\(\frac { 1 }{ \sqrt { 3 } +\sqrt { 2 } } -\frac { 2 }{ \sqrt { 5 } -\sqrt { 3 } } -\frac { 3 }{ \sqrt { 2 } +\sqrt { 5 } } \)
\(=\frac { 1 }{ \sqrt { 3 } +\sqrt { 2 } } \times \frac { \sqrt { 3 } -\sqrt { 2 } }{ \sqrt { 3 } -\sqrt { 2 } } -\frac { 2 }{ \sqrt { 5 } -\sqrt { 3 } } \times \frac { \sqrt { 5 } +\sqrt { 3 } }{ \sqrt { 5 } +\sqrt { 3 } } -\frac { 3 }{ \sqrt { 2 } +\sqrt { 5 } } \times \frac { \sqrt { 2 } +\sqrt { 5 } }{ \sqrt { 2 } +\sqrt { 5 } } \)
\(=\frac { \sqrt { 3 } -\sqrt { 2 } }{ 1 } -\frac { 2\left( \sqrt { 5 } +\sqrt { 3 } \right) }{ 2 } -\frac { 3\left( \sqrt { 2 } +\sqrt { 5 } \right) }{ -3 } \)
\(=\sqrt { 3 } -\sqrt { 2 } -\sqrt { 5 } -\sqrt { 3 } +\sqrt { 2 } +\sqrt { 5 } \)
= 0

Let larger number be x, then 5 times of larger number=5x and smaller number be y
Quotient=2 and remainder=9
So, according to the question,
5x=2y+9
5x-2y-9=0

f(x) = x³-3x²+3x-4
f(2)=(2)³-3(2)²+3(2)-4
= 8 - 12 + 6 - 4
f(2) = -2
f(-2)=(-2)³-3(-2)²+3(-2)-4
= -8-12-6-4
f(-2)=-30
f(0) = -4
\(\therefore\) f(2)+f(-2)+f(0)=-2-30-4=-36

\(y={5x+20\over4};\ (-4,0);\ (0,5)\)

\(\pi x+y=9\)
\(y=9-\pi x\)
Put x=0, we get y=9-\(\pi\)(0)=9-0=9
Put x=1, we get y=9-\(\pi\)(1)=9-\(\pi\)
Put x=-1, we get y=9-\(\pi\)(-1)=9+\(\pi\)
Put \(x={9\over \pi}\), we get \(y=9-\pi\left(9\over\pi\right)=9-9=0\)
Four solution are (o, 9), (1, 9-\(\pi\)), (-1, 9+\(\pi\)) and \(\left({9\over \pi},0\right)\)

(a) (5,4)
(b) (-5,4)
(c) (-5,-4)
(d) (5,-4)

There can be infinitely many rational numbers between 3 and 4.
\(\frac { 3+4 }{ 2 } =\frac { 7 }{ 2 } \)
\(\\ \frac { 3+\frac { 7 }{ 2 } }{ 2 } =\frac { 13 }{ 4 } \)
\(\\ \frac { 3+\frac { 13 }{ 4 } }{ 2 } =\frac { 25 }{ 8 }\)
\( \\ \frac { 3+\frac { 25 }{ 8 } }{ 2 } =\frac { 49 }{ 16 } =\frac { 3+\frac { 49 }{ 16 } }{ 2 } =\frac { 97 }{ 32 } =\frac { 3+\frac { 97 }{ 32 } }{ 2 } =\frac { 193 }{ 64 } \)
Thus, six rational numbers between 3 and 4
\(\frac { 193 }{ 64 } ,\frac { 97 }{ 32 } ,\frac { 49 }{ 16 } ,\frac { 25 }{ 8 } ,\frac { 13 }{ 4 } \)and \(\frac { 7 }{ 2 } \)
Aliter
\(3=\frac { 3 }{ 1 } =\frac { 3\times 7 }{ 1\times 7 } =\frac { 21 }{ 7 } \)
\(\\ 4=\frac { 4 }{ 1 } =\frac { 4\times 7 }{ 1\times 7 } =\frac { 28 }{ 7 } \)
6 + 1 = 7
the six rational numbers between 3 and 4 can be taken as 
\(\frac { 22 }{ 7 } ,\frac { 23 }{ 7 } ,\frac { 24 }{ 7 } ,\frac { 25 }{ 7 } ,\frac { 26 }{ 7 } \) and \(\frac { 27 }{ 7 } \)

(a+b+c)²=a²+b²+c²+2(ab+bc+ca)
(6)²=a²+b²+c²+2\(\times\)11
a²+b²+c²=36-22=14
a³+b³+c³-3abc=(a+b+c)[a²+b²+c²-(ab+bc+ca)]
= 6 \(\times\) (14-11)=6\(\times\)3=18.

