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GaryK [48]
3 years ago
10

7-4+3x0+1= How to solve this problem

Mathematics
2 answers:
zepelin [54]3 years ago
8 0
7-4+<u>3x0</u>+1=<u>7-4</u>+0+1
=<u>3+0</u>+1
=<u>3+1
</u>=4<u>
</u>
Goshia [24]3 years ago
6 0
7-4+(3*0)+1
7-4+1
7-5
2
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PLEASE HELP!!!!!!!!!!!!!!
Lelechka [254]

Answer:

Step-by-step explanation:

(8x²-18x+10)/(x²+5)(x-3)

express the expression as a partial fraction:

(8x²-18x+10)/[(x^2+5)(x-3)] =A/x-3 +bx+c/x²+5

both denominator are equal , so require only work with the nominator

(8x²-18x+10)=(x²+5)A+(x-3)(bx+c)

8x²-18x+10= x²A+5A+bx²+cx-3bx-3c

combine like terms:

x²(A+b)+x(-3b+c)+5A-3c

(8x²-18x+10)

looking at the equation

A+b=8

-3b+c=-18

5A-3c=10

solve for A,b and c (system of equation)

A=2 , B=6, and C=0

substitute in the value of A, b and c

(8x²-18x+10)/[(x^2+5)(x-3)] =A/x-3 +(bx+c)/x²+5

(8x²-18x+10)/[(x^2+5)(x-3)] = 2/x-3 + (6x+0)/(x²+5)

(8x²-18x+10)/[(x^2+5)(x-3)] =

<h2>2/(x-3)+6x/x²+5</h2>

(4x+2)/[(x²+4)(x-2)]

(4x+2)/[(x²+4)(x-2)]= A/(x-2) + bx+c/(x²-2)

(4x+2)=a(x²-2)+(bx+c)(x-2)

follow the same step in the previous answer:

the answer is :

<h2>(4x+2)/[(x²+4)(x-2)]= 5/4/(x-2) + (3/2 -5x/4)/(x²+4)</h2>

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