Class-VIII / প্রশ্নমালা -6.2/ K. C. NAG

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Class-VIII 

 

 প্রশ্নমালা -6.2

 

 K. C. NAG

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সমষ্টি নির্ণয় করোঃ 

সমষ্টি নির্ণয় করোঃ 

(i). \frac{3a-1}{10a}, \frac{4-7a}{15a} 

সমাধানঃ 

\frac{3a-1}{10a}, \frac{4-7a}{15a}  এদের সমষ্টি

\\\frac{3a-1}{10a}+ \frac{4-7a}{15a}\\\\=\frac{3(3a-1)+2(4-7a)}{30a}\\\\=\frac{9a-3+8-14a}{30a}\\\\=\frac{5-5a}{30a}\\\\=\frac{1-a}{6a}

(ii). \frac{3a+b}{a}, \frac{c-b}{2a}, \frac{c-a}{a}

সমাধানঃ 

\frac{3a+b}{a}, \frac{c-b}{2a}, \frac{c-a}{a} এদের সমষ্টি

\frac{3a+b}{a}+\frac{c-b}{2a}+ \frac{c-a}{a}\\\\=\frac{2(3a+b)+(c-b)+2(c-a)}{2a}\\\\=\frac{6a+2b+c-b+2c-2a}{2a}\\\\=\frac{4a+b+3c}{2a}

(iii). \frac{2a}{c}, -\frac{2b}{a}, \frac{2c}{b}

 

সমাধানঃ 

\frac{2a}{c}, -\frac{2b}{a}, \frac{2c}{b} এদের সমষ্টি

\frac{2a}{c}+ \left ( -\frac{2b}{a} \right )+ \frac{2c}{b}\\\\=\frac{2a^2b-2b^2c+2c^2a}{abc}\\\\=\frac{2(a^2b-b^2c+c^2a)}{abc}

(iv). \frac{x}{3}, \frac{y}{4}, \frac{z}{6}

সমাধানঃ 

\frac{x}{3}, \frac{y}{4}, \frac{z}{6} এদের সমষ্টি

\frac{x}{3}+ \frac{y}{4}+\frac{z}{6}\\\\=\frac{4x+3y+2z}{12}

(v). \frac{x-y}{xy}, \frac{y-z}{yz}, \frac{z-x}{xz}

 

সমাধানঃ

\frac{x-y}{xy}, \frac{y-z}{yz}, \frac{z-x}{xz} এদের সমষ্টি

\frac{x-y}{xy}+ \frac{y-z}{yz}+ \frac{z-x}{xz}\\\\=\frac{z(x-y)+x(y-z)+y(z-x)}{xyz}\\\\=\frac{zx-yz+xy-zx+yz-yx}{xyz}\\\\=0

সরল করোঃ 

1. \frac{5}{a+2b}-\frac{2}{a-2b} 

সমাধানঃ 

\frac{5}{a+2b}-\frac{2}{a-2b}\\\\=\frac{5(a-2b)-2(a+2b)}{(a+2b)(a-2b)}\\\\=\frac{5a-10b-2a-4b}{a^2-4b^2}\\\\=\frac{3a-14b}{a^2-4b^2}

2. \frac{2}{a+3}+\frac{5}{(a+3)(a-3)}
 
 
সমাধানঃ 
\frac{2}{a+3}+\frac{5}{(a+3)(a-3)}\\\\=\frac{2(a-3)+5}{(a+3)(a-3)}\\\\=\frac{2a-6+5}{(a+3)(a-3)}\\\\=\frac{2a-1}{a^2-9}
3. \frac{x}{(x-2y)^2}-\frac{2y}{(x+2y)(x-2y)} 
 