LCM of 3,4,6 and 12 is 12
\(\sqrt[12]{2}\)=\({ 2 }^{ \frac { 1 }{ 3 } }\)=\({ 2 }^{ \frac { 4 }{ 12 } }\) \(\sqrt[12]{16}\)
\(\sqrt [ 4 ]{ 5 } ={ 5 }^{ \frac { 1 }{ 4 } }={ 5 }^{ \frac { 3 }{ 12 } }\quad \sqrt [ 12 ]{ 125 } \)
\(\sqrt [ 6 ]{ 7 } ={ 7 }^{ \frac { 1 }{ 6 } }\quad { 7 }^{ \frac { 2 }{ 12 } }\quad \sqrt [ 12 ]{ 49 } \)
\(\sqrt [ 12 ]{ 3 } ={ 3 }^{ \frac { 1 }{ 12 } }=\sqrt [ 12 ]{ 3 } \)
Descending order is
\(\sqrt [ 12 ]{ 125 } ,\sqrt [ 12 ]{ 49 } ,\sqrt [ 12 ]{ 16 } ,\sqrt [ 12 ]{ 3 } \)
i.e., \(\sqrt [ 4 ]{ 5 } ,\sqrt [ 6 ]{ 7 } ,\sqrt [ 3 ]{ 2 } ,\sqrt [ 12 ]{ 3 } \)

\((ax+by)^2+(ay-bx)^2\)
\(=a^2x^2+b^2y^2+2abxy+a^2y^2+b^2x^2-2abxy\)
\(=a^2(x^2+y^2)+b^2(x^2+y^2)\)
\(=(x^2+y^2)(a^2+b^2)\)

Divisor = x + 3
\(x+3=0\ \Rightarrow x=-3\)
\(\therefore\)Remainder= p(-3) | By remainder theorem
\(= k{ (-3) }^{ 3 }+9{ (-3) }^{ 2 }+4(-3)-8\) 
\(=-27k+81-12-8\) 
\(=-27k+61\)
According to the question,
\(-27k+61=10(1-k)\) 
\(\Rightarrow -27k+61=10-10k\) 
\(\Rightarrow 17k=51\) 
\(\Rightarrow k=3\) 

\(\frac { \sqrt { 147 } }{ \sqrt { 75 } } =\frac { \sqrt { 3\times 7\times 7 } }{ \sqrt { 3\times 5\times 5 } } =\frac { 7\sqrt { 3 } }{ 5\sqrt { 3 } } =\frac { 7 }{ 5 } \)
which is clearly a rational number.
Hence, the given statement is false.The reason is that 'if we divide two irrationals, the result may be rational or irrational'.

\(\frac { 1 }{ 5 } =\frac { 1\times 6 }{ 5\times 6 } =\frac { 6 }{ 30 } =\frac { 6\times 10 }{ 30\times 10 } =\frac { 60 }{ 300 } \) 
\(\\ \frac { 1 }{ 6 } =\frac { 1\times 5 }{ 6\times 5 } =\frac { 5 }{ 30 } =\frac { 5\times 30 }{ 30\times 10 } =\frac { 50 }{ 300 }\)
\( \\ \because 50<51<52<53<54<60\)
\(\\ \therefore \frac { 50 }{ 300 } <\frac { 51 }{ 300 } <\frac { 52 }{ 300 } <\frac { 53 }{ 300 } <\frac { 54 }{ 300 } <\frac { 60 }{ 300 } \)
Hence, four rational numbers between \(\frac { 1 }{ 5 } \) and \(\frac { 1 }{ 6 } \) can be taken as
\(\frac { 51 }{ 300 } ,\frac { 52 }{ 300 } ,\frac { 53 }{ 300 } \)and \(\frac { 54 }{ 300 } \)
or \(\frac { 17 }{ 100 } ,\frac { 13 }{ 75 } ,\frac { 53 }{ 300 } \)and \(\frac { 9 }{ 50 } \)

Let, x²+7=p and 2x-1=q, then given expression
=12p²-8pq-15q²
=12p²-18pq+10pq-15q²
=6p(2p-3q)+5q(2p-3q)
=(2p-3q)(6p+5q)
=[2(x²+7)-3(2x-1][6(x²+7)+5(2x-1)]
=(2x²+14-6x+3)(6x²+42+10x-5)
=(2x²-6x+17)(6x²+10x+37).

(4√3+3√2)x(4√3-3√2)=(4√3)²-(3√2)²
= 48-18
= 30

x=4

3x-5y-4=0; No; a=3, b=-5, c=-4; Yes

11130

331/770, 166/385, 333/770

\(\sqrt { \frac { \left( \sqrt { 2 } -1 \right) }{ \left( \sqrt { 2 } +1 \right) } } \)is an irrational number.
\(\sqrt { \frac { \left( \sqrt { 2 } -1 \right) }{ \left( \sqrt { 2 } +1 \right) } } =\sqrt { \frac { \left( \sqrt { 2 } -1 \right) }{ \left( \sqrt { 2 } +1 \right) } \times \frac { \left( \sqrt { 2 } -1 \right) }{ \left( \sqrt { 2 } -1 \right) } } \)
\(=\sqrt { \frac { { \left( \sqrt { 2 } -1 \right) }^{ 2 } }{ 2-1 } } \)
\(=\sqrt { \frac { { \left( \sqrt { 2 } -1 \right) }^{ 2 } }{ 1 } } =\sqrt { 2 } -1\)
which is an irrational number.
Let, there is a number x such that x³ is an irrational number but x⁵ is a rational number.
Let, x =\(\sqrt[5]{7}\) be the number.
⇒ x³ = (5√7)³ = (7)^3/5
is an irrational number.
But x⁵ = (\(\sqrt[5]{7}\))⁵=(7)^5/5  = 7
=7 is a rational number.
(ii) Accepting own mistakes gracefully, co-operative learning among the classmates.