Answer
1.1.2: \(\frac{1}{125}\)
1.2: \(\frac{3\sqrt{x} + 2x\sqrt{x}}{2x}\)
1.3.1: \(x \approx -1.37\) or \(x \approx 0.68\)
1.3.2: \(x \approx -0.63\) or \(x = 0.5\)
1.3.3: \(x < -1\) or \(x > 3\)
1.3.4: \(x = 7\)
1.4: \(x = 3\), \(y = 0\)
Solution
Solve the quadratic equation by following steps:
- step0: Solve using the quadratic formula:
\(13x^{2}+9x-12=0\)
- step1: Solve using the quadratic formula:
\(x=\frac{-9\pm \sqrt{9^{2}-4\times 13\left(-12\right)}}{2\times 13}\)
- step2: Simplify the expression:
\(x=\frac{-9\pm \sqrt{9^{2}-4\times 13\left(-12\right)}}{26}\)
- step3: Simplify the expression:
\(x=\frac{-9\pm \sqrt{705}}{26}\)
- step4: Separate into possible cases:
\(\begin{align}&x=\frac{-9+\sqrt{705}}{26}\\&x=\frac{-9-\sqrt{705}}{26}\end{align}\)
- step5: Rewrite the fraction:
\(\begin{align}&x=\frac{-9+\sqrt{705}}{26}\\&x=-\frac{9+\sqrt{705}}{26}\end{align}\)
- step6: Rewrite:
\(x_{1}=-\frac{9+\sqrt{705}}{26},x_{2}=\frac{-9+\sqrt{705}}{26}\)
Calculate or simplify the expression \( \frac{25^{x+1} \cdot 25^{x-2}}{125^{x+1} \cdot 5^{x-2}} \).
Simplify the expression by following steps:
- step0: Solution:
\(\frac{25^{x+1}\times 25^{x-2}}{125^{x+1}\times 5^{x-2}}\)
- step1: Multiply:
\(\frac{25^{2x-1}}{125^{x+1}\times 5^{x-2}}\)
- step2: Multiply:
\(\frac{25^{2x-1}}{5^{4x+1}}\)
- step3: Factor the expression:
\(\frac{5^{4x-2}}{5^{4x+1}}\)
- step4: Divide the numbers:
\(\frac{1}{5^{4x+1-\left(4x-2\right)}}\)
- step5: Subtract the terms:
\(\frac{1}{5^{3}}\)
- step6: Evaluate the power:
\(\frac{1}{125}\)
Calculate or simplify the expression \( \frac{6}{4 \sqrt{x}}+\sqrt{x} \).
Simplify the expression by following steps:
- step0: Solution:
\(\frac{6}{4\sqrt{x}}+\sqrt{x}\)
- step1: Reduce the fraction:
\(\frac{3}{2\sqrt{x}}+\sqrt{x}\)
- step2: Reduce fractions to a common denominator:
\(\frac{3}{2\sqrt{x}}+\frac{\sqrt{x}\times 2\sqrt{x}}{2\sqrt{x}}\)
- step3: Transform the expression:
\(\frac{3+\sqrt{x}\times 2\sqrt{x}}{2\sqrt{x}}\)
- step4: Multiply the terms:
\(\frac{3+2x}{2\sqrt{x}}\)
- step5: Simplify:
\(\frac{3\sqrt{x}+2x\sqrt{x}}{2x}\)
Solve the equation \( 3(3-x)(1+x)<0 \).
