Question

Can u guys help me With this!9th Grade Chapter-Numbers​

Answer 1

$1)(\sqrt{3}-\sqrt{2})^2\\ Identity:(a-b)^2=a^2+b^2-2ab\\=(\sqrt{3})^2+(\sqrt{2})^2-2*\sqrt{3}*\sqrt{2}\\ =3+2-2\sqrt{6}\\ =5-2\sqrt{6}\\ 2)(\sqrt{7} +\sqrt{11})^2\\ Identity:(a+b)^2=a^2+b^2+2ab\\=(\sqrt{7}^2)+(\sqrt{11})^2 +2* \sqrt{7}*\sqrt{11} \\=7+11+2\sqrt{77}\\ =18+2\sqrt{77}\\ 3)(\sqrt{7}+5)(\sqrt{7}+ 3)\\ =7+3\sqrt{7}+5\sqrt{7}+15\\ =22+8\sqrt{7}\\ 4)(\sqrt{3} -1)(\sqrt{3} +1)\\Identity:(a+b)(a-b)=a^2-b^2\\=(\sqrt{3})^2-(1)^2\\ =3-1\\=2$

Hope it helps......   :D

Answer 2

Solution :

$1) \sf \: \: ( \sqrt{3} - \sqrt{2} ) ^{2}$

Using identity :

$\bigstar\: \boxed{ \sf \: (a - b) ^{2} = {a}^{2} - 2ab + {b}^{2} }$

$\longrightarrow \sf \: (\sqrt{3} )^{2} - 2( \sqrt{3} ) (\sqrt{2} ) + ( \sqrt{2} )^{2}$

$\longrightarrow \sf \: 3 - 2 \sqrt{3 \times 2} + 2$

$\longrightarrow \sf \: 3 - 2 \sqrt{6} + 2$

$\longrightarrow \sf \red{5- 2 \sqrt{6} }$

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$2) \sf \: \: ( \sqrt{7} + \sqrt{11} ) ^{2}$

Using identity :

$\bigstar\: \boxed{ \sf \: (a + b) ^{2} = {a}^{2} + 2ab + {b}^{2} }$

$\longrightarrow \sf \: (\sqrt{7} ) ^{2} + 2( \sqrt{7} )( \sqrt{11} ) + (\sqrt{11} ) ^{2}$

$\longrightarrow \sf \: 7 + 2 \sqrt{7 \times 11} + 11$

$\longrightarrow \sf \red{18 + 2 \sqrt{77} }$

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$3) \sf \: \: ( \sqrt{7} + 5)( \sqrt{7} + 3)$

Using identity :

$\bigstar\: \boxed{ \sf \: (x+ a)(x + b) = {x}^{2} + ( a + b)x+ ab}$

$\longrightarrow \sf \:( \sqrt{7} ) ^{2} + (5 + 3 ) \sqrt{7} + ( 5 \times 3 )$

$\longrightarrow \sf \:7 + 8( \sqrt{7} ) + 15$

$\longrightarrow \sf \red{22 + 8 \sqrt{7} }$

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$4) \sf \: \: ( \sqrt{3} - 1)( \sqrt{3} + 1)$

Using identity :

$\bigstar\: \boxed{ \sf \: (a - b)(a +b) = {a}^{2} - {b}^{2} }$

$\longrightarrow \sf (\sqrt{3} ) ^{2} - (\sqrt{1} )^{2}$

$\longrightarrow \sf ( \sqrt{3} \times \sqrt{3} ) - ( \sqrt{1} \times \sqrt{1} )$

$\longrightarrow \sf 3 - 1$

$\longrightarrow \sf \red{2 }$