Question

pls do all the sums ​

Answer 1

» Question :

Simplify :

• $\sf{(2x - 1)^{2} - (x - 1)^{2}}$

• $\sf{(6x + 3)(6x - 3)}$

• $\sf{(7 + 5x)(7 + 5x) - (2x - 3)(2x - 3)}$

• $\sf{(2m - 3n)(2m + 3n) + (2m + 3n)^{2}}$

» We Know :

• $\sf{(a + b)^{2} = a^{2} + 2ab + b^{2}}$

• $\sf{(a - b)^{2} = a^{2} - 2ab + b^{2}}$

• $\sf{a^{2} - b^{2} = (a + b)(a - b)}$

» Solution :

$\large{\sf{(2x - 1)^{2} - (x - 1)^{2}}}$

Using the identity ,

$\sf{(a - b)^{2} = a^{2} - 2ab + b^{2}}$ ,

we get :

$\sf{\Rightarrow (2x)^{2} - 2 \times 2x \times 1 + 1^{2} - (x^{2} - 2 \times x \times 1 + 1^{2})}$

$\sf{\Rightarrow 4x^{2} - 4x + 1 - x^{2} + 2x - 1}$

$\sf{\Rightarrow 3x^{2} - 2x}$

Hence , $\sf{(2x - 1)^{2} - (x - 1)^{2} = 3x^{2} - 2x}$

$\large{\sf{(6x + 3)(6x - 3)}}$

$\sf{\Rightarrow (6x + 3)(6x - 3)}$

Using the identity ,

$\sf{a^{2} - b^{2} = (a + b)(a - b)}$ ,

we get :

$\sf{\Rightarrow (6x)^{2} - 3^{2}}$

$\sf{\Rightarrow 36x^{2} - 9}$

Hence ,$\sf{(6x + 3)(6x - 3) = 36x^{2} - 9}$

$\large{\sf{(7 + 5x)(7 + 5x) - (2x - 3)(2x - 3)}}$

$\sf{\Rightarrow (7 + 5x)(7 + 5x) - (2x - 3)(2x - 3)}$

Using the identity ,

• $\sf{(a + b)^{2} = a^{2} + 2ab + b^{2}}$
• $\sf{(a - b)^{2} = a^{2} - 2ab + b^{2}}$.

We Get :

$\sf{\Rightarrow (7 + 5x)^{2} - (2x - 3)^{2}}$

$\sf{\Rightarrow 49 + 70x + 25x^{2} - [4x^{2} - 12x + 9]}$

$\sf{\Rightarrow 49 + 70x + 25x^{2} - 4x^{2} + 12x - 9}$

$\sf{\Rightarrow 40 + 82x + 21x^{2}}$

Hence , $\sf{(7 + 5x)(7 + 5x) - (2x - 3)(2x - 3) = 40 + 82x + 21x^{2}}$

$\large{\sf{(2m - 3n)(2m + 3n) + (2m + 3n)^{2}}}$

Using the identity ,

• $\sf{a^{2} - b^{2} = (a + b)(a - b)}$
• $\sf{(a + b)^{2} = a^{2} + 2ab + b^{2}}$ ,

we get :

$\sf{\Rightarrow (2m)^{2} - (3n)^{2} + (2m)^{2} + 2 \times 2m \times 3n + (3n)^{2}}$

$\sf{\Rightarrow 4m^{2} - 9n^{2} + 4m^{2} + 12mn + 9n^{2}}$

$\sf{\Rightarrow 4m^{2} - \cancel{9n^{2}} + 4m^{2} + 12mn + \cancel{9n^{2}}}$

$\sf{\Rightarrow 8m^{2} + 12mn}$

Hence ,$\sf{(2m - 3n)(2m + 3n) + (2m + 3n)^{2} = 8m^{2} + 12mn}$

» Additional information :

• $\sf{a^{2} + b^{2} = (a + b)^{2} - 2ab}$

• $\sf{\bigg(x + \dfrac{1}{x}\bigg)^{2} = x^{2} + \dfrac{1}{x^{2}} + 2}$

• $\sf{\bigg(x - \dfrac{1}{x}\bigg)^{2} = x^{2} + \dfrac{1}{x^{2}} - 2}$

Answer 2

$(2x−1)2−(x−1)2</p><p></p><p>\sf{(6x + 3)(6x - 3)}(6x+3)(6x−3)</p><p></p><p>\sf{(7 + 5x)(7 + 5x) - (2x - 3)(2x - 3)}(7+5x)(7+5x)−(2x−3)(2x−3)</p><p></p><p>\sf{(2m - 3n)(2m + 3n) + (2m + 3n)^{2}}(2m−3n)(2m+3n)+(2m+3n)2</p><p></p><p>We know</p><p> </p><p> </p><p>[tex]a+b)2=a2+2ab+b2</p><p></p><p>\sf{(a - b)^{2} = a^{2} - 2ab + b^{2}}(a−b)2=a2−2ab+b2</p><p></p><p>\sf{a^{2} - b^{2} = (a + b)(a - b)}a2−b2=(a+b)(a−b)</p><p></p><p>$

