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Anna [14]
3 years ago
6

1. 8y^2-4y+1 divided by 2y-1

Mathematics
1 answer:
____ [38]3 years ago
5 0
Q1. The answer is 4y+ \frac{1}{2y-1}

\frac{8y^{2}-4y+1 }{2y-1} = \frac{4y*2y-4y+1}{2y-1} = \frac{4y(2y-1)+1}{2y-1} = \frac{4y(2y-1)}{2y-1}+ \frac{1}{2y-1} =4y+ \frac{1}{2y-1}


Q2. The answer is 2a+3+ \frac{6}{a-1}

\frac{2 a^{2}+a+3 }{a-1} = \frac{a*2a-2a+3a+3}{a-1} = \frac{2a(a-1)+3a+3}{a-1}=  \frac{2a(a-1)}{a-1}+ \frac{3a+3}{a-1} = \\  \\ =2a+ \frac{3a+3}{a-1}=2a+ \frac{3a-3+3+3}{a-1}=2a+ \frac{3(a-1)+6}{a-1} =2a+ \frac{3(a-1)}{a-1} + \frac{6}{a-1}= \\  \\ =2a+3+ \frac{6}{a-1}
<span>

Q3. The answer is </span>2 x^{2} +5x+2<span>

</span>\frac{6 x^{3} +11 x^{2} -4x-4}{3x-2} = \frac{3x*2 x^{2}-2 x^{2} *2+15 x^{2} -4x-4  }{3x-2} = \frac{2 x^{2} (3x-2)+15 x^{2} -4x-4}{3x-2}= \\  \\ =  \frac{2 x^{2} (3x-2)}{3x-2} + \frac{15 x^{2} -4x-4}{3x-2} =2 x^{2} +\frac{15 x^{2} -4x-4}{3x-2}=2 x^{2} + \frac{15 x^{2} -10x+6x-4}{3x-2}= \\  \\ =2 x^{2} + \frac{5x*3x-5x*2+6x-4}{3x-2} =2 x^{2} + \frac{5x(3x-2)+3x*2-2*2}{3x-2} = \\  \\ =2 x^{2} + \frac{5x(3x-2)}{3x-2}  + \frac{3x*2-2*2}{3x-2} =2 x^{2} +5x+ \frac{2(3x-2)}{3x-2} =2 x^{2} +5x+2
<span>

Q4. The answer is 2x + 7

</span>\frac{6 x^{2} +11x-35}{3x-5} = \frac{6 x^{2} -10x+21x-35 }{3x-5} =  \frac{3 x *2x-5*2x+7*3x-7*5 }{3x-5} = \\  \\ = \frac{2x(3x-5)+7(3x-5)}{3x-5}= = \frac{(3x-5)(2x+7)}{3x-5} =2x+7
<span>

Q5. The answer is </span>x+1- \frac{3}{x-1}<span>
      
</span>\frac{ x^{2} -4}{x-1} = \frac{ x^{2} -x+x-1-3 }{x-1} = \frac{x*x-x+x-1-3}{x-1} = \frac{x(x-1)+(x-1)-3}{x-1} =  \\  \\ \frac{(x+1)(x-1)-3}{x-1} =  \frac{(x+1)(x-1)}{x-1}  -\frac{3}{x-1} =x+1- \frac{3}{x-1}


Q6. The answer is y^{2} -2y+3

\frac{ y^{3}-4 y^{2}+7y-6  }{y-2} = \frac{y* y^{2} -2y^{2}-2 y^{2} +7y-6 }{y-2} = \frac{y^{2}(y-2)-2 y^{2} +7y-6}{y-2}= \\  \\ = \frac{y^{2}(y-2)}{y-2}+   \frac{-2 y^{2} +7y-6}{y-2} = y^{2} + \frac{-2 y^{2} +4y + 3y-6}{y-2} =  \\  \\ =y^{2} + \frac{-2y*y-2y(-2)+3y-3*2}{y-2} = y^{2} + \frac{(-2y)(y-2)+3(y-2)}{y-2} = \\  \\ = y^{2} + \frac{(-2y+3)(y-2)}{y-2} = y^{2} +(-2y+3) =y^{2} -2y+3
<span>

Q7. The answer is </span>x^{2} +xy+ y^{2}}{x-y}<span>

</span>\frac{ x^{3} - \frac{x}{y}  y^{3} }{x-y} =  \frac{(x-y)( x^{2} +xy+ y^{2}) }{x-y} = \frac{ x^{2} +xy+ y^{2}}{x-y}
<span>

Q8. The answer is </span>(a^{2} +2ab+2b^{2})<span>

</span>\frac{a^{4} +4b^{4} }{a^{2}-2ab+2 b^{2} } = \frac{ (a^{2})^{2} +(2b)^{2}}{a^{2}-2ab+2 b^{2}} = \frac{(a^{2} -2ab+2b^{2})(a^{2} +2ab+2b^{2}) }{(a^{2} -2ab+2b^{2})} =(a^{2} +2ab+2b^{2})<span>


Q9. The answer is a^{n-8} - a^{-14}

</span>\frac{ (a^{2}) ^{n} - a^{n-6}  }{ a^{n+8} }= \frac{(a^{2}) ^{n} }{a^{n+8}}- \frac{ a^{n-6} }{a^{n+8}}    \\  \\   (x^{y}) ^{z}= x^{y*z}   \\  \\  \frac{ x^{y} }{ x^{z} } = x^{y-z}  \\  \\ &#10;\frac{(a^{2}) ^{n} }{a^{n+8}}- \frac{ a^{n-6} }{a^{n+8}}  =\frac{a^{2n}  }{a^{n+8}}- \frac{ a^{n-6} }{a^{n+8}}  =  a^{2n-(n+8)} - a^{n-6-(n+8)} = \\  \\ =a^{2n-n-8} - a^{n-6-n-8} = a^{n-8} - a^{-14}
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It's a question from real and complex numbers which I can't solve. so someone PLZ HeLp​
Semmy [17]

Answer:

\frac{1}{5}

Step-by-step explanation:

Using the rules of exponents

a^{m} × a^{n} = a^{(m+n)}, \frac{a^{m} }{a^{n} } = a^{(m-n)}, (a^m)^{n} = a^{mn}

Simplifying the product of the first 2 terms

\frac{a^{p^2+pq} }{a^{pq+q^2} } × \frac{a^{q^2+qr} }{a^{qr+r^2} }

= a^{p^2-q^2} × a^{q^2-r^2}

= a^{p^2-r^2}

Simplifying the third term

5((a^p+r)^{p-r}

= 5a^{(p+r)(p-r)} = 5a^{(p^2-r^2)}

Performing the division, that is

\frac{a^{(p^2-r^2)} }{5a^{(p^2-r^2)} } ← cancel a^{(p^2-r^2)} on numerator/ denominator leaves

= \frac{1}{5}

4 0
3 years ago
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pentagon [3]

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First you make -8 1/3 a improper fraction

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and you keep the sign of the biggest number

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