Đáp án:
Giải thích các bước giải:
Bài `6`
$M = \dfrac{ \dfrac{3}{4} - \dfrac{3}{5} + \dfrac{3}{7} + \dfrac{3}{11} }{ \dfrac{13}{4}-\dfrac{13}{5} + \dfrac{13}{7} + \dfrac{13}{11} }$
`=>`$M = \dfrac{ (\dfrac{13}{4} - \dfrac{13}{5} + \dfrac{13}{7} + \dfrac{13}{11}). \dfrac{3}{13} }{ \dfrac{13}{4}-\dfrac{13}{5} + \dfrac{13}{7} + \dfrac{13}{11} }$
`=>M=3/13`
Bài `7`
`A=(3x+2)/(x-3)`
`ĐK: x\ne 3`
`+)x=1(t``/m)`
`=>A=(3.1+2)/(1-3)=5/(-2)=(-5)/2`
`+)x=2(t``/m)`
`=>A=(3.2+2)/(2-3)=8/(-1)=(-8)/1=-8`
`+)x=5/2(t``/m)`
$A = \dfrac{3. \dfrac{5}{2} +2}{ \dfrac{5}{2}-3 }$
$A = \dfrac{ \dfrac{19}{2} }{ \dfrac{-1}{2} }$
`A=19/2: (-1)/2=19/2 .(-2)=-19`
`b)`
`A=(3x+2)/(x-3)=(3(x-3)+11)/(x-3)=3 +11/(x-3)`
Để `A in Z <=>11/(x-3) in Z`
`<=>11 \vdots (x-3)`
`=>x-3 in Ư(11)={+-1,+-11}`
`=>x in {4,2,14,-8}`
`c)`
`A=(x^2+3x-7)/(x-3)=(x(x+3)-7)/(x+3)=x- 7/(x+3)`
Để `A in Z `, mà `x in Z=>7/(x+3) in Z`
`<=>7 \vdots (x+3)`
`=>x+3 in Ư(7)={+-1,+-7}`
`=>x in {-4,-2,4,-10}`
`d)`
Nhận thấy
`x in {4,2,14,-8}` thì `A in Z`
`x in {-4,-2,4,-10}` thì `B in Z`
`=>` Để `A,B ` cùng là số nguyên `(A;B in Z)<=>x=4`