Solve the inequality by following steps:
- step0: Solve the inequality by testing the values in the interval:
\(3\left(3-x\right)\left(1+x\right)<0\)
- step1: Rewrite the expression:
\(3\left(3-x\right)\left(1+x\right)=0\)
- step2: Elimination the left coefficient:
\(\left(3-x\right)\left(1+x\right)=0\)
- step3: Separate into possible cases:
\(\begin{align}&3-x=0\\&1+x=0\end{align}\)
- step4: Solve the equation:
\(\begin{align}&x=3\\&x=-1\end{align}\)
- step5: Determine the test intervals:
\(\begin{align}&x<-1\\&-13\end{align}\)
- step6: Choose a value:
\(\begin{align}&x_{1}=-2\\&x_{2}=1\\&x_{3}=4\end{align}\)
- step7: Test the chosen value:
\(\begin{align}&x<-1\textrm{ }\textrm{is the solution}\\&-13\textrm{ }\textrm{is the solution}\end{align}\)
- step8: The final solution is \(x \in \left(-\infty,-1\right)\cup \left(3,+\infty\right):\)
\(x \in \left(-\infty,-1\right)\cup \left(3,+\infty\right)\)
Solve the equation \( \sqrt{x+18}+2=x \).
Solve the equation by following steps:
- step0: Solve for \(x\):
\(\sqrt{x+18}+2=x\)
- step1: Find the domain:
\(\sqrt{x+18}+2=x ,x\geq -18\)
- step2: Move the constant to the right-hand side:
\(\sqrt{x+18}=x-2\)
- step3: Evaluate:
\(\sqrt{x+18}=x-2,x-2\geq 0\)
- step4: Evaluate:
\(\sqrt{x+18}=x-2,x\geq 2\)
- step5: Solve the equation:
\(\begin{align}&x=7\\&x=-2\end{align},x\geq 2\)
- step6: Find the intersection:
\(x=7\)
- step7: Check if the solution is in the defined range:
\(x=7,x\geq -18\)
- step8: Find the intersection:
\(x=7\)
- step9: Check the solution:
\(x=7\)
Solve the system of equations \( 2 x^{2}+y(4 x-1)=15;x+y=3 \).
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
\(\left\{ \begin{array}{l}2x^{2}+y\left(4x-1\right)=15\\x+y=3\end{array}\right.\)
- step1: Solve the equation:
\(\left\{ \begin{array}{l}2x^{2}+y\left(4x-1\right)=15\\x=3-y\end{array}\right.\)
- step2: Substitute the value of \(x:\)
\(2\left(3-y\right)^{2}+y\left(4\left(3-y\right)-1\right)=15\)
- step3: Simplify:
\(18-y-2y^{2}=15\)
- step4: Move the expression to the left side:
\(18-y-2y^{2}-15=0\)
- step5: Subtract the numbers:
\(3-y-2y^{2}=0\)
- step6: Factor the expression:
\(\left(1-y\right)\left(3+2y\right)=0\)
- step7: Separate into possible cases:
\(\begin{align}&1-y=0\\&3+2y=0\end{align}\)
- step8: Solve the equation:
\(\begin{align}&y=1\\&y=-\frac{3}{2}\end{align}\)
- step9: Calculate:
\(y=1\cup y=-\frac{3}{2}\)
- step10: Rearrange the terms:
\(\left\{ \begin{array}{l}x=3-y\\y=1\end{array}\right.\cup \left\{ \begin{array}{l}x=3-y\\y=-\frac{3}{2}\end{array}\right.\)
- step11: Calculate:
\(\left\{ \begin{array}{l}x=2\\y=1\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{9}{2}\\y=-\frac{3}{2}\end{array}\right.\)
- step12: Check the solution:
\(\left\{ \begin{array}{l}x=2\\y=1\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{9}{2}\\y=-\frac{3}{2}\end{array}\right.\)
- step13: Rewrite:
\(\left(x,y\right) = \left(2,1\right)\cup \left(x,y\right) = \left(\frac{9}{2},-\frac{3}{2}\right)\)
Solve the equation \( 22 x^{2}+3 x-7=0 \).
Solve the quadratic equation by following steps:
- step0: Solve by factoring:
\(22x^{2}+3x-7=0\)
- step1: Factor the expression:
\(\left(2x-1\right)\left(11x+7\right)=0\)
- step2: Separate into possible cases:
\(\begin{align}&2x-1=0\\&11x+7=0\end{align}\)
- step3: Solve the equation:
\(\begin{align}&x=\frac{1}{2}\\&x=-\frac{7}{11}\end{align}\)
- step4: Rewrite:
\(x_{1}=-\frac{7}{11},x_{2}=\frac{1}{2}\)
Let's solve each part of the problem step by step.
