bài 1:Tìm số nguyên x
25-(25-x)=-8+3(x-4)
5+|x+3|+9
3-(7+x)=-16+2.(5-x)
x-(-25+x)=13-x
15-(30+x)=x-(27-|-8|)
bài 2:Tìm số nguyên n để 2n+3 chia hết cho n-3
bài 1:Tìm số nguyên x
25-(25-x)=-8+3(x-4)
5+|x+3|+9
3-(7+x)=-16+2.(5-x)
x-(-25+x)=13-x
15-(30+x)=x-(27-|-8|)
bài 2:Tìm số nguyên n để 2n+3 chia hết cho n-3
`B1:`
`25-(25-x)=-8+3(x-4)`
`=>25-25+x=-8+3x-12`
`=>x=3x-20`
`=>2x=20`
`=>x=10`
Vậy `x=10` thì `25-(25-x)=-8+3(x-4)`.
`5+|x+3|=9`
`=>|x+3|=9-5=4`
`=>` \(\left[ \begin{array}{l}x+3=4\\x+3=-4\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=1\\x=-7\end{array} \right.\)
Vậy `x=1` hoặc `x=-7` thì `5+|x+3|=9`.
`3-(7+x)=-16+2(5-x)`
`=>3-7-x=-16+10-2x`
`=>-4-x=-2x-6`
`=>x=-2`
Vậy `x=-2` thì `3-(7+x)=-17+2(5-x)`.
`x-(-25+x)=13-x`
`=>x+25-x=13-x`
`=>25=13-x`
`=>x=13-25=-12`
Vậy `x=-12` thì `x-(-25+x)=13-x`.
`15-(30+x)=x-(27-|-8|)`
`=>15-30-x=x-(27-8)`
`=>-15-x=x-19`
`=>2x=4`
`=>x=2`
Vậy `x=2` thì `15-(30+x)=x-(27-|-8|)`.
`B2:`
`2n+3 vdots n-3(n in ZZ)`
`=>2n-6+9 vdots n-3`
`=>2(n-3)+9 vdots n-3`
`=>9 vdots n-3`(do `2(n-3) vdots n-3`)
`=>n-3 in Ư(9)={+-1,+-3,+-9}`
Ta có bảng sau:
$\begin{array}{|c|c|c|}\hline n-3&1&-1&3&-3&9&-9 \\\hline n&4&2&6&0&12&-6 \\\hline KL&TM&TM&TM&TM&TM&TM \\\hline\end{array}$
Vậy với `n in {4,2,6,0,12,-6}` thì `2n+3 vdots n-3`.
Đáp án:
Bài `1`
` 25 – (25-x) = -8 + 3(x-4)`
` 25 – 25 + x = -8 + 3x – 12`
` x = 3x – 20`
` -2x = -20`
` x = 10`
——–
` 5 + |x+3| = 9`
` |x+3| = 4`
` => x + 3 = 4` or ` x+3 = -4`
` => x =1` or ` x = -7`
————-
` 3 -(7+x) = -16 + 2.(5-x)`
` 3 – 7 – x = -16 + 10 – 2x`
` -4 – x= -6 – 2x`
` – 4 – x +6 + 2x = 0`
` x + 2 = 0`
` x = -2`
————-
`x – (-25+x) = 13 – x`
` x +25 – x = 13 – x`
` 13 – x = 25`
` x = 13 -25`
` x = -12`
Bài `2`
`=> 2(n-3) + 3 \vdots n – 3`
` => 3 \vdots n -3`
Ta có bảng
n -3 -3 -1 1 3
n 0 2 4 6