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2T
4 tháng 9 2019
a) \(25^{n+1}-25^n=25^n\left(25-1\right)=25^n.4⋮25.4=100\)
b) \(n^2\left(n-1\right)-2n\left(n-1\right)=\left(n^2-2n\right)\left(n-1\right)\)
\(=n\left(n-1\right)\left(n-2\right)\)
Tích 3 số tự nhiên liên tiếp chia hết cho 6 nên \(n^2\left(n-1\right)-2n\left(n-1\right)⋮6\)
c) \(n^3-n=n\left(n^2-1\right)=\left(n-1\right)n\left(n+1\right)\)
Tích 3 số tự nhiên liên tiếp chia hết cho 6 nên \(n^3-n⋮6\)
17 tháng 8 2019
Bài 1:
a) \(x^3-5x^2+8x-4\)
\(=x^3-4x^2+4x-x^2+4x-4\) \(=x\left(x^2-4x+4\right)-\left(x^2-4x+4\right)\)\(=\left(x-1\right)\left(x-2\right)^2\)
b) Ta có: \(\frac{A}{M}=\frac{10x^2-7x-5}{2x-3}=5x+4+\frac{7}{2x-3}\)
Với \(x\in Z\)thì \(A⋮M\)khi \(\frac{7}{2x-3}\in Z\)\(\Rightarrow7⋮\left(2x-3\right)\)\(\Rightarrow2x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
\(\Rightarrow=\left\{1;5;\pm2\right\}\)thì khi đó \(A⋮M\)
R
17 tháng 8 2019
Các bài làm này có đúng ko ạ, ai đó duyệt giúp em, em cảm ơn.
Bài 1:
a)x3-5x2+8x-4=x3-4x2+4x-x2+4x-4
=x(x2-4x-4)-(x2-4x+4)
=(x-1) (x-2)2
b)Xét:
\(\frac{a}{b}-\frac{10x^2-7x-5}{2x-3}\)
=\(5x+4+\frac{7}{2x-3}\)
Với x thuộc Z thì A /\ B khi \(\frac{7}{2x-3}\) thuộc Z => 7 /\ (2x-3)
Mà Ư(7)={-1;1;-7;7} => x=5;-2;2;1 thì A /\ B
c)Biến đổi \(\frac{x}{y^3-1}-\frac{x}{x^3-1}=\frac{x^4-x-y^4+y}{\left(y^3-1\right)\left(x^3-1\right)}\)
=\(\frac{\left(x^4-y^4\right)\left(x-y\right)}{xy\left(y^2+y+1\right)\left(x^2+x+1\right)}\)(do x+y=1=>y-1=-x và x-1=-y)
=\(\frac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)-\left(x-y\right)}{xy\left[x^2y^2+y^2x+y^2+xy^2+xy+y+x^2+x+1\right]}\)
=\(\frac{\left(x-y\right)\left(x^2+y^2-1\right)}{xy\left[x^2y^2+xy\left(x+y\right)+x^2+y^2+xy+2\right]}\)
=\(\frac{\left(x-y\right)\left(x^2-x+y^2-y\right)}{xy\left[x^2y^2+\left(x+y\right)^2+2\right]}=\frac{\left(x-y\right)\left[x\left(x-1\right)+y\left(y-1\right)\right]}{xy\left(x^2y^2+3\right)}\)
=\(\frac{\left(x-y\right)\left[x\left(-y\right)+y\left(-x\right)\right]}{xy\left(x^2y^2+3\right)}=\frac{\left(x-y\right)\left(-2xy\right)}{xy\left(x^2y^2+3\right)}\)
=\(\frac{-2\left(x-y\right)}{x^2y^2+3}\)Suy ra điều phải chứng minh
Bài 2 )
a)(x2+x)2+4(x2+x)=12 đặt y=x2+x
y2+4y-12=0 <=>y2+6y-2y-12=0
<=>(y+6)(y-2)=0 <=> y=-6;y=2
>x2+x=-6 vô nghiệm vì x2+x+6 > 0 với mọi x
>x2+x=2 <=> x2+x-2=0 <=> x2+2x-x-2=0
<=>x(x+2)-(x+2)=0 <=>(x+2)(x-1) <=> x=-2;x-1
Vậy nghiệm của phương trình x=-2;x=1
b)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}+\frac{x+4}{2005}+\frac{x+5}{2004}\)\(+\frac{x+6}{2003}\)
=\(\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)+\left(\frac{x+4}{2005}+1\right)\)\(+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)
<=>\(\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}\)\(+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)
<=>\(\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}\)\(-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)
Nhờ OLM xét giùm em vs ạ !