সমাধানঃ 
\frac{x}{(x-2y)^2}-\frac{2y}{(x+2y)(x-2y)}\\\\=\frac{x(x+2y)-2y(x-2y)}{(x-2y)^2(x+2y)}\\\\=\frac{x^2+2xy-2xy+4y^2}{(x-2y)^2(x+2y)}\\\\=\frac{x^2+4y^2}{(x-2y)^2(x+2y)}
4. \frac{7}{a^2+2a-15}-\frac{7}{a^2+3a-10} 
 
সমাধানঃ 
\frac{7}{a^2+2a-15}-\frac{7}{a^2+3a-10}\\\\=\frac{7}{a^2+5a-3a-15}-\frac{7}{a^2+5a-2a-10}\\\\=\frac{7}{a(a+5)-3(a+5)}-\frac{7}{a(a+5)-2(a+5)}\\\\=\frac{7}{(a+5)(a-3)}-\frac{7}{(a+5)(a-2)}\\\\=\frac{7(a-2)-7(a-3)}{(a+5)(a-3)(a-2)}\\\\=\frac{7a-14-7a+21}{(a+5)(a-3)(a-2)}\\\\=\frac{7}{(a+5)(a-3)(a-2)}
5. \frac{2a+1}{3}+\frac{3}{2a+1}+(a-1) 
 
সমাধানঃ
\frac{2a+1}{3}+\frac{3}{2a+1}+(a-1)\\\\=\frac{(2a+1)^2+3\times 3+3(a-1)(2a+1)}{3(2a+1)}\\\\=\frac{4a^2+4a+1+9+3(2a^2+a-2a-1)}{3(2a+1)}\\\\=\frac{4a^2+4a+10+3(2a^2-a-1)}{3(2a+1)}\\\\=\frac{4a^2+4a+10+6a^2-3a-3}{3(2a+1)}\\\\=\frac{10a^2+a+7}{3(2a+1)}
6. \frac{x}{x-y}+\frac{y}{x+y}+\frac{2xy}{y^2-x^2} 
 
সমাধানঃ 
\frac{x}{x-y}+\frac{y}{x+y}+\frac{2xy}{y^2-x^2}\\\\= \frac{x}{x-y}+\frac{y}{x+y}+\frac{2xy}{(y-x)(y+x)}\\\\=\frac{x}{x-y}+\frac{y}{x+y}-\frac{2xy}{(x-y)(y+x)}\\\\=\frac{x(x+y)+y(x-y)-2xy}{(x+y)(x-y)}\\\\=\frac{x^2+xy+yx-y^2-2xy}{(x+y)(x-y)}\\\\=\frac{x^2-y^2}{(x+y)(x-y)}\\\\=\frac{(x-y)(x+y)}{(x+y)(x-y)}=1
7. 1+\frac{2a}{2a-1}-\frac{8a^2}{4a^2-1} 
 
সমাধানঃ 
1+\frac{2a}{2a-1}-\frac{8a^2}{4a^2-1}\\\\=1+\frac{2a}{2a-1}-\frac{8a^2}{(2a-1)(2a+1)}\\\\=\frac{4a^2-1+2a(2a+1)-8a^2}{(2a-1)(2a+1)}\\\\=\frac{4a^2-1+4a^2+2a-8a^2}{(2a-1)(2a+1)}\\\\=\frac{2a-1}{(2a-1)(2a+1)}\\\\=\frac{1}{2a+1}
8. \frac{1}{x^2-8x+15}+\frac{1}{x^2-4x+3}-\frac{2}{x^2-6x+5} 
 
সমাধানঃ 
\frac{1}{x^2-8x+15}+\frac{1}{x^2-4x+3}-\frac{2}{x^2-6x+5}\\\\=\frac{1}{x^2-5x-3x+15}+\frac{1}{x^2-3x-x+3}-\frac{2}{x^2-5x-x+5}\\\\=\frac{1}{x(x-5)-3(x-5)}+\frac{1}{x(x-3)-1(x-3)}-\frac{2}{x(x-5)-1(x-5)}\\\\=\frac{1}{(x-3)(x-5)}+\frac{1}{(x-3)(x-1)}-\frac{2}{(x-5)(x-1)}\\\\=\frac{x-1+x-5-2(x-3)}{(x-3)(x-5)(x-1)}\\\\=\frac{2x-6-2x+6}{(x-3)(x-5)(x-1)}=0
9. \frac{a+b}{a-b}+\frac{a-b}{a+b}-\frac{2(a^2+b^2)}{a^2-b^2} 
 
