Algebraic Expressions

If $x+\frac{1}{x}=5$, then the value of $x^8+\frac{1}{x^8}$ is:

If $\frac{4x}{3}+2p=10$ for what value of $p, x=12?$

If $\frac{1}{x+1}+ \frac{2}{y+2} +\frac{1009}{z+1009} =1$, then find the value of  $\frac{x}{x+1}+\frac{y}{y+2}+\frac{z}{z+1009}=$?

Find the value of the followings

$\left(1+\frac{1}{x}\right)\left(1+\frac{1}{x+1}\right)\left(1+\frac{1}{x+2}\right)\left(1+\frac{1}{x+3}\right)$

If $x+\frac{a}{x }=b$, then find the value of $\frac{x^2+bx+a }{b{x}^2-x^3}$

The value of $\left(1+\frac{1}{\mathrm{a}}\right)\left(1+\frac{1}{\mathrm{a}+1}\right)\left(1+\frac{1}{\mathrm{a}+2}\right)\left(1+\frac{1}{\mathrm{a}+3}\right)\left(1+\frac{1}{\mathrm{a}+4}\right)$ _____.

If $a\times b=a+b+\frac{a}{b}$ then the value of $12\times 4$ is :

If $$\left( { x }^{ 2 }+\cfrac { 1 }{ { 5 } } x+{ a }^{ 2 } \right) $$ is a perfect square, then a is -

If $\mathrm{a}(\mathrm{a} + 2) = \mathrm{a} + \mathrm{b} + \mathrm{c}, \mathrm{b}(\mathrm{b} + 2) = \mathrm{a} + \mathrm{b} + \mathrm{c}$ and $\mathrm{c}(\mathrm{c}+2)=\mathrm{a}+\mathrm{b}+\mathrm{c}$, then what is the value of $\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}$?

$a \times b = b \times a$is

If $x+y=2x$, then the value of $\frac{x}{x-z}=\frac{z}{y-z}$ is-

The simplest form of the expression:

$\frac{p^2-p}{2p^3+6p^2}+\frac{p^2-1}{p^2+3p}+\frac{p^2}{p+1}$

$xy$ is a number that is divided by ab where$xy<ab$  and gives a result $0.xyxyxy$..... then$ab$  equals:

Multiplication of '$ab$' and '$a-b$' is:

$\frac{x}{b-c}=\frac{y}{c-a}=\frac{z}{a-b}$, then $x+y+z$=?

If $a = 3, b = -1, c = 2$which is true
$$\begin{gathered} \left(I\right) a+b+c = 4 \\ \left(II\right) 3a + 5b = 4 \\ \left(III\right) 2c - 5a = 6 \end{gathered}$$

If ${\mathrm{a}}^{\mathrm{x}}={(\mathrm{x}+\mathrm{y}+\mathrm{z})}^{\mathrm{y}} , {\mathrm{a}}^{\mathrm{y}}={(\mathrm{x}+\mathrm{y}+\mathrm{z})}^{\mathrm{z}} , {\mathrm{a}}^{\mathrm{z}}={(\mathrm{x}+\mathrm{y}+\mathrm{z})}^{\mathrm{x}}$ , then what will be the value of $x+y+z$?

Number of terms in the expression $xyz+1$ Is:

If $a, b, c$are non zero$a+\frac{1}{b}=1$ and $b+\frac{1}{c}=1$ then the value of $abc$ is:

If $x=\frac{4ab}{a+b} (a\ne b)$ then the value of $\frac{x+2a}{x-2a}+\frac{x+2b}{x-2b}$ is :