Solve number 5-7 na please

Question

Solve number 5-7 na please

Asked on March 28, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-03-28T14:34:14.968Z

Here are the solutions for numbers 5, 6, and 7. Number 5: Step 1: Simplify the complex fraction. $$ \frac{\frac{y^3}{x^4}}{\frac{y^4-y^3}{x^4}} = \frac{y^3}{x^4} \cdot \frac{x^4}{y^4-y^3} $$ Cancel $x^4$: $$ = \frac{y^3}{y^4-y^3} $$ Factor out $y^3$ from the denominator: $$ = \frac{y^3}{y^3(y-1)} $$ Cancel $y^3$ (assuming $y \neq 0$): $$ = \frac{1}{y-1} $$ Step 2: Substitute the simplified fraction back into the original expression. $$ 2 - \frac{1}{y-1} $$ Step 3: Combine into a single fraction. $$ = \frac{2(y-1)}{y-1} - \frac{1}{y-1} $$ $$ = \frac{2y-2-1}{y-1} $$ $$ = \frac{2y-3}{y-1} $$ The simplified expression is $\boxed{\frac{2y-3}{y-1}}$. Number 6: Step 1: Set up the partial fraction decomposition. $$ \frac{1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2} $$ Step 2: Multiply both sides by $(x-1)(x+2)$. $$ 1 = A(x+2) + B(x-1) $$ Step 3: Solve for $A$ and $B$. Set $x=1$: $$ 1 = A(1+2) + B(1-1) $$ $$ 1 = 3A \implies A = \frac{1}{3} $$ Set $x=-2$: $$ 1 = A(-2+2) + B(-2-1) $$ $$ 1 = -3B \implies B = -\frac{1}{3} $$ Step 4: Write the partial fraction decomposition. $$ \frac{1}{(x-1)(x+2)} = \frac{\frac{1}{3}}{x-1} + \frac{-\frac{1}{3}}{x+2} $$ $$ = \frac{1}{3(x-1)} - \frac{1}{3(x+2)} $$ The partial fraction decomposition is $\boxed{\frac{1}{3(x-1)} - \frac{1}{3(x+2)}}$. Number 7: Step 1: Set up the partial fraction decomposition. $$ \frac{3}{(x-2)^2(x+1)} = \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{C}{x+1} $$ Step 2: Multiply both sides by $(x-2)^2(x+1)$. $$ 3 = A(x-2)(x+1) + B(x+1) + C(x-2)^2 $$ Step 3: Solve for $A$, $B$, and $C$. Set $x=2$: $$ 3 = A(2-2)(2+1) + B(2+1) + C(2-2)^2 $$ $$ 3 = 0 + 3B + 0 \implies 3B = 3 \implies B = 1 $$ Set $x=-1$: $$ 3 = A(-1-2)(-1+1) + B(-1+1) + C(-1-2)^2 $$ $$ 3 = 0 + 0 + C(-3)^2 \implies 3 = 9C \implies C = \frac{3}{9} = \frac{1}{3} $$ Set $x=0$ and substitute $B=1$, $C=\frac{1}{3}$: $$ 3 = A(0-2)(0+1) + B(0+1) + C(0-2)^2 $$ $$ 3 = -2A + B + 4C $$ $$ 3 = -2A + 1 + 4\left(\frac{1}{3}\right) $$ $$ 3 = -2A + 1 + \frac{4}{3} $$ $$ 3 - 1 - \frac{4}{3} = -2A $$ $$ 2 - \frac{4}{3} = -2A $$ $$ \frac{6}{3} - \frac{4}{3} = -2A $$ $$ \frac{2}{3} = -2A $$ $$ A = \frac{2}{3} \cdot \left(-\frac{1}{2}\right) = -\frac{1}{3} $$ Step 4: Write the partial fraction decomposition. $$ \frac{3}{(x-2)^2(x+1)} = \frac{-\frac{1}{3}}{x-2} + \frac{1}{(x-2)^2} + \frac{\frac{1}{3}}{x+1} $$ $$ = -\frac{1}{3(x-2)} + \frac{1}{(x-2)^2} + \frac{1}{3(x+1)} $$ The partial fraction decomposition is $\boxed{-\frac{1}{3(x-2)} + \frac{1}{(x-2)^2} + \frac{1}{3(x+1)}}$.

