Question

solve for x 1 upon 1 + x - 1 upon x - 1 = 6 upon x + 1 square​

Answer 1

Answer:

areh bosdi ke jada math aata h kya bur mh danda ghused dunga

Answer 2

Step-by-step explanation:

• 1/(1+x)-1/(x-1)=6/(x+1)²
• {x-1-(x+1)}/(x+1)(x-1)=6/(x+1)²
• now we will open the square and it will open by the formula *(x+y)²=x²+2xy+y²
• {x-1-x-1}/x²-x+x-1=6/x²+2(x)(1)+1²
• now in curly bracket {} x with plus will cancel -x and outside the bracket -x will also cancel x with plus
• (-2)/x²-1=6/x²+2x+1
• as the rule x²-1 is getting Divide on left side will go to the right side and would multiply with 6
• -2=6(x²-1)/x²+2x+1
• and now on the right side x²+2x+1 is getting Divide will go to the left side and would multiply with -2
• -2(x²+2x+1)=6(x²-1)
• -2x²-4x-2=6x²-6
• now we will shift 6x²-6 from right side to left side
• -6x²-2x²-4x-2+6=0
• -8x²-4x+4=0
• now we will shift all the numbers to right side because we can't have any value of square with subtraction sign like -8x² by shifting the value from one side to another the sign of value gets change
• 0=8x²+4x-4
• now we will use the method of middle term break
• 0=8x²+8x-4x-4
• 0=8x(x+1)-4(x+1)
• 0=(x+1)(8x-4)
• and this is the answer

I HOPE IT WILL HELP