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\(\dfrac{A}{B}=\dfrac{3x^4+3x^2+x^3+x-3x^2-3+5x-2}{x^2+1}=3x^2+x-3+\dfrac{5x-2}{x^2+1}\)
Để A chia hết cho B thì \(\left(5x-2\right)\left(5x+2\right)⋮x^2+1\)
\(\Leftrightarrow25x^2-4⋮x^2+1\)
\(\Leftrightarrow25x^2+25-29⋮x^2+1\)
\(\Leftrightarrow x^2+1\in\left\{1;29\right\}\)
hay \(x\in\left\{0;2\sqrt{7};-2\sqrt{7}\right\}\)
Bài 3:
a) ta có: \(A=x^2+4x+9\)
\(=x^2+4x+4+5=\left(x+2\right)^2+5\)
Ta có: \(\left(x+2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+2\right)^2+5\ge5\forall x\)
Dấu '=' xảy ra khi
\(\left(x+2\right)^2=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)
Vậy: GTNN của đa thức \(A=x^2+4x+9\) là 5 khi x=-2
b) Ta có: \(B=2x^2-20x+53\)
\(=2\left(x^2-10x+\frac{53}{2}\right)\)
\(=2\left(x^2-10x+25+\frac{3}{2}\right)\)
\(=2\left[\left(x-5\right)^2+\frac{3}{2}\right]\)
\(=2\left(x-5\right)^2+2\cdot\frac{3}{2}\)
\(=2\left(x-5\right)^2+3\)
Ta có: \(\left(x-5\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-5\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-5\right)^2+3\ge3\forall x\)
Dấu '=' xảy ra khi
\(2\left(x-5\right)^2=0\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)
Vậy: GTNN của đa thức \(B=2x^2-20x+53\) là 3 khi x=5
c) Ta có : \(M=1+6x-x^2\)
\(=-x^2+6x+1\)
\(=-\left(x^2-6x-1\right)\)
\(=-\left(x^2-6x+9-10\right)\)
\(=-\left[\left(x-3\right)^2-10\right]\)
\(=-\left(x-3\right)^2+10\)
Ta có: \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x-3\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-3\right)^2+10\le10\forall x\)
Dấu '=' xảy ra khi
\(-\left(x-3\right)^2=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy: GTLN của đa thức \(M=1+6x-x^2\) là 10 khi x=3
Bài 2:
a) \(\left(x+y\right)^2+\left(x^2-y^2\right)\)
\(=\left(x+y\right)^2+\left(x-y\right).\left(x+y\right)\)
\(=\left(x+y\right).\left(x+y+x-y\right)\)
\(=\left(x+y\right).2x\)
c) \(x^2-2xy+y^2-z^2+2zt-t^2\)
\(=\left(x^2-2xy+y^2\right)-\left(z^2-2zt+t^2\right)\)
\(=\left(x-y\right)^2-\left(z-t\right)^2\)
\(=\left[x-y-\left(z-t\right)\right].\left(x-y+z-t\right)\)
\(=\left(x-y-z+t\right).\left(x-y+z-t\right)\)
Chúc bạn học tốt!
a: \(=35^{2018}\left(35-1\right)=35^{2018}\cdot34⋮17\)
b: \(=43^{2018}\left(43+1\right)=43^{2018}\cdot44⋮11\)
d: \(=6mn-4m-9n+6-6mn+9m+4n-6\)
=5m-5n=5(m-n) chia hết cho 5
a) \(a^2-a=a\left(a-1\right)⋮2\) ( Tích 2 số nguyên liên tiếp ⋮ 2 )
b) \(a^3-a=a\left(a^2-1\right)=a\left(a-1\right)\left(a+1\right)⋮3\)( Tích 3 số nguyên liên tiếp ⋮ 3)
c) \(a^5-a=a\left(a^4-1\right)=a\left(a^2-1\right)\left(a^2+1\right)\)
\(=a\left(a-1\right)\left(a+1\right)\left(a^2+1\right)\)
\(=a\left(a-1\right)\left(a+1\right)\left(a^2+5-4\right)\)
\(=a\left(a-1\right)\left(a+1\right)\left(a^2-4\right)+5a\left(a-1\right)\left(a+1\right)\)
\(=a\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)+5a\left(a-1\right)\left(a+1\right)\)
Ta có:
\(a\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)\) tích 5 số nguyên liên tiếp ⋮ 5
5a (a-1)(a+1) ⋮ 5
Suy ra: a5 - a ⋮ 5
Câu d : Ta có :
\(a^7-a\)
\(=a\left(a^6-1\right)\)
\(=a\left(a^3-1\right)\left(a^3+1\right)\)
\(=a\left(a-1\right)\left(a+1\right)\left(a^2+a+1\right)\left(a^2-a+1\right)\)
Nếu : \(a=7k\) thì \(a\) chia hết cho 7
Nếu : \(a=7k-1\) thì \(a+1\) chia hết cho 7
Nếu : \(a=7k+1\) thì \(a-1\) chia hết cho 7
Nếu : \(a=7k+2\) thì \(a^2+a+1=49k^2+35k+7\) chia hết cho 7
Nếu : \(a=7k+3\) thì \(a^2-a+1=49k^2+35k+7\) chia hết cho 7
Vì mọi trường hợp đều chia hết cho 7 .
\(\Rightarrow a^7-a⋮7\left(đpcm\right)\)
Bài 1 :
b, Ta có : \(4x^2-25-\left(2x-5\right)\left(2x+7\right)\)
\(=\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)\)
\(=\left(2x-5\right)\left(2x+5-2x-7\right)\)
\(=-2\left(2x-5\right)\)
c, Ta có : \(x^3+27+\left(x+3\right)\left(x-9\right)\)
\(=\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)\)
\(=\left(x+3\right)\left(x^2-3x+9+x-9\right)\)
\(=x\left(x+3\right)\left(x-2\right)\)
Bài 2 :
a, Để \(x^3+3x^2+3x-2⋮x+1\)
<=> \(x^3+1+3x^2+3x-3⋮x+1\)
<=> \(\left(x+1\right)^3-3⋮x+1\)
Ta thấy : \(\left(x+1\right)^3⋮x+1\)
<=> \(-3⋮x+1\)
<=> \(x+1\inƯ_{\left(3\right)}\)
<=> \(x+1=\left\{1,-1,3,-3\right\}\)
<=> \(x=\left\{0,-2,2,-4\right\}\)
Vậy ...
b, Để \(2x^2+x-7⋮x-2\)
<=> \(2x^2-8x+8+9x-15⋮x-2\)
<=> \(2\left(x-2\right)^2+9x-15⋮x-2\)
Ta thấy : \(2\left(x-2\right)^2⋮x-2\)
<=> \(9x-15⋮x-2\)
<=> \(9x-18+3⋮x-2\)
Ta thấy : \(8\left(x-2\right)⋮x-2\)
<=> \(3⋮x-2\)
<=> \(x-2\inƯ_{\left(3\right)}\)
<=> \(x-2=\left\{1,-1,3,-3\right\}\)
<=> \(x=\left\{3,1,5,-1\right\}\)
Vậy ...
Tham khảo nha bạn : http://lazi.vn/edu/exercise/xac-dinh-cac-hang-so-a-va-b-sao-cho-x4-ax-b-chia-het-cho-x2-4-x4-ax-bx-1-chia-het-cho-x2-1
a=5k+3
b=5c+2
ab=(5k+3)(5c+2)
=25kc+10k+15c+6
=5(5kc+2k+3c+1)+1 chia 5 dư 1