\(\frac{x-1}{x}^2+\frac{x-1}{x-2}^2=\frac{40}{49}\)
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a. 60%x + 0,4x + x : 3 = 2
0.6x + 0,4x + x : 3 = 2
x(0,6 + 0,4 : 3 ) = 2
\(x.\frac{1}{3}=2=>x=2:\frac{1}{3}=\frac{1}{6}\)
câu B tự làm nha .
a) \(\frac{5-x}{4x^2-8x}\) + \(\frac{7}{8x}\) = \(\frac{x-1}{2x\left(x-2\right)}\) +\(\frac{1}{8x-16}\) ĐKXĐ : x #0, x#2, x#-2
<=> \(\frac{5-x}{4x\left(x-2\right)}\) + \(\frac{7}{8x}=\frac{x-1}{2x\left(x-2\right)}\) + \(\frac{1}{8\left(x-2\right)}\)
<=> \(\frac{2\left(5-x\right)}{8x\left(x-2\right)}+\frac{7\left(x-2\right)}{8x\left(x-2\right)}=\frac{4\left(x-1\right)}{8x\left(x-2\right)}+\frac{x}{8x\left(x-2\right)}\)
=> 10 - 2x + 7x - 14 = 4x - 4 + x
<=>-2x + 7x - 4x + x = -4 - 10 + 14
<=>x=-14
x | 7 | 9 | |||
x2 | 49 | 81 | |||
x2-49 | - | 0 | + | + | + |
x2-81 | - | - | - | 0 | + |
A | + | 0 | - | 0 | + |
dựa vào bảng ta có khi 7<x<9 thì A<0 vậy 7<x<9
b, ta có : \(\frac{2015}{1}\)+\(\frac{2014}{2}\)+\(\frac{2013}{3}\)+......+\(\frac{1}{2015}\)
=1+1+1+1......+1+\(\frac{2014}{2}\)+\(\frac{2013}{3}\)+.......+\(\frac{1}{2015}\)
(2015 số 1)
=1+(1+\(\frac{2014}{2}\))+(1+\(\frac{2013}{3}\))+........+(1+\(\frac{1}{2015}\))
=\(\frac{2016}{2016}\)+\(\frac{2016}{2}\)+\(\frac{2016}{3}\)+.........+\(\frac{2016}{2015}\)
=2016(\(\frac{1}{2016}\)+\(\frac{1}{2}\)+\(\frac{1}{3}\)+.........+\(\frac{1}{2015}\))
=2016(\(\frac{1}{2}\)+\(\frac{1}{3}\)+.......+\(\frac{1}{2015}\)+\(\frac{1}{2016}\))vậy x= 2016
\(\frac{\left(x-1\right)^2}{x}\)+ \(\frac{\left(x-1\right)^2}{x-2}\)= \(\frac{40}{49}\)
= \(\frac{\left(x-2\right)\left(x^2-2x+1\right)}{x\left(x-2\right)}\)+ \(\frac{x\left(x^2-2x+1\right)}{x\left(x-2\right)}\)= \(\frac{40}{49}\)
= \(\frac{\left(x-2\right)\left(x^2-2x+1\right)+x\left(x^2-2x+1\right)}{x\left(x-2\right)}\)= \(\frac{40}{49}\)
= \(\frac{\left(x-2+x\right)\left(x^2-2x+1\right)}{x\left(x-2\right)}\)= \(\frac{40}{49}\)
= > nhân pahs bn có thể tự làm tiếp
x\(-1^2\) khác với