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Solution: If the walking speed is x, the bicycle speed is 4x
According to the title, 7 x + (19-7) 4 x = 2
7+12/4=2x
x=5 km.
Answer: Slightly.
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If you can walk at a speed of x kilometers per hour, then you can ride a bicycle at a speed of 4 kilometers per hour! The distance traveled by the bicycle is (19 7) kilometers, and according to the question: 7 x (19 7) 4 x 2, and the solution is x 5 4 x 20, so the speed of walking is 5 kilometers per hour, and the speed of cycling is 20 kilometers per hour.
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Let the walking speed be x km h, then the bicycle speed is 4x km h x 7 + 4x 12 2 Multiply 4x on both sides of the equation to get x=5, so 4x=20 so the walking speed is 5 km h The bicycle speed is 20 km h
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Let the walking speed be x, then 7 x+12 4x=2, and the solution is: x=5
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If the walk is x, then the bicycle is 4x 19-7=12
7 divided by x) + (12 divided by 4x) = 2
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If the speed of walking is x, the speed of the bicycle is 4x.
7--- plus --12-- equals 2 solution; x is equal to 5Check, when x is equal to 5, 4x is not 0Then x is equal to 5 and is the original fractional equation.
x 4x.
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It can only be said to be a half-pair, first of all, in the process of "multiply both sides by (x-3), get x-3=-2(x-3)" This error may be that you made a mistake (you should get x a=-2 (x-3)).
And then you have to say it at the end because of X-3≠0
So x≠3, so (a+6) 3≠3
Therefore, a≠3 so the value range of a is a -6, and a ≠ 3
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No, make sure that the numerator is not 0, i.e. x≠3
That is, (a+6) 3≠3 gives a≠3
In summary, a -6 and a ≠3
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1.Assuming that the original plan is to repair x meters per day, then the equation is solved to get x=80
2.Let A process x pieces per day and B process y pieces per day, then 1200 x-10 = 1200 y, y =, solve the equation to obtain x=40, y=60;
3.Let A spread x kilometers per week and B spread y kilometers per week, then x+1=y, 18 x-3=18 y solve the equation to get x=2, y=3
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Set the normal trace key, highway speed, attitude, and block x
600/x=2*480/(x+45)
600x+600*45=960x
360x=600*45
x=75600 plus 75=8 hours.
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Put 1 generation into -x square + 3 into the calculation When you encounter a constant, you will bring it, it should be calculated like this, I counted two parts, but I have to sleep.
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It's very unclear, give you a clear answer.
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5x=-y-z
So x 2=(y+z) 2
So y 2+z 2-x 2=-2yz
y^2=(x+z)^2
So z 2+x 2-y 2=-2xz
z^2=(x+y)^2
So x 2 + y 2-z 2 = -2xy
So the original formula =1 (-2yz)+1 (-2xz)+1 (-2xy) =x (-2xyz)+y (-2xyz)+z (-2xyz) =(x+y+z) (-2xyz) =0
6x/3y=y/2x-5y
2x^2-5xy-3y^2=0
x-3y)(2x+y)=0①
x/3y=6x-15y/x
x^2-18xy+45y^2=0
x-3y)(x-15y)=0②
y/2x-5y=6x-15y/x
xy=3(2x-5y)^2
x-3y)(12x-25)=0③
If both are true, x-3y=0, x=3y
4x^2-5xy+6y^2)/(x^2-2xy+3y^2)=[4(3y)^2-5*(3y)*y+6y^2]/[(3y)^2-2*(3y)*y+3y^2]
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1.From the known, x+y=-z, (x+y) 2=z 2, can be reduced to: x 2+y 2-z 2=2xy, the same way is y 2+z 2-x 2=2yz, z 2+x 2-y 2=2xz, the original formula =1 2yz+1 2xz+1 2xy=(x+y+z) 2xyz=0
2.From the known x=3y, the original formula = [4(3y) 2-5 3y·y+6y 2] [(3y) 2-2 3y·y+3y 2]=9 2
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