43868.$\left(\dfrac{x^{b}}{x^{c}}\right)^{b+c-a}.\left(\dfrac{x^{c}}{x^{a}}\right)^{c+a-b}.\left(\dfrac{x^{a}}{x^{b}}\right)^{a+b-c}=?$
xabc
1
xab + bc + ca
xa + b + c
Explanation:
Given Exp.
= x(b - c)(b + c - a) . x(c - a)(c + a - b) . x(a - b)(a + b - c)
= x(b - c)(b + c) - a(b - c) . x(c - a)(c + a) - b(c - a) . x(a - b)(a + b) - c(a - b)
= x(b2 - c2 + c2 - a2 + a2 - b2) . x-a(b - c) - b(c - a) - c(a - b)
= (x0 x x0)
= (1 x 1) = 1.
Given Exp.
= x(b - c)(b + c - a) . x(c - a)(c + a - b) . x(a - b)(a + b - c)
= x(b - c)(b + c) - a(b - c) . x(c - a)(c + a) - b(c - a) . x(a - b)(a + b) - c(a - b)
= x(b2 - c2 + c2 - a2 + a2 - b2) . x-a(b - c) - b(c - a) - c(a - b)
= (x0 x x0)
= (1 x 1) = 1.
43869.$\dfrac{(243)^{n/5} \times 3^{2n+1}}{9^{n}\times 3^{n-1}}$=?
1
2
9
3n
Explanation:
Given Expression = $\dfrac{(243)^{n/5} \times 3^{2n+1}}{9^{n}\times 3^{n-1}}$
=$\dfrac{(3^{5})^{(n/5)} \times 3^{2n+1}}{(3^{2})^{n}\times 3^{n-1}}$
=$\dfrac{(3^{5\times(n/5)} \times 3^{2n+1})}{(3^{2n}\times 3^{n-1})}$
=$\dfrac{(3^{n} \times 3^{2n+1})}{(3^{2n}\times 3^{n-1})}$
=$\dfrac{3^{(n+2n+1)}}{3^{(2n+n-1)}}$
=$\dfrac{3^{(3n+1)}}{3^{(3n-1)}}$
= 3(3n + 1 - 3n + 1) = 32 = 9.
Given Expression = $\dfrac{(243)^{n/5} \times 3^{2n+1}}{9^{n}\times 3^{n-1}}$
=$\dfrac{(3^{5})^{(n/5)} \times 3^{2n+1}}{(3^{2})^{n}\times 3^{n-1}}$
=$\dfrac{(3^{5\times(n/5)} \times 3^{2n+1})}{(3^{2n}\times 3^{n-1})}$
=$\dfrac{(3^{n} \times 3^{2n+1})}{(3^{2n}\times 3^{n-1})}$
=$\dfrac{3^{(n+2n+1)}}{3^{(2n+n-1)}}$
=$\dfrac{3^{(3n+1)}}{3^{(3n-1)}}$
= 3(3n + 1 - 3n + 1) = 32 = 9.
43870.(25)7.5 x (5)2.5 ÷ (125)1.5 = 5?
8.5
13
16
17.5
Explanation:
Let (25)7.5 x (5)2.5 ÷ (125)1.5 = 5x
Then,$\dfrac{(5^{2})^{7.5}\times(5)^{2.5}}{(5^{3})^{1.5}}=5^{x}$
$\Rightarrow\dfrac{5^{(2\times7.5)}\times5^{2.5}}{5^{(3\times1.5)}}= 5^{x}$
$\Rightarrow\dfrac{5^{15}\times5^{2.5}}{5^{4.5}}= 5^{x}$
$\Rightarrow 5^{x}$=5(15+2.5-4.5)
$\Rightarrow 5^{x}$=5(13)
$\therefore$ x = 13.
Let (25)7.5 x (5)2.5 ÷ (125)1.5 = 5x
Then,$\dfrac{(5^{2})^{7.5}\times(5)^{2.5}}{(5^{3})^{1.5}}=5^{x}$
$\Rightarrow\dfrac{5^{(2\times7.5)}\times5^{2.5}}{5^{(3\times1.5)}}= 5^{x}$
$\Rightarrow\dfrac{5^{15}\times5^{2.5}}{5^{4.5}}= 5^{x}$
$\Rightarrow 5^{x}$=5(15+2.5-4.5)
$\Rightarrow 5^{x}$=5(13)
$\therefore$ x = 13.
