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# 1.Use a suitable identity to get each of the following products: (i) $$(x + 3) (x + 3)$$ (ii) $$(2y + 5) (2y + 5)$$ (iii) $$(2a – 7) (2a – 7)$$ (iv)$$(3a – \frac { 1 }{ 2 }) (3a – \frac { 1 }{ 2 })$$ (v)$$(1.1m – 0.4) (1.1m + 0.4)$$ (vi)$$(a2 + b2) (-a2 + b2)$$ (vii) $$(6x – 7) (6x + 7)$$ (viii) $$(-a + c) (-a + c)$$ (ix)$$(\frac { x }{ 2 } + \frac { 3y }{ 4 }) (\frac { x }{ 2 } + \frac { 3y }{ 4 })$$ (x) $$(7a – 9b) (7a – 9b)$$

(i)Given: $$(x + 3) (x + 3)$$
$$=(x+3)^2$$

$$=(x)^2+2\times x \times 3+(3)^2\quad$$[∵$$(a+b)^2=a^2+2ab+b^2$$]

$$=x^2+6x+9$$

(ii)Given: $$(2y + 5) (2y + 5)$$
$$=(2y + 5)^2$$

$$=(2y)^2+2\times 2y \times 5+(5)^2\quad$$[∵$$(a+b)^2=a^2+2ab+b^2$$]

$$=4y^2+20y+25$$

(iii)Given: $$(2a - 7) (2a - 7)$$
$$=(2a - 7)^2$$

$$=(2a)^2-2\times (2a) \times 7+(7)^2\quad$$[∵$$(a-b)^2=a^2-2ab+b^2$$]

$$=4a^2-28a+49$$

(iv)Given: $$(3a - \frac{1}2) (3a - \frac{1}2)$$
$$=(3a - \frac{1}2)^2$$

$$=(3a)^2-2\times (3a) \times (\frac{1}2)+(\frac{1}2)^2\quad$$[∵$$(a-b)^2=a^2-2ab+b^2$$]

$$=9a^2-3a+\frac{1}4$$

(v)Given: $$(1.1m - 0.4) (1.1m + 0.4)$$
$$=((1.1m)^2 - (0.4)^2\quad$$[∵$$(a-b)(a+b)=a^2-b^2$$]

$$=1.21a^2-0.16$$

(vi)Given: $$(a^2 + b^2) (-a^2 + b^2)$$
$$=(b^2 + a^2)(b^2 -a^2)$$

$$=(b^2)^2-(a^2)^2\quad$$[∵$$(a-b)(a+b)=a^2-b^2$$]

$$=(b^4 - a^4)$$

(vii)Given: $$(6x - 7) (6x + 7)$$
$$=(6x)^2 - (7)^2\quad$$[∵$$(a-b)(a+b)=a^2-b^2$$]

$$=36x^2-49$$

(viii)Given: $$(-a+c) (-a+c)$$
$$=[(-a)+c]^2$$

$$=(-a)^2+2\times (-a) \times c+(c)^2\quad$$[∵$$(a+b)^2=a^2+2ab+b^2$$]

$$=a^2-2ac+c^2$$

(ix)Given:$$(\frac { x }{ 2 } + \frac { 3y }{ 4 }) (\frac { x }{ 2 } + \frac { 3y }{ 4 })$$

$$=(\frac { x }{ 2 } + \frac { 3y }{ 4 })^2$$

$$=(\frac { x }{ 2 })^2+2\times (\frac { x }{ 2 }) \times (\frac { 3y }{ 4 })+(\frac { 3y }{ 4 })^2\quad$$[∵$$(a+b)^2=a^2+2ab+b^2$$]

$$=\frac { x^2 }{ 4 }+\frac { 3 }{ 4 }xy+\frac { 9y^2 }{ 16 }$$

(x)Given: $$(7a - 9b) (7a - 9b)$$
$$=(7a - 9b)^2$$

$$=(7a)^2-2\times (7a) \times (9b)+(9b)^2\quad$$[∵$$(a-b)^2=a^2-2ab+b^2$$]

$$=49a^2-126ab+81b^2$$