<em>Given - a+b+c = 0</em>
<em>To prove that- </em>
<em>a²/bc + b²/ac + c²/ab = 3</em>
<em>Now we know that</em>
<em>when x+y+z = 0,</em>
<em>then x³+y³+z³ = 3xyz</em>
<em>that means</em>
<em> (x³+y³+z³)/xyz = 3 ---- eq 1)</em>
<em>Lets solve for LHS</em>
<em>LHS = a²/bc + b²/ac + c²/ab</em>
<em>we can write it as LHS = a³/abc + b³/abc + c</em><em>³</em><em>/abc</em>
<em>by multiplying missing denominators,</em>
<em>now take common abc from denominator and you'll get,</em>
<em>LHS = (a³+b³+c³)/abc --- eq (2)</em>
<em>Comparing one and two we can say that</em>
<em>(a³+b³+c³)/abc = 3</em>
<em>Hence proved,</em>
<em>a²/bc + b²/ac + c²/ab = 3</em>
Let one integer be x, the consecutive odd integer is x+2
reciprocals are 1/x and 1/(x+2)
1/x +1/(x+2)=24/143
make the denominators the same:
(x+2)/[x(x+2)] +x/[x(x+2)]=24/143
143(2x+2)=24(x)(x+2)
286x+286=24x²+48x
24x²-238x-286=0
use the quadratic formula: x=22 this doesn't work because 22 is not an odd number.
or x=-2.1666666666 (this is not an integer)
weird.
Answer:
79%
Step-by-step explanation:
1.145 cup you’re welcome ! :)
Mark me as brainlist