43872.If x = 3 + 2$\sqrt{2}$, then the value of $\left( \sqrt{x} -\dfrac{1}{\sqrt{x}} \right)$ is:
1
2
2$\sqrt{2}$
3$\sqrt{3}$
Explanation:
$\left( \sqrt{x} -\dfrac{1}{\sqrt{x}} \right)^{2}=x+\dfrac{1}{x}-2$
$= (3 + 2\sqrt{2}) +\dfrac{1}{(3 + 2\sqrt{2})}-2$
$= (3 + 2\sqrt{2}) +\dfrac{1}{(3 + 2\sqrt{2})}\times\dfrac{(3 - 2\sqrt{2})}{(3 - 2\sqrt{2})}-2$
here the denominator part is 1,
(a+b) (a-b) = a2 - b2
(3 + 2$\sqrt{2}$) \times (3 - 2$\sqrt{2}$) = 3 2 - (2$\sqrt{2})^{2}$
=9-4*2(square of $\sqrt{2}$ is 2)
=9-8
=1
= (3 + 2$\sqrt{2}$) + (3 - 2$\sqrt{2}$) - 2
= 4.
$ \therefore \left( \sqrt{x} -\dfrac{1}{\sqrt{x}} \right)$=2.
$\left( \sqrt{x} -\dfrac{1}{\sqrt{x}} \right)^{2}=x+\dfrac{1}{x}-2$
$= (3 + 2\sqrt{2}) +\dfrac{1}{(3 + 2\sqrt{2})}-2$
$= (3 + 2\sqrt{2}) +\dfrac{1}{(3 + 2\sqrt{2})}\times\dfrac{(3 - 2\sqrt{2})}{(3 - 2\sqrt{2})}-2$
here the denominator part is 1,
(a+b) (a-b) = a2 - b2
(3 + 2$\sqrt{2}$) \times (3 - 2$\sqrt{2}$) = 3 2 - (2$\sqrt{2})^{2}$
=9-4*2(square of $\sqrt{2}$ is 2)
=9-8
=1
= (3 + 2$\sqrt{2}$) + (3 - 2$\sqrt{2}$) - 2
= 4.
$ \therefore \left( \sqrt{x} -\dfrac{1}{\sqrt{x}} \right)$=2.
43879.$\dfrac{1}{1+x^{(b-a)}+x^{(c-a)}}+\dfrac{1}{1+x^{(a-b)}+x^{(c-b)}}+\dfrac{1}{1+x^{(b-c)}+x^{(a-c)}}$
0
1
xa - b - c
None of these
Explanation:
Given Exp. = $\dfrac{1}{\left(1+\dfrac{x^{b}}{x^{a}}+\dfrac{x^{c}}{x^{a}}\right)} + \dfrac{1}{\left(1+\dfrac{x^{a}}{x^{b}}+\dfrac{x^{c}}{x^{b}}\right)}+\dfrac{1}{\left(1+\dfrac{x^{b}}{x^{c}}+\dfrac{x^{a}}{x^{c}}\right)}$
=$ \dfrac{x^{a}}{x^{a}+x^{b}+x^{c}}+\dfrac{x^{b}}{x^{a}+x^{b}+x^{c}}+\dfrac{x^{c}}{x^{a}+x^{b}+x^{c}}$
=$\dfrac{x^{a}+x^{b}+x^{c}}{x^{a}+x^{b}+x^{c}}$
= 1.
Given Exp. = $\dfrac{1}{\left(1+\dfrac{x^{b}}{x^{a}}+\dfrac{x^{c}}{x^{a}}\right)} + \dfrac{1}{\left(1+\dfrac{x^{a}}{x^{b}}+\dfrac{x^{c}}{x^{b}}\right)}+\dfrac{1}{\left(1+\dfrac{x^{b}}{x^{c}}+\dfrac{x^{a}}{x^{c}}\right)}$
=$ \dfrac{x^{a}}{x^{a}+x^{b}+x^{c}}+\dfrac{x^{b}}{x^{a}+x^{b}+x^{c}}+\dfrac{x^{c}}{x^{a}+x^{b}+x^{c}}$
=$\dfrac{x^{a}+x^{b}+x^{c}}{x^{a}+x^{b}+x^{c}}$
= 1.
44469.Which is larget √2 or $\sqrt[3]{3}$?
√2
$\sqrt[3]{3}$
Both are equal
None of these
Explanation:
Given surds are of order 2 and 3, Their LCM is 6.
