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Lấy biểu thức A chia cho n-5
ta được số dư là 3 để A chia hết cho n-5 thì n-5 E Ứ(3)
=> n-5 E {-1;-3;1;3}
=> n E {-6;-7;-4;-2}
bn shitbo oy, mk ko bk bn lm tek nào mak ra kq ý, nhưng mk lại lm ra #, mong bn xem lại !!! :)
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ta có: A = n^4 - 5n^3 - 3n^2 + 17n + 13 chia hết cho n - 5
=> n^4 - 5n^3 - 3n^2 + 15n + 2n - 10 + 23 chia hết cho n - 5
n^3.(n-5) - 3n.(n-5) + 2.(n-5) + 23 chia hết cho n - 5
(n-5).(n^3 - 3n+2) + 23 chia hết cho n - 5
mà (n-5).(n^3 - 3n+2) chia hết cho n - 5
=> 23 chia hết cho n - 5
=>...
bn tự làm tiếp nha
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\(n^3+17n=n^3-n+18n=n\left(n^2-1\right)+18n\)
\(=\left(n-1\right)n\left(n+1\right)+18n\)
Mà \(\left(n-1\right)n\left(n+1\right)\) là tích 3 số nguyên liên tiếp nên luôn chia hết cho 6
Vậy biểu thức đã cho chia hết cho 6
\(D=x^2-12x+36+1=\left(x-6\right)^2+1\ge1\)
\(D_{min}=1\) khi \(x=6\)
\(E=-\left(x^2-2x+1\right)-\left(y^2-4y+4\right)+42\)
\(E=-\left(x-1\right)^2-\left(y-2\right)^2+42\le42\)
\(E_{min}=42\) khi \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{4n^2}{17n^4}\cdot\dfrac{-7n^2}{12n}\) \(\left(n\ne0\right)\)
\(=\dfrac{4n^2\cdot-7n^2}{17n^4\cdot12n}\)
\(=\dfrac{-28n^4}{204n^5}\)
\(=\dfrac{-7}{51n}\)
b) \(\dfrac{3x-1}{10x^2+2x}\cdot\dfrac{25x^2+10x+1}{1-9x^2}\) \(\left(x\ne\pm\dfrac{1}{3};x\ne0;x\ne-\dfrac{1}{5}\right)\)
\(=\dfrac{3x-1}{2x\left(5x+1\right)}\cdot\dfrac{\left(5x+1\right)^2}{\left(1-3x\right)\left(3x+1\right)}\)
\(=\dfrac{-\left(1-3x\right)\left(5x+1\right)^2}{2x\left(5x+1\right)\left(1-3x\right)\left(1+3x\right)}\)
\(=\dfrac{-\left(5x+1\right)}{2x\left(1+3x\right)}\)
\(=-\dfrac{5x+1}{6x^2+2x}\)
c) \(\dfrac{27-a^3}{5a+10}:\dfrac{a-3}{3a+6}\) \(\left(a\ne-2;a\ne3\right)\)
\(=\dfrac{\left(3-a\right)\left(9+3a+a^2\right)}{5\left(a+2\right)}\cdot\dfrac{3\left(a+2\right)}{a-3}\)
\(=\dfrac{-\left(a-3\right)\left(a^2+3a+9\right)\cdot3\left(a+2\right)}{5\left(a+2\right)\left(a-3\right)}\)
\(=\dfrac{-3\left(a^2+3x+9\right)}{5}\)
\(=-\dfrac{3x^2+9x+27}{5}\)
d) \(\dfrac{x^2-1}{x^2+2x-15}:\dfrac{x^2+5x+4}{x^2-10x+21}\) \(\left(x\ne3;x\ne-5;x\ne-1;x\ne-4\right)\)
\(=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x+5\right)}:\dfrac{\left(x+1\right)\left(x+4\right)}{\left(x-3\right)\left(x-7\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x+5\right)}\cdot\dfrac{\left(x-3\right)\left(x-7\right)}{\left(x+1\right)\left(x+4\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-7\right)}{\left(x+5\right)\left(x+4\right)}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Cho mình sửa lại câu b nha!
\(\frac{\left(x^2+a\right)\left(1+a\right)+a^2x^2+1}{\left(x^2-a\right)\left(1-a\right)+a^2x^2+1}\)
a/\(n^3+17n=n^3-n+18n=n\left(n-1\right)\left(n+1\right)+18n\)
Có n(n-1)(n+1) vừa chia hết cho 2,3 nên chia hết cho 6 (2,3 nguyên tố cùng nhau)
Và 18n chia hết 6
Nên có ĐPCM