a(b+5)}{b(3b+4a)}\)
= \( a\) [উত্তর]
(viii) \(\frac{b+c}{(a-b)(a-c)} + \frac{c+a}{(b-a)(b-c)} + \frac{a+b}{(c-a)(c-b)}\)
সমাধান –
\(\frac{b+c}{(a-b)(a-c)} + \frac{c+a}{(b-a)(b-c)} + \frac{a+b}{(c-a)(c-b)}\)= \( -\frac{b+c}{(a-b)(c-a)} – \frac{c+a}{(a-b)(b-c)} – \frac{a+b}{(c-a)(b-c)}\)
= \( \frac{-(b+c)(b-c)-(c+a)(c-a)-(a+b)(a-b)}{(a-b)(b-c)(c-a)}\)
= \( \frac{-(b^{2}-c^{2})-(c^{2}-a^{2})-(a^{2}-b^{2})}{(a-b)(b-c)(c-a)}\)
= \( \frac{-b^{2}+c^{2}-c^{2}+a^{2}-a^{2}+b^{2}}{(a-b)(b-c)(c-a)}\)
= \( \frac{0}{(a-b)(b-c)(c-a)}\)
= \( 0\) [উত্তর]
(ix) \(\frac{b+c-a}{(a-b)(a-c)} + \frac{c+a-b}{(b-c)(b-a)} + \frac{a+b-c}{(c-a)(c-b)}\)
সমাধান –
\(\frac{b+c-a}{(a-b)(a-c)} + \frac{c+a-b}{(b-c)(b-a)} + \frac{a+b-c}{(c-a)(c-b)}\)= \( -\frac{b+c-a}{(a-b)(c-a)} – \frac{c+a-b}{(b-c)(a-b)} – \frac{a+b-c}{(c-a)(b-c)}\)
= \( \frac{-(b-c)(b+c-a)-(c-a)(c+a-b)-(a-b)(a+b-c)}{(a-b)(b-c)(c-a)}\)
= \( \frac{-(b^{2}+bc-ab-bc-c^{2}+ac)-(c^{2}+ca-cb-ac-a^{2}+ab)-(a^{2}+ab-ac-ba-b^{2}+bc)}{(a-b)(b-c)(c-a)}\)
= \( \frac{-b^{2}-bc+ab+bc+c^{2}-ac-c^{2}-ca+cb+ac+a^{2}-ab-a^{2}-ab+ac+ba+b^{2}-bc}{(a-b)(b-c)(c-a)} \)
= \( \frac{-b^{2}+b^{2}+a^{2}-a^{2}+c^{2}-c^{2}+2bc-2bc+2ac-2ac+2ab-2ab}{(a-b)(b-c)(c-a)}\)
= \( \frac{0}{(a-b)(b-c)(c-a)}\)
= \( 0\) [উত্তর]
(x) \(\frac{\frac{a^{2}}{x-a} + \frac{b^{2}}{x-b} + \frac{c^{2}}{x-c} + a + b + c}{\frac{a}{x-a} + \frac{b}{x-b} + \frac{c}{x-c}}\)
সমাধান –
\(\frac{\frac{a^{2}}{x-a} + \frac{b^{2}}{x-b} + \frac{c^{2}}{x-c} + a + b + c}{\frac{a}{x-a} + \frac{b}{x-b} + \frac{c}{x-c}}\)= \( \frac{\frac{a^{2}}{x-a} + a + \frac{b^{2}}{x-b} + b + \frac{c^{2}}{x-c} + c}{\frac{a}{x-a} + \frac{b}{x-b} + \frac{c}{x-c}}\)
= \( \frac{\frac{a^{2}+ax-a^{2}}{(x-a)} + \frac{b^{2}+bx-b^{2}}{(x-b)} + \frac{c^{2}+cx-c^{2}}{(x-c)}}{\frac{a}{x-a} + \frac{b}{x-b} + \frac{c}{x-c}}\)
= \( \frac{\frac{ax}{(x-a)} + \frac{bx}{(x-b)} + \frac{cx}{(x-c)}}{\frac{a}{x-a} + \frac{b}{x-b} + \frac{c}{x-c}}\)
= \( \frac{x \left[ \frac{a}{(x-a)} + \frac{b}{(x-b)} + \frac{c}{(x-c)} \right]}{\frac{a}{x-a} + \frac{b}{x-b} + \frac{c}{x-c}}\)
= \( x\) [উত্তর]
(xi) \(\left(\frac{a^{2}+b^{2}}{a^{2}-b^{2}} – \frac{a^{2}-b^{2}}{a^{2}+b^{2}}\right) \div \left(\frac{a+b}{a-b} – \frac{a-b}{a+b}\right) \times \left(\frac{a}{b} + \frac{b}{a}\right)\)
সমাধান –