Changing each to a surd of order 6, we get :
√2=$2^{\dfrac{1}{2}}$
=$2^{\dfrac{1}{2}\times\dfrac{3}{3} }$
=$2^{\dfrac{3}{6}}$
=$(2^{3})^{\dfrac{1}{6}}$
=$(8)^{\dfrac{1}{6}}$
=$\sqrt[6]{8}$
$\sqrt[3]{3}=3^{\dfrac{1}{3}}$
=$3^{\dfrac{1}{3}\times\dfrac{2}{2} }$
=$3^{\dfrac{2}{6}}$
=$(3^{2})^{\dfrac{1}{6}}$
=$(9)^{\dfrac{1}{6}}$
=$\sqrt[6]{9}$
Clearly, $\sqrt[6]{9}$ > $\sqrt[6]{8}$ and hence $\sqrt[3]{3}$ > √2
Given surds are of order 2 and 3, Their LCM is 6.
Changing each to a surd of order 6, we get :
√2=$2^{\dfrac{1}{2}}$
=$2^{\dfrac{1}{2}\times\dfrac{3}{3} }$
=$2^{\dfrac{3}{6}}$
=$(2^{3})^{\dfrac{1}{6}}$
=$(8)^{\dfrac{1}{6}}$
=$\sqrt[6]{8}$
$\sqrt[3]{3}=3^{\dfrac{1}{3}}$
=$3^{\dfrac{1}{3}\times\dfrac{2}{2} }$
=$3^{\dfrac{2}{6}}$
=$(3^{2})^{\dfrac{1}{6}}$
=$(9)^{\dfrac{1}{6}}$
=$\sqrt[6]{9}$
Clearly, $\sqrt[6]{9}$ > $\sqrt[6]{8}$ and hence $\sqrt[3]{3}$ > √2
44470.Number of prime numbers in $\dfrac{6^{12}\times 35^{28}\times 15^{16}}{14^{12}\times 21^{11}}is:$
56
66
112
none of this
Explanation:
$\dfrac{6^{12}\times 35^{28}\times 15^{16}}{14^{12}\times 21^{11}}=$ $\dfrac{\left(2\times 3\right)^{12}\times \left(5\times7\right)^{28}\times \left(3\times 5\right)^{16}}{\left(2\times 7\right)^{12}\times \left(3 \times 7\right)^{11}}$
$=\dfrac{2^{12}\times 3^{12}\times 5^{28}\times 7^{28}\times 3^{16}\times 5^{16}}{2^{12}\times 7^{12}\times 3^{11}\times 7^{11}}$
$=2^{\left(12-12\right)}\times 3^{\left(12+16-11\right)}\times5^{\left(28+16\right)}\times7^{\left(28-12-11\right)}$
$=2^{0} \times3^{17} \times5^{44} \times7^{-5} $
$=\dfrac{3^{17}\times 5^{44} }{7^{5}}$
Number of prime factors = 17 + 44 + 5 = 66.
44471.$\dfrac{243^{\dfrac{n}{5} \times} 3^{2n+1}}{9^{n} \times 3^{n-1}}=?$
1
3
9
27
Explanation:
Given Expression $=\dfrac{243^{\dfrac{n}{5} \times} 3^{2n+1}}{9^{n} \times 3^{n-1}}$
$=\dfrac{\left(3^{5}\right)^{\dfrac{n}{5} }\times 3^{2n+1}}{\left(3^{2}\right)^{n} \times 3^{n-1}}$
$=\dfrac{3^{\left(5 \times \dfrac{n}{5} \right)} \times 3^{2n+1}}{3^{2n}\times 3^{n-1}}$
$=\dfrac{3^{n} \times 3^{2n+1}}{3^{2n}\times 3^{n-1}}$
$=\dfrac{ 3^{n+2n+1}}{ 3^{2n+n-1}}$
$=\dfrac{ 3^{3n+1}}{ 3^{3n-1}}$
$=3^{\left(3n+1-3n+1\right)}$
$=3^{2}=9$
Given Expression $=\dfrac{243^{\dfrac{n}{5} \times} 3^{2n+1}}{9^{n} \times 3^{n-1}}$
$=\dfrac{\left(3^{5}\right)^{\dfrac{n}{5} }\times 3^{2n+1}}{\left(3^{2}\right)^{n} \times 3^{n-1}}$
$=\dfrac{3^{\left(5 \times \dfrac{n}{5} \right)} \times 3^{2n+1}}{3^{2n}\times 3^{n-1}}$
$=\dfrac{3^{n} \times 3^{2n+1}}{3^{2n}\times 3^{n-1}}$
$=\dfrac{ 3^{n+2n+1}}{ 3^{2n+n-1}}$
$=\dfrac{ 3^{3n+1}}{ 3^{3n-1}}$
$=3^{\left(3n+1-3n+1\right)}$