3 tháng 9 2021
a) (5x - 2y) (x2 - xy + 1)
=5x^3 − 5x^2y + 5x − 2x^2y +2xy^2 − 2y
=5x^3 − 7x^2y + 2xy^2 + 5x − 2y
b) (x - 1) (x + 1) (x + 2)
=(x^2−1)(x+2)
=x^3+2x^2−x−2
phần c) mình ko biết nha
3 tháng 9 2021
a) (5x - 2y) (x2 - xy +1)
= 5x3-5x2y+5x-2x2y+2xy2+2y
= 5x3 - 7x2y+2xy2+5x+2y
b) (x - 1) (x + 1) (x + 2)
= (x\(^2\) - 1)(x + 2)
= x3 +2x2 - x - 2
c) \(\frac{1}{2}\)x2y2 (2x+y)(2x-y)
= \(\frac{1}{2}\)x2y2 (4x2 - y2)
= 2x4y2 - \(\frac{1}{2}\)x2y4
22 tháng 12 2019
ban oi a^2+b^2+c^2= a^2+b^2+c^2 là chuyện đương nhiên mà bạn
A
3
18 tháng 8 2019
x2 - 6x + 10
= x2 - 2.x.3 + 32 + 1
= ( x - 3 )2 + 1
Vì \(\left(x-3\right)^2\ge0\forall x\)
1 > 0
=> \(\left(x-3\right)^2+1\ge0\forall x\) ( đpcm )
Study well
22 tháng 12 2019
bạn kiếm kiểu gì cx ko có ai giải đâu, đề này sai r, nãy mình sửa mới đúng
a7-a chứ nhỉ :))
a^7 - a = a.(a^6 - 1)= a.(a^3 -1).(a^3+1)=a.(a-1).(a+1).(a^2-a+1).(a^2+a+1)
Đến đây xét các TH a= 7k , 7k+1.... thay vào một trong mấy thừa số vừa tách để CM chia hết cho 7
có một cách phân tích ra thành tích 7 số nguyên liên tiếp nhưng tui ngại đánh máy :v
Có ai làm được ko ?
Tui làm rồi nè -_-
Sorry
ko có đâu sai rồi
\(a^7-a=a.\left(a^6-1\right)=a.\left(a^3-1\right).\left(a^3+1\right)\)
Th1: a chia hết cho 7 \(\Rightarrow a^7-a⋮7\)
Th2: a chia 7 dư 1, dư 2, dư 4
\(a\equiv1\left(mod7\right)\Rightarrow a^3\equiv1\left(mod7\right)\)(1)
\(a\equiv2\left(mod7\right)\Rightarrow a^3\equiv2^3=7+1\left(mod7\right)\Rightarrow a^3\equiv1\left(mod7\right)\)(2)
\(a\equiv4\left(mod7\right)\Rightarrow a^3\equiv4^3=64=63+1\left(mod7\right)\Rightarrow a^3\equiv1\left(mod7\right)\)(3)
Từ (1), (2), (3) =>a3-1 chia hết cho 7(với mọi a chia 7 dư 1,2,4)
=> a.(a3-1).(a3+1) chia het cho 7 <=> a7-a chia hết cho 7
Th3: a chia 7 dư 3,5,6
\(a\equiv3\left(mod7\right)\Rightarrow a^3\equiv3^3=27=28-1\left(mod7\right)\Rightarrow a^3\equiv-1\left(mod7\right)\)(4)
\(a\equiv5\left(mod7\right)\Rightarrow a^3\equiv5^3=125=126-1\left(mod7\right)\Rightarrow a^3\equiv-1\left(mod7\right)\)(5)
\(a\equiv6\left(mod7\right)\Rightarrow a^3\equiv6^3=216=217-1\left(mod7\right)\Rightarrow a^3\equiv-1\left(mod7\right)\)(6)
=> Từ (4), (5), (6) =>a3+1 chia hết cho 7(với mọi a chia 7 dư 5,6,7)
=> a.(a3-1).(a3+1) chia het cho 7 <=> a7-a chia hết cho 7
vậy với mọi a thuộc Z thì a7-a chia hết cho 7
cách khác gọn hơn ne`:
\(a^7-a=a.\left(a^6-1\right)=a.\left(a^3-1\right).\left(a^3+1\right)\)
\(=a.\left(a-1\right).\left(a^2+a+1\right).\left(a^2-a+1\right).\left(a+1\right)\)
\(=a.\left(a-1\right).\left(a+1\right).\left(a^4+a^2+1\right)=a.\left(a-1\right).\left(a+1\right).\left(a^4-4a^2-9a^2+36+14a^2-35\right)\)
\(=a.\left(a-1\right).\left(a+1\right).\left(a-2\right).\left(a+2\right).\left(a-3\right).\left(a+3\right)+\left(14a^2-35\right).a.\left(a-1\right).\left(a+1\right)\)
Vì trong 7 số nguyên liên tiếp có 1 số chia hết cho 7 => tích của 7 số nguyên liên tiếp chia hết cho 7(1)
Vì 14a2-35 chia hết cho 7 => (14a2-35).a.(a-1).(a+1) chia hết cho 7(2)
từ (1) và (2) => a7-a chia hết cho 7
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