a,( x-1)^7 = (1-x)^10
b,(2x-3)^5 = (3-2x)^9
c,2^n+2 + 2^n+1-2^n = 56
d, 2^m + 2^n = 2^m+n
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4.a)n2(n+1)+2n(n+1)=(n+1)(n2+2n)=n(n+1)(n+2)
n,(n+1),(n+2) là ba số nguyên liên tiếp nên chia hết cho 2 và 3
\(\Rightarrow\)n(n+1)(n+2) chia hết cho 6
4 Chứng minh rằng:
a)\(n^2+\left(n+1\right)+2n\left(n+1\right)\) chia hết cho 6
Ta có:
\(n^2\left(n+1\right)+2n\left(n+1\right)\)
\(=n^3+3n^2+2n\)
\(=n\left(n^2+3n+2\right)\)
\(=n\left(n+1\right)\left(n+2\right)\)
Ta thấy n , n+1 và n+2 là ba số tự nhiên liên tiếp
=> n(n+1) (n+2)\(⋮\)6
=> đpcm
b)\(\left(2n-1\right)^3-\left(2n-1\right)\) chia hết cho 8
Ta có:
\(\left(2n-1\right)^3-\left(2n-1\right)\)
\(=\left(2n-1\right)\left[\left(2n-1\right)^2-1\right]\)
\(=\left(2n-1\right)\left[\left(2n-1\right)^2-1^2\right]\)
\(=\left(2n-1\right)\left(2n-1-1\right)\left(2n-1+1\right)\)
\(=\left(2n-1\right).2\left(n-1\right).2n\)
\(=4n\left(2n-1\right)\left(n-1\right)\)
=>\(4n\left(2n-1\right)\left(n-1\right)⋮4\left(1\right)\)
Mà(2n-1)(n-1)=(n+n-1)(n-1)
=>\(\left(2n-1\right)\left(n-1\right)⋮2\left(2\right)\)
Từ (1) và (2)=> Đpcm
c)\(\left(n+7\right)^2-\left(n-5\right)^2\) chia hết cho 24
Câu hỏi của Ngoc An Pham - Toán lớp 8 | Học trực tuyến
Chúc bạn học tốt!^^
Bài 1 :
a) \(x\left(x+1\right)\left(x-1\right)-\left(x^2-1\right)\left(x+1\right)\)
\(=\left(x^3-x\right)-\left(x^3+x^2-x-1\right)\)
\(=x^3-x-x^3-x^2+x+1\)
\(=1-x^2\)
b) \(\left(x+1\right)\left(x-2\right)-\left(2x-1\right)\left(x+2\right)+2x\left(x-1\right)\)
\(=\left(x^2-x+2\right)-\left(2x^2+3x-2\right)+\left(2x^2-2x\right)\)
\(=x^2-x+2-2x^3-3x+2+2x^3+2x\)
\(=x^2-2x+4\)
\(=\left(x^2-2x.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{15}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{15}{4}\)
c) \(\left(x^2+2x-1\right)\left(x+2\right)-\left(x-1\right)\left(2x+1\right)\)
\(=\left(x^3+4x^2+3x-2\right)-\left(2x^2-x-1\right)\)
\(=x^3+4x^2+3x-2-2x^3+x+1\)
\(=-x^3+4x^2+4x-1\)
Bài 1
\(a)x\left(x+1\right)\left(x-1\right)-\left(x^2-1\right)\left(x+1\right)\\ =\left(x+1\right)\left[x\left(x-1\right)-\left(x^2-1\right)\right]\\ =\left(1+x\right)\left(x^2-x-x^2+1\right)\\ =\left(1+x\right)\left(1-x\right)\\ =1-x^2\)
\(b)\left(x+1\right)\left(x-2\right)-\left(2x-1\right)\left(x+2\right)+2x\left(x-1\right)\\ =x^2-2x+x-2-\left(2x^2+4x-x-2\right)+2x^2-2x\\ =x^2-2x+x-2-(2x^2+3x-2)+2x^2-2x\\ =x^2-2x+x-2-2x^2-3x+2+2x^2-2x\\ =x^2-6x\)
\(c)\left(x^2+2x-1\right)\left(x+2\right)-\left(x-1\right)\left(2x+1\right)\\ =x^3+2x^2+2x^2+4x-x-2-\left(2x^2+x-2x-1\right)\\ =x^3+2x^2+2x^2+4x-x-2-\left(2x^2-x-1\right)\\ =x^3+2x^2+2x^2+4x-x-2-2x^2+x+1\\ =x^3+2x^2+4x-1\)
a/Ta có: M(x)+N(x) = (2x5 - 4x3 + 2x2 + 10x - 1) + (-2x5 + 2x4 + 4x3 + x2 + x - 10)
= 2x5 - 2x5 - 4x3 + 4x3 + 2x4 + 2x2 + x2 + 10x + x -1 - 10
= 2x4 + 3x2 + 11x - 11
b/ Ta có: A(x) = N(x)-M(x) = (-2x5 + 2x4 + 4x3 + x2 + x - 10) - (2x5 - 4x3 + 2x2 + 10x - 1)
= -2x5 - 2x5 + 2x4 + 4x3 + 4x3 + x2 - 2x2 + x - 10x -10 + 1
= -2x5 + 2x4 + 8x3 - x2 - 9x -9
a: Ta có: \(\left(x^2-2x+2\right)\left(x^2-2\right)\left(x^2+2x+2\right)\left(x^2+2\right)\)
\(=\left(x^4-4\right)\left[\left(x^2+2\right)^2-4x^2\right]\)
\(=\left(x^4-4\right)\left(x^4+4x^2+4-4x^2\right)\)
\(=\left(x^4-4\right)\cdot\left(x^4+4\right)\)
\(=x^8-16\)
b: Ta có: \(\left(x+1\right)^2-\left(x-1\right)^2+3x^2-3x\left(x+1\right)\left(x-1\right)\)
\(=x^2+2x+1-x^2+2x-1+3x^2-3x\left(x^2-1\right)\)
\(=3x^2+4x-3x^3+3x\)
\(=-3x^3+3x^2+7x\)
a) 5.(x-y)-y(x-y)
= (x-y)(5-y)
b) y.(x-2)+7(2-x)
= y.(x-2) -7(x-2)
=(x-2)(y-7)
c) sửa lại đề
272.(y-1) -9x3(1-y)
=272.(y-1) + 9x3(y-1)
= (y-1)(272+9x3)
d) 4x(x-2016-x+2016)
= 4x.0
= 0
e) (2x-1)3 - 2x+1
= (2x-1)3 - (2x-1)
= (2x-1)[(2x-1)2-1]
=(2x-1)(2x-1-1)(2x-1+1)
=2x.(2x-1)(2x-2)
=4x.(x-1)(2-1)
m) 25-(3-x)2
= [5-(3-x)][5+(3-x)]
=(5-3+x)(5+3-x)
n) (x-5)2=16
=> (x-5)2=42 hoặc (x-5)2=(-4)2
=> (x-5)=4 hoặc x-5=-4
trường hợp 1
x-5=4
x= 4+5
x=9
trường hợp 2
x-5=-4
x=-4+5
x=1
vậy x=9 hoặc x=1
a) \(5.\left(x-y\right)-y.\left(x-y\right)=\left(x-y\right)\left(5-y\right)\)
b) \(y.\left(x-2\right)+7.\left(2-x\right)=\left(x-2\right)\left(y+7\right)\)
Cho mk sửa lại ở câu c là 2^n+2 + 2^n+1 - 2^n = 56 nha!