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Bài 1:
a) Ta có: (a-b)+(c-d)-(a+c)
=a-b+c-d-a-c
=-b-d(1)
Ta lại có: -(b+d)=-b-d(2)
Từ (1) và (2) suy ra (a-b)+(c-d)-(a+c)=-(b+d)
b) Ta có: (a-b)-(c-d)+(b+c)
=a-b-c+d+b+c
=a+d(đpcm)
c) Ta có: a(b-c)-b(a-c)
=ab-ac-ab+cb
=cb-ca
=c(b-a)(đpcm)
d) Ta có: b(c-a)+a(b-c)
=bc-ba+ab-ac
=bc-ac
=c(b-a)(đpcm)
e) Ta có: -c(-a+b)+b(c-a)
=ca-cb+bc-ba
=ca-ba
=a(c-b)(đpcm)
g) Ta có: a(c-b)-b(-a-c)
=ac-ab+ba+bc
=ac+bc
=c(a+b)(đpcm)
a) \(n+1\inƯ\left(n^2+2n-3\right)\)
\(\Leftrightarrow n^2+2n-3⋮n+1\)
\(\Leftrightarrow n\left(n+1\right)+n-3⋮n+1\)
Vì \(n\left(n+1\right)⋮n+1\Rightarrow n-3⋮n+1\)
\(\Leftrightarrow n+1-4⋮n+1\)
Vì \(n+1⋮n+1\Rightarrow-4⋮n+1\Rightarrow n+1\inƯ\left(-4\right)=\left\{-1;1;-2;2;-4;4\right\}\)
Ta có bảng sau:
| \(n+1\) | \(-1\) | \(1\) | \(-2\) | \(2\) | \(-4\) | \(4\) |
| \(n\) | \(-2\) | \(0\) | \(-3\) | \(1\) | \(-5\) | \(3\) |
Vậy...
b) \(n^2+2\in B\left(n^2+1\right)\)
\(\Leftrightarrow n^2+2⋮n^2+1\)
\(\Leftrightarrow n^2+1+1⋮n^2+1\)
Vì \(n^2+1⋮n^2+1\) nên \(1⋮n^2+1\Rightarrow n^2+1\inƯ\left(1\right)=\left\{-1;1\right\}\)
Ta có bảng sau:
| \(n^2+1\) | \(-1\) | \(1\) |
| \(n\) | \(\sqrt{-2}\) (vô lý, vì 1 số ko âm mới có căn bậc hai) |
\(0\) (tm) |
Vậy \(n=0\)
c) \(2n+3\in B\left(n+1\right)\)
\(\Leftrightarrow2n+3⋮n+1\)
\(\Leftrightarrow2n+2+1⋮n+1\)
\(\Leftrightarrow2\left(n+1\right)+1⋮n+1\)
Vì \(2\left(n+1\right)⋮n+1\) nên \(1⋮n+1\Rightarrow n+1\inƯ\left(1\right)=\left\{-1;1\right\}\)
Ta có bảng sau:
| \(n+1\) | \(-1\) | \(1\) |
| \(n\) | \(-2\) | \(0\) |
Vậy...
a) n+1∈Ư(n2+2n−3)n+1∈Ư(n2+2n−3)
⇔n2+2n−3⋮n+1⇔n2+2n−3⋮n+1
⇔n(n+1)+n−3⋮n+1⇔n(n+1)+n−3⋮n+1
Vì n(n+1)⋮n+1⇒n−3⋮n+1n(n+1)⋮n+1⇒n−3⋮n+1
⇔n+1−4⋮n+1⇔n+1−4⋮n+1
Vì n+1⋮n+1⇒−4⋮n+1⇒n+1∈Ư(−4)={−1;1;−2;2;−4;4}n+1⋮n+1⇒−4⋮n+1⇒n+1∈Ư(−4)={−1;1;−2;2;−4;4}
Ta có bảng sau:
| n+1n+1 | −1−1 | 11 | −2−2 | 22 | −4−4 | 44 |
| nn | −2−2 | 00 | −3−3 | 11 | −5−5 | 33 |
Vậy...