 
সমাধানঃ 
\frac{a+b}{a-b}+\frac{a-b}{a+b}-\frac{2(a^2+b^2)}{a^2-b^2}\\\\=\frac{a+b}{a-b}+\frac{a-b}{a+b}-\frac{2(a^2+b^2)}{(a-b)(a+b)}\\\\=\frac{(a+b)^2+(a-b)^2-2(a^2+b^2)}{(a+b)(a-b)}\\\\=\frac{2(a^2+b^2)-2(a^2+b^2)}{(a+b)(a-b)}=0
10. \frac{x+y}{y}-\frac{x}{x+y}-\frac{x^3-x^2y}{x^2y-y^3} 
 
সমাধানঃ
\frac{x+y}{y}-\frac{x}{x+y}-\frac{x^3-x^2y}{x^2y-y^3}\\\\=\frac{x+y}{y}-\frac{x}{x+y}-\frac{x^2(x-y)}{y(x^2-y^2)}\\\\=\frac{x+y}{y}-\frac{x}{x+y}-\frac{x^2(x-y)}{y(x-y)(x+y)}\\\\=\frac{x+y}{y}-\frac{x}{x+y}-\frac{x^2}{y(x+y)}\\\\=\frac{(x+y)^2-xy-x^2}{y(x+y)}=\frac{x^2+2xy+y^2-xy-x^2}{y(x+y)}\\\\=\frac{xy+y^2}{xy+y^2}=1
11. 1-\frac{x-2y}{2y}-\frac{2x}{x-2y}+\frac{x^2}{2y(x-2y)} 
 
সমাধানঃ 
1-\frac{x-2y}{2y}-\frac{2x}{x-2y}+\frac{x^2}{2y(x-2y)}\\\\=\frac{2y(x-2y)-(x-2y)^2-4xy+x^2}{2y(x-2y)}\\\\=\frac{2yx-4y^2-x^2+4xy-4y^2-4xy+x^2}{2y(x-2y)}\\\\=\frac{2yx-4y^2-x^2+4xy-4y^2-4xy+x^2}{2y(x-2y)}\\\\=\frac{2y(x-4y)}{2y(x-2y)}=\frac{(x-4y)}{(x-2y)}
12. \frac{2x-1}{2(x-1)(2x-3)}-\frac{1}{2x-2}-\frac{2}{2x-3} 
 
সমাধানঃ 
\frac{2x-1}{2(x-1)(2x-3)}-\frac{1}{2x-2}-\frac{2}{2x-3}\\\\=\frac{2x-1}{2(x-1)(2x-3)}-\frac{1}{2(x-1)}-\frac{2}{2x-3}\\\\=\frac{2x-1-(2x-3)-4(x-1)}{2(x-1)(2x-3)}\\\\=\frac{2x-1-2x+3-4x+4}{2(x-1)(2x-3)}\\\\=\frac{6-4x}{2(x-1)(2x-3)}\\\\=\frac{-2(2x-3)}{2(x-1)(2x-3)}\\\\=\frac{1}{1-x}
13. \frac{a-1}{a-2}-\frac{a+1}{a+2}-\frac{4}{4-a^2}+\frac{2}{2-a} 
 
