WBBSE Class 7 Mathematics Solutions - নিজে করি 12.1

সমাধান :

(i) $(x+3)^2 = x^2+2 \cdot x \cdot 3 + 3^2 = x^2+6x+9$

এখানে, $a = x$, $b = 3$

(ii) $(p+9)^2 = p^2+2 \cdot p \cdot 9 + 9^2 = p^2+18p+81$

এখানে, $a = p$, $b = 9$

(iii) $(6-x)^2 = (6)^2-2 \cdot 6 \cdot x + x^2 = 36-12x+x^2$

এখানে, $a = 6$, $b = x$ এবং সূত্রটি হলো $(a-b)^2 = a^2-2ab+b^2$

(iv) $(y-2)^2 = y^2-2 \cdot y \cdot 2 + 2^2 = y^2-4y+4$

এখানে, $a = y$, $b = 2$ এবং সূত্রটি হলো $(a-b)^2 = a^2-2ab+b^2$

(v) $(mn+l)^2 = (mn)^2+2 \cdot mn \cdot l + l^2 = m^2n^2+2mnl+l^2$

এখানে, $a = mn$, $b = l$

(vi) $(6x+3)^2 = (6x)^2+2 \cdot 6x \cdot 3 + 3^2 = 36x^2+36x+9$

এখানে, $a = 6x$, $b = 3$

(vii) $(4x+5y)^2 = (4x)^2+2 \cdot 4x \cdot 5y + (5y)^2 = 16x^2+40xy+25y^2$

এখানে, $a = 4x$, $b = 5y$

(viii) $(pqc+2)^2 = (pqc)^2+2 \cdot pqc \cdot 2 + 2^2 = p^2q^2c^2+4pqc+4$

এখানে, $a = pqc$, $b = 2$

(ix) $(\frac{5}{k}+3)^2 = (\frac{5}{k})^2+2 \cdot \frac{5}{k} \cdot 3 + 3^2 = \frac{25}{k^2}+\frac{30}{k}+9$

এখানে, $a = \frac{5}{k}$, $b = 3$

(x) $(x+\frac{3}{r})^2 = x^2+2 \cdot x \cdot \frac{3}{r} + (\frac{3}{r})^2 = x^2+\frac{6x}{r}+\frac{9}{r^2}$

এখানে, $a = x$, $b = \frac{3}{r}$

(xi) $(\frac{p}{q}+\frac{m}{n})^2 = (\frac{p}{q})^2+2 \cdot \frac{p}{q} \cdot \frac{m}{n} + (\frac{m}{n})^2 = \frac{p^2}{q^2}+\frac{2pm}{qn}+\frac{m^2}{n^2}$

এখানে, $a = \frac{p}{q}$, $b = \frac{m}{n}$

(xii) $(m^2+n^2)^2 = (m^2)^2+2 \cdot m^2 \cdot n^2 + (n^2)^2 = m^4+2m^2n^2+n^4$

এখানে, $a = m^2$, $b = n^2$

(xiii) $(3xy+4z)^2 = (3xy)^2+2 \cdot 3xy \cdot 4z + (4z)^2 = 9x^2y^2+24xyz+16z^2$

এখানে, $a = 3xy$, $b = 4z$

(xiv) $(2x+3y+z)^2 = ((2x+3y)+z)^2 = (2x+3y)^2+2(2x+3y)z+z^2 = (4x^2+12xy+9y^2)+4xz+6yz+z^2 = 4x^2+9y^2+z^2+12xy+4xz+6yz$

এখানে, $a = (2x+3y)$, $b = z$

(xv) $102^2 = (100+2)^2 = 100^2+2 \cdot 100 \cdot 2 + 2^2 = 10000+400+4 = 10404$

এখানে, $a = 100$, $b = 2$

(xvi) $(p+q+r+s)^2 = ((p+q)+(r+s))^2 = (p+q)^2+2(p+q)(r+s)+(r+s)^2 = (p^2+2pq+q^2)+2(pr+ps+qr+qs)+(r^2+2rs+s^2) = p^2+q^2+r^2+s^2+2pq+2pr+2ps+2qr+2qs+2rs$

এখানে, $a = (p+q)$, $b = (r+s)$