b) n2+2∈B(n2+1)n2+2∈B(n2+1)
⇔n2+2⋮n2+1⇔n2+2⋮n2+1
⇔n2+1+1⋮n2+1⇔n2+1+1⋮n2+1
Vì n2+1⋮n2+1n2+1⋮n2+1 nên 1⋮n2+1⇒n2+1∈Ư(1)={−1;1}1⋮n2+1⇒n2+1∈Ư(1)={−1;1}
Ta có bảng sau:
| n2+1n2+1 | −1−1 | 11 |
| nn | √−2−2 (vô lý, vì 1 số ko âm mới có căn bậc hai) |
00 (tm) |
Vậy n=0n=0
c) 2n+3∈B(n+1)2n+3∈B(n+1)
⇔2n+3⋮n+1⇔2n+3⋮n+1
⇔2n+2+1⋮n+1⇔2n+2+1⋮n+1
⇔2(n+1)+1⋮n+1⇔2(n+1)+1⋮n+1
Vì 2(n+1)⋮n+12(n+1)⋮n+1 nên 1⋮n+1⇒n+1∈Ư(1)={−1;1}1⋮n+1⇒n+1∈Ư(1)={−1;1}
Ta có bảng sau:
| n+1n+1 | −1−1 | 11 |
| nn | −2−2 | 00 |
2)
a. \(n+5⋮n-2\Rightarrow n-2+7⋮n-2\)
\(\Rightarrow7⋮n-2\Rightarrow n-2\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
\(\Rightarrow n\in\left\{-5;1;3;9\right\}\)
b. \(2n+1⋮n-5\Rightarrow2n-10+11⋮n-5\)
\(\Rightarrow11⋮n-5\Rightarrow n-5\inƯ\left(11\right)=\left\{\pm1;\pm7\right\}\)
\(\Rightarrow n\in\left\{-6;4;6;16\right\}\)
c. \(n^2+3n-13⋮n+3\)
\(\Rightarrow n.n+3n-13⋮n+3\)
\(\Rightarrow n.\left(n+3\right)-13⋮n+3\)
\(\Rightarrow-13⋮n+3\Rightarrow n+3\inƯ\left(-13\right)=\left\{\pm1;\pm13\right\}\)
\(\Rightarrow n\in\left\{-16;-4;-2;10\right\}\)
d. \(n^2+3⋮n-1\Rightarrow n^2-n+n+3⋮n-1\)
\(\Rightarrow n.\left(n-1\right)+\left(n-1\right)+4⋮n-1\)
\(\Rightarrow4⋮n-1\Rightarrow n-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
\(\Rightarrow n\in\left\{-3;-1;0;2;3;5\right\}\)
Bài 2:
a: Để E là số nguyên thì \(3n+5⋮n+7\)
\(\Leftrightarrow3n+21-16⋮n+7\)
\(\Leftrightarrow n+7\in\left\{1;-1;2;-2;4;-4;8;-8;16;-16\right\}\)
hay \(n\in\left\{-6;-8;-5;-9;-3;-11;1;-15;9;-23\right\}\)
b: Để F là số nguyên thì \(2n+9⋮n-5\)
\(\Leftrightarrow2n-10+19⋮n-5\)
\(\Leftrightarrow n-5\in\left\{1;-1;19;-19\right\}\)
hay \(n\in\left\{6;4;29;-14\right\}\)
Bài 2:
a: \(\Leftrightarrow n-10+9⋮n-10\)
\(\Leftrightarrow n-10\in\left\{1;-1;3;-3;9;-9\right\}\)
hay \(n\in\left\{11;9;13;7;19;1\right\}\)
b: \(\Leftrightarrow3n+9⋮3n-3\)
\(\Rightarrow3n-3\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)
hay \(n\in\left\{\dfrac{4}{3};\dfrac{2}{3};\dfrac{5}{3};\dfrac{1}{3};2;0;\dfrac{7}{3};-\dfrac{1}{3};3;-1;5;-3\right\}\)
c: \(\Leftrightarrow12n+2⋮4n-3\)
\(\Leftrightarrow4n-3\in\left\{1;-1;11;-11\right\}\)
hay \(n\in\left\{1;\dfrac{1}{2};\dfrac{7}{2};-2\right\}\)
a) Ta có:
\(2n+1⋮n-3\)
\(\Rightarrow\left(2n-6\right)+7⋮n-3\)
\(\Rightarrow2\left(n-3\right)+7⋮n-3\)
\(\Rightarrow7⋮n-3\)
\(\Rightarrow n-3\in\left\{1;7\right\}\) ( Vì \(n\in N\) )
\(\Rightarrow\left\{{}\begin{matrix}n-3=1\Rightarrow n=4\\n-3=7\Rightarrow n=10\end{matrix}\right.\)
Vậy n=4 hoặc n=10
b) Ta có:
\(n^2+3n-13⋮n+3\)
\(\Rightarrow n\left(n+3\right)-13⋮n+3\)
\(\Rightarrow-13⋮n+3\)
\(\Rightarrow n+3\in\left\{1;13\right\}\) ( Vì \(n\in N\) )
\(\Rightarrow\left\{{}\begin{matrix}n+3=1\Rightarrow n=-2\left(loai\right)\\n+3=13\Rightarrow n=10\end{matrix}\right.\)
Vậy n=10
c) Ta có:
\(n^2+3⋮n-1\)
\(\Rightarrow n^2-1+4⋮n-1\)
\(\Rightarrow\left(n-1\right)\left(n+1\right)+4⋮n-1\)
\(\Rightarrow n+1+4⋮n-1\)
\(\Rightarrow n+5⋮n-1\)
\(\Rightarrow\left(n-1\right)+6⋮n-1\)
\(\Rightarrow6⋮n-1\)
\(\Rightarrow n-1\in\left\{1;2;3;6\right\}\) ( Vì \(n\in N\) )
\(\Rightarrow\left\{{}\begin{matrix}n-1=1\Rightarrow n=2\\n-1=2\Rightarrow n=3\\n-1=3\Rightarrow n=4\\n-1=6\Rightarrow n=7\end{matrix}\right.\)
Vậy n=2 hoặc n=3 hoặc n=4 hoặc n=7
a,\(2n+1=2n-6+7=2\left(n-3\right)+7\)
Do \(2\left(n-3\right)⋮n-3\)
\(\Rightarrow n-3\in\left\{\pm1;\pm7\right\}\)
\(\Leftrightarrow\left[{}\begin{matrix}n-3=1\\n-3=-1\\n-3=7\\n-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}n=4\\n=2\\n=10\\n=-4\end{matrix}\right.\)