2021 WB Math Suggestion | একচলবিশিষ্ট দ্বিঘাত সমীকরণ

WB Mathematics Suggestion with Answers | Class 10th Math Suggestion With Solution | একচলবিশিষ্ট দ্বিঘাত সমীকরণ | কষে দেখি – 1.2

(x) $\frac{x}{x+1}+\frac{x+1}{x}=2\frac{1}{12}$

$\Rightarrow \frac{x^{2}+\left ( x+1 \right )^{2}}{x\left ( x+1 \right )}=\frac{25}{12}$

$\Rightarrow \frac{x^{2}+x^{2}+x+x+1}{x^{2}+x}=\frac{25}{12}$
$\Rightarrow 25x^{2}+25x=24x^{2}+24x+12$
$\Rightarrow 25x^{2}-24x^{2}+25x-24x-12=0$
$\Rightarrow x^{2}+x-12=0$
$\Rightarrow x^{2}+4x-3x-12=0$
$\Rightarrow x\left ( x+4 \right )-3\left ( x+4 \right )=0$
$\Rightarrow \left ( x+4 \right )\left ( x-3 \right )=0$

∴ হয় $x+4=0$

$\therefore x=-4$

অথবা,
$x-3=0$
$\therefore x=3$

নির্ণেয় দ্বিঘাত সমীকরণের সমাধান, x= -4, 3 Ans.

(xiii) $\frac{x+1}{2}+\frac{2}{x+1}=\frac{x+1}{3}+\frac{3}{x+1}-\frac{5}{6},\quad x\ne -1$
$\Rightarrow \frac{x+1}{2}-\frac{x+1}{3}+\frac{5}{6}=\frac{3}{x+1}-\frac{2}{x+1}$
$\Rightarrow \frac{3\left ( x+1 \right )-2\left ( x+1 \right )+5}{6}=\frac{3-2}{x+1}$
$\Rightarrow \frac{3x+3-2x-2+5}{6}=\frac{1}{x+1}$
$\Rightarrow \frac{x+6}{6}=\frac{1}{x+1}$
$\Rightarrow \left ( x+6 \right )\left ( x+1 \right )=6$
$\Rightarrow x^{2}+x+6x+6-6=0$
$\Rightarrow x^{2}+7x=0$
$\Rightarrow x\left ( x+7 \right )=0$

$\therefore x=0$
অথবা,
$x+7=0$
$\therefore x=-7$

∴ নির্ণেয় দ্বিঘাত সমীকরণের সমাধান, x= 0, -7 Ans.

(xvi) $\frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x},\quad x\ne 0,-\left( a+b \right)$
$\Rightarrow \frac{1}{a+b+x}-\frac{1}{x}=\frac{1}{a}+\frac{1}{b}$
$\Rightarrow \frac{x-a-b-x}{x\left ( a+b+x \right )}=\frac{b+a}{ab}$
$\Rightarrow \frac{-\left ( a+b \right )}{ax+bx+x^{2}}=\frac{a+b}{ab}$
$\Rightarrow \frac{-1}{ax+bx+x^{2}}=\frac{1}{ab}$
$\Rightarrow ax+bx+x^{2}=-ab$
$\Rightarrow x^{2}+ax+bx+ab=0$
$\Rightarrow x\left ( x+a \right )+b\left ( x+a \right )=0$
$\Rightarrow \left ( x+a \right )\left ( x+b \right )=0$

হয় $x+a=0$
$\therefore x=-a$
অথবা,
$x+b=0$
$\therefore x=-b$

∴ নির্ণেয় দ্বিঘাত সমীকরণের সমাধান, x= -a, -b Ans.

