| Ok so I got them all now: ---1st Problem x(4/x-x) / x(x+4/x+4) Since both the first part and last part got an x outside of the parenthesis we can strike them out. And x is the same as x/1, so we��ll put that in instead so it will be easier to get a view of what we��re doing. (4/x-x/1) / (x/1+4/x+x/1) Next thing is making the denominator the same, so multiply all the x/1 with x so it will become x^2/x. Now we have (4/x-x^2/x) / (x^2/x+4/x+x^2/x) Which is the same as [ (4-x^2)/x ] / [ (x^2+4+x^2)/x ] When you divide with fractions you flip the last one and multiply, so that��s what we��ll be doing [ (4-x^2)/x ] * [ x/(x^2+4+x^2) ] Multiply, the numerators with each other, and denominators with each other [ (4-x^2)*x ] / [ x(x^2+4+x^2) ] And we��ll get (4x-x^3) / (x^3+4x+x^3) Simplify it again, by putting x outside the parenthesis: [ x (4-x^2) ] / [ x ( x^2+4+x^2 ) ] Strike out the x��s, both in the numerator and denominator, and this is what you will have left (4-x^2) / (x^2+4+x^2) That��s the answer. ---Moving on, 2nd problem: x(x-6+9/x) / x(3/x-1) It��s pretty much the same as the first problem, so I won��t be giving much description of what I��m doing If you got any questions, ask :p x(x-6+9/x) / x(3/x-1) <- did nothing yet (x-6+9/x) / (3/x-1) (x/1-6/1+9/x) / (3/x-1/1) (x^2/x-6x/x+9/x) / (3/x-x/x) <- multiply with x. and 1x is the same as x, so I put that. [ (x^2-6x+9x)/x ] / [ (3-x)/x ] <- The same Multiply [ (x^2-6x+9)x ] / [ x(3-x) ] (x^3-6x^2+9x)/(3x-x^2) [ x (x^2-6x+9) ] / [ x (3-x) ] (x^2-6x+9) / (3-x) <- final answer ---3rd problem: x(y+yx^-1) / x(1/1x) This one was actually tricky with the ^-1 thing xD Yx^1 is the same as y*1/x Because, X^-1 is the same as ( x : x : x ), and x : x is the same as 1, which leaves us with 1:x, which again is the same as 1/x So yx^-1 is the same as y*1/x Y is the same as y/1, so when you multiply that with 1/x we get y/x. Put that in ! And with the 1/1x part, I��ll count it as 1 over 1x, and since 1x is the same as x, I��ll put that instead, 1/x. Do the same as what I did earlier, cross the x��s So, that leaves us with ( y/1+y/x ) / (1/x) Multiply y/1 with x to get the same denominator. y/1*x/x = yx/x, put that in, shall we ? ( yx/x+yx ) / ( 1/x ) Same as with before, flip before you multiply [ (Yx+y) * x ] / [ x*1] (Yx^2+yx) / x Remember, yx^2 is different from (yx)^2. We got yx^2 which means y*x*x Moving on, we can put x outside the parenthesis [ x (yx+y) ] / [ x ( 1 ) ] You can cross x up and down. (yx+y)/1 Which is the same as just Yx+y <- final answer ---4th problem x^2-10x+25/25-x^2 Lol this one was actually kind of weird, easy almost compared to the ones before. 25/25 is the same as 1, done with that X^2-x^2 is the same as zero, so you can scratch that That leaves us with -10x + 1 <- final answer ---5th math problem 5x-125x^3/10x^2-2x About the 125x^3/10x^2 part, we can simplify the x��s so we have 125x/10. Then we can simplify once again with 5, so that leaves us with 25x/2, put that in. And again, 5 |