\(\left(\frac{a^{2}+b^{2}}{a^{2}-b^{2}} – \frac{a^{2}-b^{2}}{a^{2}+b^{2}}\right) \div \left(\frac{a+b}{a-b} – \frac{a-b}{a+b}\right) \times \left(\frac{a}{b} + \frac{b}{a}\right)\)= \( \left\{ \frac{(a^{2}+b^{2})^{2} – (a^{2}-b^{2})^{2}}{(a^{2}-b^{2})(a^{2}+b^{2})} \right\} \div \left\{ \frac{(a+b)^{2} – (a-b)^{2}}{(a-b)(a+b)} \right\} \times \left(\frac{a^{2}+b^{2}}{ab}\right)\)
= \( \frac{4a^{2}b^{2}}{(a^{2}-b^{2})(a^{2}+b^{2})} \div \frac{4ab}{(a-b)(a+b)} \times \left(\frac{a^{2}+b^{2}}{ab}\right)\)
= \( \frac{4a^{2}b^{2}}{(a^{2}-b^{2})(a^{2}+b^{2})} \times \frac{(a-b)(a+b)}{4ab} \times \left(\frac{a^{2}+b^{2}}{ab}\right)\)
= \( \frac{4a^{2}b^{2}}{(a^{2}-b^{2})(a^{2}+b^{2})} \times \frac{a^{2}-b^{2}}{4ab} \times \frac{a^{2}+b^{2}}{ab}\)
= \( 1\) [উত্তর]
(xii) \(\frac{b+c}{bc}(b+c-a) + \frac{c+a}{ca}(c+a-b) + \frac{a+b}{ab}(a+b-c)\)
সমাধান –
\(\frac{b+c}{bc}(b+c-a) + \frac{c+a}{ca}(c+a-b) + \frac{a+b}{ab}(a+b-c)\)= \( \frac{(b+c)(b+c-a)}{bc} + \frac{(c+a)(c+a-b)}{ca} + \frac{(a+b)(a+b-c)}{ab}\)
= \( \frac{a(b+c)(b+c-a) + b(c+a)(c+a-b) + c(a+b)(a+b-c)}{abc}\)
= \( \frac{a(b^{2}+bc-ab+bc+c^{2}-ca) + b(c^{2}+ca-bc+ac+a^{2}-ab) + c(a^{2}+ab-ac+ba+b^{2}-bc)}{abc}\)
= \( \frac{ab^{2}+abc-a^{2}b+abc+ac^{2}-ca^{2} + bc^{2}+abc-b^{2}c+abc+a^{2}b-ab^{2} + a^{2}c+abc-ac^{2}+abc+b^{2}c-bc^{2}}{abc}\)
= \( \frac{6abc}{abc}\)
= \( 6\) [উত্তর]
(xiii) \(\frac{y^{2}+yz+z^{2}}{(x-y)(x-z)} + \frac{z^{2}+zx+x^{2}}{(y-z)(y-x)} + \frac{x^{2}+xy+y^{2}}{(z-x)(z-y)}\)
সমাধান –
\(\frac{y^{2}+yz+z^{2}}{(x-y)(x-z)} + \frac{z^{2}+zx+x^{2}}{(y-z)(y-x)} + \frac{x^{2}+xy+y^{2}}{(z-x)(z-y)}\)= \( – \frac{(y^{2}+yz+z^{2})}{(x-y)(z-x)} – \frac{(z^{2}+zx+x^{2})}{(y-z)(x-y)} – \frac{(x^{2}+xy+y^{2})}{(z-x)(y-z)}\)
= \( \frac{-(y-z)(y^{2}+yz+z^{2}) – (z-x)(z^{2}+zx+x^{2}) – (x-y)(x^{2}+xy+y^{2})}{(x-y)(y-z)(z-x)}\)
= \( \frac{-(y^{3}-z^{3}) – (z^{3}-x^{3}) – (x^{3}-y^{3})}{(x-y)(y-z)(z-x)}\)
= \( \frac{-y^{3}+z^{3} – z^{3}+x^{3} – x^{3}+y^{3}}{(x-y)(y-z)(z-x)}\)
= \( \frac{0}{(x-y)(y-z)(z-x)}\)
= \( 0\) [উত্তর]
এই আর্টিকেলে অষ্টম শ্রেণির গণিতের ‘বীজগাণিতিক সংখ্যামালার সরলীকরণ’ অধ্যায়ের ‘কষে দেখি – 15’-এর সমস্ত গাণিতিক সমস্যার সমাধান তুলে ধরেছি। আশা করি, এই পোস্টটি আপনাদের বা শিক্ষার্থীদের পরীক্ষার প্রস্তুতিতে সহায়ক হবে। কোনো প্রশ্ন, মতামত বা সাহায্যের প্রয়োজন হলে নিচে কমেন্ট করতে পারেন অথবা সরাসরি আমাদের টেলিগ্রাম চ্যানেলে যোগাযোগ করতে পারেন। আমরা আপনাদের সব প্রশ্নের উত্তর দেওয়ার জন্য সর্বদা প্রস্তুত।
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