(xviii) $\frac{1}{x}-\frac{1}{x+b}=\frac{1}{a}-\frac{1}{a+b},\quad x\ne 0,-b$

$\Rightarrow \frac{x+b-x}{x\left ( x+b \right )}=\frac{a+b-a}{a\left ( a+b \right )}$
$\Rightarrow \frac{b}{x\left ( x+b \right )}=\frac{b}{a\left ( a+b \right )}$
$\Rightarrow \frac{1}{\left ( x^{2}+bx \right )}=\frac{1}{\left ( a^{2}+ab \right )}$
$\Rightarrow x^{2}+bx=a^{2}+ab$
$\Rightarrow x^{2}-a^{2}+bx-ab=0$
$\Rightarrow \left ( x+a \right )\left ( x-a \right )+b\left ( x-a \right )=0$
$\Rightarrow \left ( x-a \right )\left ( x+a+b \right )=0$
হয় $x-a=0$
$\therefore x=a$
অথবা,
$x+a+b=0$
$\therefore x=-a-b=-\left ( a+b \right )$
∴ নির্ণেয় দ্বিঘাত সমীকরণের সমাধান x= a, -a(a+b) Ans.

(xix) $\frac{1}{\left( x-1 \right)\left( x-2 \right)}+\frac{1}{\left( x-2 \right)\left( x-3 \right)}+\frac{1}{\left( x-3 \right)\left( x-4 \right)}=\frac{1}{6}$
$\Rightarrow \frac{1}{x-2}-\frac{1}{x-1}+\frac{1}{x-3}-\frac{1}{x-2}+\frac{1}{x-4}-\frac{1}{x-3}=\frac{1}{6}$
$\Rightarrow \frac{1}{x-4}-\frac{1}{x-1}=\frac{1}{6}$
$\Rightarrow \frac{x-1-x-4}{\left ( x-4 \right )\left ( x-1 \right )}=\frac{1}{6}$
$\Rightarrow \frac{3}{x^{2}-x-4x+4}=\frac{1}{6}$
$\Rightarrow \frac{3}{x^{2}-5x+4}=\frac{1}{6}$
$\Rightarrow x^{2}-5x+4=18$
$\Rightarrow x^{2}-5x+4-18=0$
$\Rightarrow x^{2}-5x-14=0$
$\Rightarrow x^{2}-7x+2x-14=0$
$\Rightarrow x\left ( x-7 \right )+2\left ( x-7 \right )=0$
$\Rightarrow \left ( x-7 \right )\left ( x+2 \right )=0$
হয় $x-7=0$
$\therefore x=7$
অথবা,
$x+2=0$
$\therefore x=-2$
∴ নির্ণেয় দ্বিঘাত সমীকরণের সমাধান x= 7,-2 Ans.

(xx) $\frac{a}{x-a}+\frac{b}{x-b}=\frac{2c}{x-c},\quad x\ne a,b,c$

$\Rightarrow \frac{a}{x-a}-\frac{b}{x-b}=\frac{c}{x-c}+\frac{c}{x-c}$
$\Rightarrow \frac{a}{x-a}-\frac{c}{x-c}=\frac{c}{x-c}-\frac{b}{x-b}$
$\Rightarrow \frac{a\left ( x-c \right )-c\left ( x-a \right )}{\left ( x-a \right )\left ( x-c \right )}=\frac{c\left ( x-b \right )-b\left ( x-c \right )}{\left ( x-c \right )\left ( x-b \right )}$
$\Rightarrow \frac{ax-ac-cx+ac}{\left ( x-a \right )\left ( x-c \right )}=\frac{cx-bc-bx+bc}{\left ( x-c \right )\left ( x-b \right )}$
$\Rightarrow \frac{ax-cx}{x-a}=\frac{cx-bx}{x-b}$
$\Rightarrow \left ( x-a \right )\left ( cx-bx \right )=\left ( ax-cx \right )\left ( x-b \right )$
$\Rightarrow cx^{2}-bx^{2}-acx+abx=ax^{2}-abx-cx^{2}-bcx$
$\Rightarrow cx^{2}+cx^{2}-bx^{2}-ax^{2}-acx+abx+abx-bcx=0$
$\Rightarrow 2cx^{2}-bx^{2}-ax^{2}-acx+2abx-bcx=0$
$\Rightarrow x\left ( 2cx-bx-ax-ac+2ab-bc \right )=0$