What is (x/2+1-nx)(1-x+(x^2)/2) ?

The solution to (x/2+1-nx)(1-x+(x^2)/2) is ((x+2-2nx)(2-2x+x^2))/4

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simplify (x/2+1-nx)(1-x+(x^2)/2)

simplify

(\frac{x}{2} + 1 - n x) (1 - x + \frac{x ^{2}}{2})

\frac{( x + 2 - 2 n x ) ( 2 - 2 x + x ^{2} )}{4}

Solution steps

(\frac{x}{2} + 1 - n x) (1 - x + \frac{x ^{2}}{2})

Simplify

\frac{x}{2} + 1 - n x : \frac{x + 2 - 2 n x}{2}

= \frac{x + 2 - 2 n x}{2} (1 - x + \frac{x ^{2}}{2})

Apply the fraction rule:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( x + 2 - 2 n x ) ( 1 - x + \frac{x ^{2}}{2} )}{2}

Simplify

(x + 2 - 2 n x) (1 - x + \frac{x ^{2}}{2}) : \frac{( x + 2 - 2 n x ) ( 2 - x \cdot 2 + x ^{2} )}{2}

= \frac{\frac{( x + 2 - 2 n x ) ( 2 - x \cdot 2 + x ^{2} )}{2}}{2}

Apply the fraction rule:

\frac{\frac{a}{b}}{c} = \frac{a}{b \cdot c}

= \frac{( x + 2 - 2 n x ) ( 2 - x \cdot 2 + x ^{2} )}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{( x + 2 - 2 n x ) ( 2 - x \cdot 2 + x ^{2} )}{4}

= \frac{( x + 2 - 2 n x ) ( 2 - 2 x + x ^{2} )}{4}

\frac{\partial}{\partial x} ((y) (5 x + 7 z))

(4 x^{2} - 2 x + 1)^{7}

\frac{5 x - 1}{2 x - 3}

x^{2} + y^{2} - 5 - (y + x y) \frac{d y}{d x}

(x - 3) (x + i) (x + 2 i)

What is (x/2+1-nx)(1-x+(x^2)/2) ?
The solution to (x/2+1-nx)(1-x+(x^2)/2) is ((x+2-2nx)(2-2x+x^2))/4