সমাধানঃ
\frac{a-1}{a-2}-\frac{a+1}{a+2}-\frac{4}{4-a^2}+\frac{2}{2-a}\\\\=\frac{a-1}{a-2}-\frac{a+1}{a+2}-\frac{4}{(2-a)(2+a)}+\frac{2}{2-a}\\\\=\frac{a-1}{a-2}-\frac{a+1}{a+2}+\frac{4}{(a-2)(2+a)}-\frac{2}{a-2}\\\\=\frac{(a-1)(a+2)-(a+1)(a-2)+4-2(a+2)}{(a-2)(a+2)}\\\\=\frac{a^2+a-2-(a^2-a-2)+4-2a-4}{(a-2)(a+2)}\\\\=\frac{a^2+a-2-a^2+a+2+4-2a-4}{(a-2)(a+2)}=0
14. \frac{1}{x-1}+\frac{1}{x+1}+\frac{2x}{x^2+1}+\frac{4x^3}{x^4+1}+\frac{8x^7}{x^8+1} 
সমাধানঃ 
\frac{1}{x-1}+\frac{1}{x+1}+\frac{2x}{x^2+1}+\frac{4x^3}{x^4+1}+\frac{8x^7}{x^8+1}\\\\=\frac{x+1+x-1}{(x-1)(x+1)}+\frac{2x}{x^2+1}+\frac{4x^3}{x^4+1}+\frac{8x^7}{x^8+1}\\\\=\frac{2x}{x^2-1}+\frac{2x}{x^2+1}+\frac{4x^3}{x^4+1}+\frac{8x^7}{x^8+1}\\\\=\frac{2x(x^2+1)+2x(x^2-1)}{(x^2-1)(x^2+1)}+\frac{4x^3}{x^4+1}+\frac{8x^7}{x^8+1}\\\\=\frac{2x^3+2x+2x^3-2x}{x^4-1}+\frac{4x^3}{x^4+1}+\frac{8x^7}{x^8+1}\\\\=\frac{4x^3}{x^4-1}+\frac{4x^3}{x^4+1}+\frac{8x^7}{x^8+1}\\\\=\frac{4x^3(x^4+1)+4x^3(x^4-1)}{(x^4-1)(x^4+1)}+\frac{8x^7}{x^8+1}\\\\=\frac{4x^7+4x^3+4x^7-4x^3}{(x^4-1)(x^4+1)}+\frac{8x^7}{x^8+1}\\\\=\frac{8x^7}{x^8-1}+\frac{8x^7}{x^8+1}\\\\=\frac{8x^7(x^8+1)+8x^7(x^8-1)}{(x^8+1)(x^8-1)}\\\\=\frac{8x^{15}+8x^7+8x^{15}-8x^7}{x^{16}-1}\\\\=\frac{16x^{15}}{x^{16}-1}

15. \frac{1-x}{1-x+x^2}+\frac{1+x}{1+x+x^2}-\frac{1}{1+x^2+x^4}  

সমাধানঃ 

\frac{1-x}{1-x+x^2}+\frac{1+x}{1+x+x^2}-\frac{1}{1+x^2+x^4}\\\\=\frac{\left ( 1-x \right )\left ( 1+x+x^2 \right )+(1+x)(1-x+x^2)}{\left \{ \left ( x^2+1 \right )-x \right \}\left \{ \left ( x^2+1 \right )+x \right \}}-\frac{1}{1+x^2+x^4}\\\\=\frac{1-x^3+1+x^3}{\left ( 1+x^2 \right )^2-x^2}-\frac{1}{1+x^2+x^4}\\\\=\frac{2}{1+x^4+2x^2-x^2}-\frac{1}{1+x^2+x^4}\\\\=\frac{2}{x^4+x^2+1}-\frac{1}{1+x^2+x^4}=\frac{1}{1+x^2+x^4}

16. \frac{a}{(b-a)(a-c)}+\frac{b}{(c-b)(b-a)}+\frac{c}{(a-c)(c-b)} 

সমাধানঃ 

\frac{a}{(b-a)(a-c)}+\frac{b}{(c-b)(b-a)}+\frac{c}{(a-c)(c-b)}\\\\=\frac{a(c-b)+b(a-c)+c(b-a)}{\left ( b-a \right )\left ( a-c \right )\left ( c-b \right )}\\\\=\frac{ac-ab+ab-bc+bc-ac}{\left ( b-a \right )\left ( a-c \right )\left ( c-b \right )}=0