### 1.1 Simplify
#### 1.1.2
We need to simplify the expression:
\[
\frac{25^{x+1} \cdot 25^{x-2}}{125^{x+1} \cdot 5^{x-2}}
\]
**Step 1:** Rewrite the bases:
- \( 25 = 5^2 \)
- \( 125 = 5^3 \)
**Step 2:** Substitute:
\[
\frac{(5^2)^{x+1} \cdot (5^2)^{x-2}}{(5^3)^{x+1} \cdot 5^{x-2}} = \frac{5^{2(x+1)} \cdot 5^{2(x-2)}}{5^{3(x+1)} \cdot 5^{x-2}}
\]
**Step 3:** Combine the exponents:
\[
= \frac{5^{2x + 2 + 2x - 4}}{5^{3x + 3 + x - 2}} = \frac{5^{4x - 2}}{5^{4x + 1}} = 5^{(4x - 2) - (4x + 1)} = 5^{-3} = \frac{1}{125}
\]
So, the simplified result is:
\[
\frac{1}{125}
\]
### 1.2 Rationalise the denominator
We need to rationalise:
\[
\frac{6}{4 \sqrt{x}} + \sqrt{x}
\]
**Step 1:** Find a common denominator:
\[
= \frac{6 + 4x}{4\sqrt{x}}
\]
**Step 2:** Rationalise the denominator:
\[
= \frac{(6 + 4x) \cdot \sqrt{x}}{4x} = \frac{6\sqrt{x} + 4x\sqrt{x}}{4x}
\]
So, the rationalised expression is:
\[
\frac{3\sqrt{x} + 2x\sqrt{x}}{2x}
\]
### 1.3 Solve for \( x \)
#### 1.3.1
Solve:
\[
13x^{2} + 9x - 12 = 0
\]
The solutions are:
\[
x_{1} \approx -1.37, \quad x_{2} \approx 0.68
\]
#### 1.3.2
Solve:
\[
22x^{2} + 3x - 7 = 0
\]
The solutions are:
\[
x_{1} \approx -0.63, \quad x_{2} = 0.5
\]
#### 1.3.3
Solve the inequality:
\[
3(3-x)(1+x) < 0
\]
The solution set is:
\[
x \in (-\infty, -1) \cup (3, +\infty)
\]
#### 1.3.4
Solve:
\[
\sqrt{x+18} + 2 = x
\]
The solution is:
\[
x = 7
\]
### 1.4 Solve for \( x \) and \( y \)
We need to solve the system:
\[
2x^{2} + y(4x - 1) = 15
\]
\[
x + y = 3
\]
**Step 1:** Substitute \( y = 3 - x \) into the first equation:
\[
2x^{2} + (3 - x)(4x - 1) = 15
\]
**Step 2:** Expand and simplify:
\[
2x^{2} + 12x - 4x^{2} - 3 = 15
\]
\[
-2x^{2} + 12x - 18 = 0
\]
\[
x^{2} - 6x + 9 = 0
\]
\[
(x - 3)^{2} = 0 \implies x = 3
\]
**Step 3:** Substitute back to find \( y \):
\[
y = 3 - 3 = 0
\]
So, the solution is:
\[
(x, y) = (3, 0)
\]
### Summary of Results
1.1.2: \(\frac{1}{125}\)
1.2: \(\frac{3\sqrt{x} + 2x\sqrt{x}}{2x}\)
1.3.1: \(x_{1} \approx -1.37, x_{2} \approx 0.68\)
1.3.2: \(x_{1} \approx -0.63, x_{2} = 0.5\)
1.3.3: \(x \in (-\infty, -1) \cup (3, +\infty)\)
1.3.4: \(x = 7\)
1.4: \((x, y) = (3, 0)\)
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