$=3^{2}=9$
44472.If $\dfrac{9^{n} \times 3^{5} \times \left(27\right)^{3}}{3 \times \left(81\right)^{4}}=27$, then the value of n is:
0
2
3
4
Explanation:
$\dfrac{9^{n} \times 3^{5} \times \left(27\right)^{3}}{3 \times \left(81\right)^{4}}=27$
$\Leftrightarrow \dfrac{\left(3^{2}\right)^{n} \times 3^{5} \times \left(3^{3}\right)^{3}}{3 \times \left(3\right)^{4\times 4}}=3^{3}$
$\Leftrightarrow \dfrac{\left(3\right)^{2n} \times 3^{5} \times \left(3\right)^{3\times 3}}{3 \times \left(3\right)^{4\times 4}}=3^{3}$
$\Leftrightarrow\dfrac{3^{2n+5+9}}{3 \times 3^{16}}=3^{3}$
$\Leftrightarrow\dfrac{3^{2n+14}}{ 3^{17}}=3^{3}$
$\Leftrightarrow 3^{\left(2n+14-17\right)}=3^{3}$
$\Leftrightarrow 3^{\left(2n-3\right)}=3^{3}$
$=2n-3 = 3 $
$\Leftrightarrow 2n=6$
$\Leftrightarrow n=3$
$\dfrac{9^{n} \times 3^{5} \times \left(27\right)^{3}}{3 \times \left(81\right)^{4}}=27$
$\Leftrightarrow \dfrac{\left(3^{2}\right)^{n} \times 3^{5} \times \left(3^{3}\right)^{3}}{3 \times \left(3\right)^{4\times 4}}=3^{3}$
$\Leftrightarrow \dfrac{\left(3\right)^{2n} \times 3^{5} \times \left(3\right)^{3\times 3}}{3 \times \left(3\right)^{4\times 4}}=3^{3}$
$\Leftrightarrow\dfrac{3^{2n+5+9}}{3 \times 3^{16}}=3^{3}$
$\Leftrightarrow\dfrac{3^{2n+14}}{ 3^{17}}=3^{3}$
$\Leftrightarrow 3^{\left(2n+14-17\right)}=3^{3}$
$\Leftrightarrow 3^{\left(2n-3\right)}=3^{3}$
$=2n-3 = 3 $
$\Leftrightarrow 2n=6$
$\Leftrightarrow n=3$
44473.If $\left(\sqrt{3}\right)^{5} \times 9^{2}=3^{n}\times 3\sqrt{3}$, then the value of n is:
2
3
4
5
Explanation:
$\left(\sqrt{3}\right)^{5} \times 9^{2}=3^{n}\times 3\sqrt{3}$
$\Leftrightarrow \left(3^{\dfrac{1}{2}}\right)^{5} \times \left(3^{2}\right)^{2}$ $ = 3^{n}\times3 \times 3^{\dfrac{1}{2}}$
$\Leftrightarrow 3^{\left(\dfrac{1}{2}\times 5\right)}\times 3^{\left(2\times 2\right)}$ $=3^{\left(n+1+\dfrac{1}{2}\right)}$
$\Leftrightarrow3^{\left(\dfrac{5}{2}+4\right)}$ $=3^{\left(n+\dfrac{3}{2}\right)}$
$\Leftrightarrow n+\dfrac{3}{2}=\dfrac{13}{2}$
$\Leftrightarrow n=\left(\dfrac{13}{2}-\dfrac{3}{2}\right)$ $=\dfrac{10}{2}=5$
$\left(\sqrt{3}\right)^{5} \times 9^{2}=3^{n}\times 3\sqrt{3}$
$\Leftrightarrow \left(3^{\dfrac{1}{2}}\right)^{5} \times \left(3^{2}\right)^{2}$ $ = 3^{n}\times3 \times 3^{\dfrac{1}{2}}$
$\Leftrightarrow 3^{\left(\dfrac{1}{2}\times 5\right)}\times 3^{\left(2\times 2\right)}$ $=3^{\left(n+1+\dfrac{1}{2}\right)}$
$\Leftrightarrow3^{\left(\dfrac{5}{2}+4\right)}$ $=3^{\left(n+\dfrac{3}{2}\right)}$
$\Leftrightarrow n+\dfrac{3}{2}=\dfrac{13}{2}$
$\Leftrightarrow n=\left(\dfrac{13}{2}-\dfrac{3}{2}\right)$ $=\dfrac{10}{2}=5$