17.  \frac{2x}{(x+y)^2(x-y)}+\frac{2y}{(x+y)(x-y)^2}-\frac{1}{(x+y)^2}-\frac{1}{(x-y)^2} 

সমাধানঃ 

\frac{2x}{(x+y)^2(x-y)}+\frac{2y}{(x+y)(x-y)^2}-\frac{1}{(x+y)^2}-\frac{1}{(x-y)^2}\\\\=\frac{2x(x-y)+2y(x+y)-(x-y)^2-(x+y)^2}{(x+y)^2(x-y)^2}\\\\=\frac{2x^2-2xy+2xy+2y^2-x^2+2xy-y^2-x^2-2xy-y^2}{(x+y)^2(x-y)^2}\\\\=\frac{0}{(x+y)^2(x-y)^2}=0

18.  \frac{x^2+y^2}{xy}-\frac{x^2}{xy+y^2}-\frac{y^2}{x^2+xy}-\frac{1}{xy(x+y)} 

সমাধানঃ 

\frac{x^2+y^2}{xy}-\frac{x^2}{xy+y^2}-\frac{y^2}{x^2+xy}-\frac{1}{xy(x+y)}\\\\=\frac{x^2+y^2}{xy}-\frac{x^2}{y(x+y)}-\frac{y^2}{x(x+y)}-\frac{1}{xy(x+y)}\\\\=\frac{(x^2+y^2)(x+y)-x^3-y^3-1}{xy(x+y)}\\\\=\frac{x^3+x^2y+y^2x+y^3-x^3-y^3-1}{xy(x+y)}\\\\=\frac{x^2y+y^2x-1}{xy(x+y)}

19. \frac{b-c}{a^2-(b-c)^2}+\frac{c-a}{b^2-(c-a)^2}+\frac{a-b}{c^2-(a-b)^2} 
 
সমাধানঃ
\frac{b-c}{a^2-(b-c)^2}+\frac{c-a}{b^2-(c-a)^2}+\frac{a-b}{c^2-(a-b)^2}\\\\=\frac{b-c}{(a+b-c)(a-b+c)}+\frac{c-a}{(b+c-a)(b-c+a)}+\frac{a-b}{(c+a-b)(c-a+b)}\\\\=\frac{(b-c)(b+c-a)+(c-a)(c+a-b)+(a-b)(a+b-c)}{(b+c-a)(c+a-b)(a+b-c)}\\\\=\frac{b^2+bc-ab-bc-c^2+ac+c^2+ac-bc-ac-a^2+ab+a^2+ab-ac-ab-b^2+bc}{(b+c-a)(c+a-b)(a+b-c)}\\\\=\frac{0}{(b+c-a)(c+a-b)(a+b-c)}=0

20. \frac{a+b}{ab}\left ( a+b-c \right )+\frac{b+c}{bc}\left ( b+c-a \right )+\frac{c+a}{ca}\left ( c+a-b \right ) 

সমাধানঃ

\frac{a+b}{ab}\left ( a+b-c \right )+\frac{b+c}{bc}\left ( b+c-a \right )+\frac{c+a}{ca}\left ( c+a-b \right )\\\\=\frac{c(a+b)(a+b-c )+a(b+c)(b+c-a )+b(c+a)(c+a-b)}{abc}\\\\=\frac{c(a^2+2ab+b^2-ac-bc)+a(b^2+2bc+c^2-ab-ac)+b(c^2+2ac+a^2-bc-ab)}{abc}\\\\=\frac{a^2c+2abc+b^2c-ac^2-bc^2+b^2a+2bca+c^2a-a^2b-a^2c+c^2b+2abc+a^2b-b^2c-ab^2}{abc}\\\\=\frac{6abc}{abc}=6

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