Accepted Answer · 2026-03-28T14:34:14.968Z

Here are the solutions for numbers 5, 6, and 7. Number 5: Step 1: Simplify the complex fraction. $$ \frac{\frac{y^3}{x^4}}{\frac{y^4-y^3}{x^4}} = \frac{y^3}{x^4} \cdot \frac{x^4}{y^4-y^3} $$ Cancel $x^4$: $$ = \frac{y^3}{y^4-y^3} $$ Factor out $y^3$ from the denominator: $$ = \frac{y^3}{y^3(y-1)} $$ Cancel $y^3$ (assuming $y \neq 0$): $$ = \frac{1}{y-1} $$ Step 2: Substitute the simplified fraction back into the original expression. $$ 2 - \frac{1}{y-1} $$ Step 3: Combine into a single fraction. $$ = \frac{2(y-1)}{y-1} - \frac{1}{y-1} $$ $$ = \frac{2y-2-1}{y-1} $$ $$ = \frac{2y-3}{y-1} $$ The simplified expression is $\boxed{\frac{2y-3}{y-1}}$. Number 6: Step 1: Set up the partial fraction decomposition. $$ \frac{1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2} $$ Step 2: Multiply both sides by $(x-1)(x+2)$. $$ 1 = A(x+2) + B(x-1) $$ Step 3: Solve for $A$ and $B$. Set $x=1$: $$ 1 = A(1+2) + B(1-1) $$ $$ 1 = 3A \implies A = \frac{1}{3} $$ Set $x=-2$: $$ 1 = A(-2+2) + B(-2-1) $$ $$ 1 = -3B \implies B = -\frac{1}{3} $$ Step 4: Write the partial fraction decomposition. $$ \frac{1}{(x-1)(x+2)} = \frac{\frac{1}{3}}{x-1} + \frac{-\frac{1}{3}}{x+2} $$ $$ = \frac{1}{3(x-1)} - \frac{1}{3(x+2)} $$ The partial fraction decomposition is $\boxed{\frac{1}{3(x-1)} - \frac{1}{3(x+2)}}$. Number 7: Step 1: Set up the partial fraction decomposition. $$ \frac{3}{(x-2)^2(x+1)} = \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{C}{x+1} $$ Step 2: Multiply both sides by $(x-2)^2(x+1)$. $$ 3 = A(x-2)(x+1) + B(x+1) + C(x-2)^2 $$ Step 3: Solve for $A$, $B$, and $C$. Set $x=2$: $$ 3 = A(2-2)(2+1) + B(2+1) + C(2-2)^2 $$ $$ 3 = 0 + 3B + 0 \implies 3B = 3 \implies B = 1 $$ Set $x=-1$: $$ 3 = A(-1-2)(-1+1) + B(-1+1) + C(-1-2)^2 $$ $$ 3 = 0 + 0 + C(-3)^2 \implies 3 = 9C \implies C = \frac{3}{9} = \frac{1}{3} $$ Set $x=0$ and substitute $B=1$, $C=\frac{1}{3}$: $$ 3 = A(0-2)(0+1) + B(0+1) + C(0-2)^2 $$ $$ 3 = -2A + B + 4C $$ $$ 3 = -2A + 1 + 4\left(\frac{1}{3}\right) $$ $$ 3 = -2A + 1 + \frac{4}{3} $$ $$ 3 - 1 - \frac{4}{3} = -2A $$ $$ 2 - \frac{4}{3} = -2A $$ $$ \frac{6}{3} - \frac{4}{3} = -2A $$ $$ \frac{2}{3} = -2A $$ $$ A = \frac{2}{3} \cdot \left(-\frac{1}{2}\right) = -\frac{1}{3} $$ Step 4: Write the partial fraction decomposition. $$ \frac{3}{(x-2)^2(x+1)} = \frac{-\frac{1}{3}}{x-2} + \frac{1}{(x-2)^2} + \frac{\frac{1}{3}}{x+1} $$ $$ = -\frac{1}{3(x-2)} + \frac{1}{(x-2)^2} + \frac{1}{3(x+1)} $$ The partial fraction decomposition is $\boxed{-\frac{1}{3(x-2)} + \frac{1}{(x-2)^2} + \frac{1}{3(x+1)}}$.