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Find the square root of $$\frac{{\left( {0.064 - 0.008} \right)\left( {0.16 - 0.04} \right)}}{{\left( {0.016 + 0.08 + 0.04} \right){{\left( {0.4 + 0.2} \right)}^3}}}$$

Correct Answer: $$\frac{1}{3}$$
Square root
$$ \Rightarrow \sqrt {\frac{{\left( {0.064 - 0.008} \right)\left( {0.16 - 0.04} \right)}}{{\left( {0.016 + 0.08 + 0.04} \right){{\left( {0.4 + 0.2} \right)}^3}}}} $$
$$ \Rightarrow \sqrt {\frac{{\left\left}}{{\left{{\left}^3}}}} $$
$$ \Rightarrow {\text{Let }}0.4{\text{ }} = {\text{ }}a,{\text{ }}0.2{\text{ }} = {\text{ }}b$$
$$ \Rightarrow \sqrt {\frac{{\left( {{a^3} - {b^3}} \right)\left( {{a^2} - {b^2}} \right)}}{{\left( {{a^2} + {b^2} + ab} \right){{\left( {a + b} \right)}^3}}}} $$
$$ \Rightarrow \sqrt {\frac{{\left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)\left( {a + b} \right)\left( {a - b} \right)}}{{{{\left( {a + b} \right)}^3}\left( {{a^2} + {b^2} + ab} \right)}}} $$
$$\eqalign{ & \Rightarrow \sqrt {\frac{{{{\left( {a - b} \right)}^2}}}{{{{\left( {a + b} \right)}^2}}}} \cr & \Rightarrow \frac{{a - b}}{{a + b}} \cr & \Rightarrow \frac{{0.4 - 0.2}}{{0.4 + 0.2}} \cr & \Rightarrow \frac{{0.2}}{{0.6}} \cr & \Rightarrow \frac{1}{3} \cr} $$

If a + b + c + d = 4, then find the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$     + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$     + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$     + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$     is?

If a + b + c + d = 4, then the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$     + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$     + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$     + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$     is?

$$\sqrt {\frac{{0.009 \times 0.036 \times 0.016 \times 0.08}}{{0.002 \times 0.0008 \times 0.0002}}} $$       is equal to = ?

০.২, ০.০৪, ০.০০৮, ০.০০১৬ ধারাটির পরবর্তী সংখ্যাটি কত?

The value of the expression $$\frac{{{{\left( {a - b} \right)}^2}}}{{\left( {b - c} \right)\left( {c - a} \right)}} + $$   $$\frac{{{{\left( {b - c} \right)}^2}}}{{\left( {a - b} \right)\left( {c - a} \right)}} + $$    $$\frac{{{{\left( {c - a} \right)}^2}}}{{\left( {a - b} \right)\left( {b - c} \right)}}$$   = ?

The Hamiltonian of a particle is given by $$H = \frac{{{p^2}}}{{2m}} + V\left( {\left| {\overrightarrow {\bf{r}} } \right|} \right) + \phi \left( { + \left| {\overrightarrow {\bf{r}} } \right|} \right)\overrightarrow {\bf{L}} .\overrightarrow {\bf{S}} ,$$        where $$\overrightarrow {\bf{S}} $$ is the spin, $$V\left( {\left| {\overrightarrow {\bf{r}} } \right|} \right)$$  and $$\phi \left( {\left| {\overrightarrow {\bf{r}} } \right|} \right)$$  are potential functions and $$\overrightarrow {\bf{L}} \left( { = \overrightarrow {\bf{r}} \times \overrightarrow {\bf{p}} } \right)$$   is the angular momentum. The Hamiltonian does not commute with

$$\frac{{{{\left( {4.53 - 3.07} \right)}^2}}}{{\left( {3.07 - 2.15} \right)\left( {2.15 - 4.53} \right)}} + \, $$     $$\frac{{{{\left( {3.07 - 2.15} \right)}^2}}}{{\left( {2.15 - 4.53} \right)\left( {4.53 - 3.07} \right)}} + \,\, $$     $$\frac{{{{\left( {2.15 - 4.53} \right)}^2}}}{{\left( {4.53 - 3.07} \right)\left( {3.07 - 2.15} \right)}}$$     is simplified to :

The length of each side of square A is increased by 100 percent to make square B. Again the length of each side of square B is increased by 50 percent to make square C. By what percent is the area of square C greater than the sum of the areas of square A and B?

(3x - 2y) : (2x + 3y) = 5 : 6, then one of the value of $${\left( {\frac{{\root 3 \of x + \root 3 \of y }}{{\root 3 \of x - \root 3 \of y }}} \right)^2}{\text{ is = ?}}$$

(3x - 2y) : (2x + 3y) = 5 : 6, then one of the value of $${\left( {\frac{{\root 3 \of x + \root 3 \of y }}{{\root 3 \of x - \root 3 \of y }}} \right)^2}$$   is?