$2x−1)2−(x−1)2</p><p></p><p>Using the identity ,</p><p></p><p>\sf{(a - b)^{2} = a^{2} - 2ab + b^{2}}</p><p></p><p>$

$(2x)2−2×2x×1+12−(x2−2×x×1+12)</p><p></p><p>\sf{\Rightarrow 4x^{2} - 4x + 1 - x^{2} + 2x - 1}⇒4x2−4x+1−x2+2x−1</p><p></p><p>\sf{\Rightarrow 3x^{2} - 2x}⇒3x2−2x</p><p></p><p>Hence , \sf{(2x - 1)^{2} - (x - 1)^{2} = 3x^{2} - 2x}(2x−1)2−(x−1)2=3x2−2x</p><p></p><p>\\\\end</p><p></p><p>\large{\sf{(6x + 3)(6x - 3)}}</p><p></p><p>\sf{\Rightarrow (6x + 3)(6x - 3)}</p><p></p><p>Using the identity ,</p><p></p><p>\sf{a^{2} - b^{2} = (a + b)(a - b)} ,</p><p></p><p>we get :</p><p></p><p>\sf{\Rightarrow (6x)^{2} - 3^{2}}</p><p></p><p>\sf{\Rightarrow 36x^{2} - 9}</p><p></p><p>Hence ,\sf{(6x + 3)(6x - 3) = 36x^{2} - 9}</p><p></p><p>\\\\end</p><p></p><p>\large{\sf{(7 + 5x)(7 + 5x) - (2x - 3)(2x - 3)}}</p><p></p><p>\sf{\Rightarrow (7 + 5x)(7 + 5x) - (2x - 3)(2x - 3)}</p><p></p><p>Using the identity ,</p><p></p><p>\sf{(a + b)^{2} = a^{2} + 2ab + b^{2}}</p><p></p><p>\sf{(a - b)^{2} = a^{2} - 2ab + b^{2}} .</p><p></p><p>We Get :</p><p></p><p>\sf{\Rightarrow (7 + 5x)^{2} - (2x - 3)^{2}}</p><p></p><p>\sf{\Rightarrow 49 + 70x + 25x^{2} - [4x^{2} - 12x + 9]}</p><p></p><p>\sf{\Rightarrow 49 + 70x + 25x^{2} - 4x^{2} + 12x - 9}</p><p></p><p>\sf{\Rightarrow 40 + 82x + 21x^{2}}</p><p></p><p>Hence , \sf{(7 + 5x)(7 + 5x) - (2x - 3)(2x - 3) = 40 + 82x + 21x^{2}}</p><p>\\\end</p><p></p><p>\large{\sf{(2m - 3n)(2m + 3n) + (2m + 3n)^{2}}}</p><p></p><p>Using the identity ,</p><p></p><p>\sf{a^{2} - b^{2} = (a + b)(a - b)}</p><p></p><p>\sf{(a + b)^{2} = a^{2} + 2ab + b^{2}} ,</p><p></p><p></p><p>we get :</p><p></p><p></p><p>\sf{\Rightarrow (2m)^{2} - (3n)^{2} + (2m)^{2} + 2 \times 2m \times 3n + (3n)^{2}}</p><p></p><p></p><p>\sf{\Rightarrow 4m^{2} - 9n^{2} + 4m^{2} + 12mn + 9n^{2}}</p><p></p><p></p><p>\sf{\Rightarrow 4m^{2} - \cancel{9n^{2}} + 4m^{2} + 12mn + \cancel{9n^{2}}}</p><p></p><p></p><p>\sf{\Rightarrow 8m^{2} + 12mn}</p><p></p><p></p><p>Hence ,\sf{(2m - 3n)(2m + 3n) + (2m + 3n)^{2} = 8m^{2} + 12mn}</p><p></p><p></p><p>» Additional information :</p><p></p><p></p><p>\sf{a^{2} + b^{2} = (a + b)^{2} - 2ab}</p><p></p><p></p><p>\sf{\bigg(x + \dfrac{1}{x}\bigg)^{2} = x^{2} + \dfrac{1}{x^{2}} + 2}</p><p></p><p></p><p>\sf{\bigg(x - \dfrac{1}{x}\bigg)^{2} = x^{2} + \dfrac{1}{x^{2}} - 2}</p><p></p><p>$