Đáp án + Giải thích các bước giải:
Bài `5` :
`a)` `A=7/15+(-2)/11+8/15+(-9)/11`
`=(7/15+8/15)+((-2)/11+(-9)/11)`
`=15/15+(-11)/11=1+(-1)=0`
`b)` `B=1/6+(-2/3)^2-5/18`
`=1/6+4/9-5/18`
`=3/18+8/18-5/18`
`=6/18=1/3`
Bài `6` :
`a)` `2x+3/4=1/36`
`->2x=1/36-3/4`
`->2x=1/36-27/36`
`->2x=-26/36`
`->2x=-13/18`
`->x=(-13/18):2`
`->x=-13/36`
Vậy `x = -13/36`
`b)` `(4/5)^8*2x=(4/5)^9`
`->2x=(4/5)^9:(4/5)^8`
`->2x=4/5`
`->x=4/5:2=4/10=2/5`
Vậy `x = 2/5`
`c)` `(2x-1)^6=(2x-1)^5`
`->(2x-1)^6-(2x-1)^5=0`
`->(2x-1)^5[(2x-1)-1]=0`
`->(2x-1)^5(2x-1-1)=0`
`->(2x-1)^5(2x-2)=0`
`->[((2x-1)^5=0),(2x-2=0):}`
`->[(2x-1=0),(2x=2):}`
`->`\(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=1\end{array} \right.\)
Vậy `x in {1/2;1}`
`d)` `|2x-3,6|-1,4=0`
`->|2x-3,6|=1,4`
TH1 : `2x-3,6=1,4`
`->2x=5`
`->x=2,5`
TH2 : `2x-3,6=-1,4`
`->2x=-1,4+3,6`
`->2x=2,2`
`->x=1,1`
Vậy `x in{2,5;1,1}`.
