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a. A = (a + b)3 - (a - b)3
A = \(\left[\left(a+b\right)-\left(a-b\right)\right]\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
A = (a + b - a + b)\(\left[a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right]\)
A = 2b(a2 + a2 + a2 + 2ab - 2ab + b2 - b2 + b2)
A = 2b(3a2 + b2)
A = 6a2b + 2b3
a) \(\cdot\left(m+n\right)^2-\left(m-n\right)^2+\left(m+n\right)\left(m-n\right)\)
\(=\left(m+n+m-n\right)\left(m+n-m+n\right)+\left(m+n\right)\left(m-n\right)\)
\(=\left(2m\cdot2n\right)+m^2-n^2\)
\(=4mn+m^2-n^2\)
b) \(\left(a+b\right)^2-\left(a-b\right)^2-2a^3\)
\(=\left(a+b+a-b\right)\left(a+b-a+b\right)-2a^3\)
\(=2ab-2a^3\)
c) \(\left(2x+1\right)^2+\left(2x-1\right)^2+2\left(4x^2-1\right)\)
\(=\left(2x+1\right)^2+2\left(2x+1\right)\left(2x-1\right)+\left(2x-1\right)^2\)
\(=\left(2x+1+2x-1\right)^2\)
\(=\left(4x\right)^2=16x^2\)
d) \(\left(a+b+c\right)^2-2\left(a+b+c\right)\left(b+c\right)+\left(b+c\right)^2\)
\(=\left(a+b+c-b-c\right)^2=a^2\)
xin lỗi mk ghi sai đề ở bài :d) (a+b+c)^2-2(a+b+c)(b+c)+(b+c)^2
a: TH1: x>-2
\(D=\dfrac{x\left(x+2\right)\left(x-1\right)}{x\left(x+2\right)-\left(x-2\right)\left(x+2\right)}=\dfrac{x\left(x+2\right)\left(x-1\right)}{\left(x+2\right)\cdot2}=\dfrac{x\left(x-1\right)}{2}\)
TH2: x<-2
\(D=\dfrac{x\left(x+2\right)\left(x-1\right)}{-x\left(x+2\right)-\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{\left(x+2\right)\cdot x\cdot\left(x-1\right)}{-\left(x+2\right)\left(x+x-2\right)}=\dfrac{-x\left(x-1\right)}{2\left(x-1\right)}=\dfrac{-x}{2}\)
b: TH1: x>-2
D=x(x-1)/2
Vì x(x-1) chia hết cho 2
nên D luôn là số nguyên nếu x>-2
TH2: x<-2
Để D nguyên thì -x chia hết cho 2
=>x chia hết cho 2
c: Khi x=6 thì \(D=\dfrac{6\left(6-1\right)}{2}=3\cdot5=15\)
\(1,\)
\(a,\) Với \(n=1\Leftrightarrow5+2\cdot1+1=8⋮8\left(đúng\right)\)
Giả sử \(n=k\left(k\ge1\right)\Leftrightarrow5^k+2\cdot3^{k-1}+1⋮8\)
Với \(n=k+1\)
\(5^n+2\cdot3^{n-1}+1=5^{k+1}+2\cdot3^k+1\\ =5^k\cdot5+2\cdot3^k+1\\ =5^k\cdot2+2\cdot3^k+5^k\cdot3+1\\ =2\left(5^k+3^k\right)+5^k+2\cdot5^{k-1}+1+2\cdot3^{k-1}-2\cdot3^{k-1}\\ =2\left(5^k+3^k\right)+\left(5^k+2\cdot3^{k-1}+1\right)-2\left(3^{k-1}+5^{k-1}\right)\)
Vì \(5^k+3^k⋮\left(5+3\right)=8;5^{k-1}+3^{k-1}⋮\left(5+3\right)=8;5^k+2\cdot3^{k-1}+1⋮8\) nên \(5^{k+1}+2\cdot3^k+1⋮8\)
Theo pp quy nạp ta được đpcm
\(b,\) Với \(n=1\Leftrightarrow3^3+4^3=91⋮13\left(đúng\right)\)
Giả sử \(n=k\left(k\ge1\right)\Leftrightarrow3^{k+2}+4^{2k+1}⋮13\)
Với \(n=k+1\)
\(3^{n+2}+4^{2n+1}=3^{k+3}+4^{2k+3}\\ =3^{k+2}\cdot3+16\cdot4^{2k+1}\\ =3^{k+2}\cdot3+3\cdot4^{2k+1}+13\cdot4^{2k+1}\\ =3\left(3^{k+2}+4^{2k+1}\right)+13\cdot4^{2k+1}\)
Vì \(3^{k+2}+4^{2k+1}⋮13;13\cdot4^{2k+1}⋮13\) nên \(3^{k+3}+4^{2k+3}⋮13\)
Theo pp quy nạp ta được đpcm
\(1,\)
\(c,C=6^{2n}+3^{n+2}+3^n\\ C=36^n+3^n\cdot9+3^n\\ C=\left(36^n-3^n\right)+\left(3^n\cdot9+2\cdot3^n\right)\\ C=\left(36^n-3^n\right)+3^n\cdot11\)
Vì \(36^n-3^n⋮\left(36-3\right)=33⋮11;3^n\cdot11⋮11\) nên \(C⋮11\)
\(d,D=1^n+2^n+5^n+8^n\)
Vì \(1^n+2^n+5^n⋮\left(1+2+5\right)=8;8^n⋮8\) nên \(D⋮8\)
Bài 1:
b:
x=9 nên x+1=10
\(M=x^{10}-x^9\left(x+1\right)+x^8\left(x+1\right)-x^7\left(x+1\right)+...-x\left(x+1\right)+x+1\)
\(=x^{10}-x^{10}-x^9+x^9+x^8-x^8-x^7+...-x^2-x+x+1\)
=1
c: \(N=\left(1+2+2^2+2^3+2^4\right)+2^5\left(1+2+2^2+2^3+2^4\right)+2^{10}\left(1+2+2^2+2^3+2^4\right)\)
\(=31\left(1+2^5+2^{10}\right)⋮31\)
1) A=4*\(\frac{10^{2n}-1}{9}\) B=\(2\cdot\frac{10^{n+1}-1}{9}\) C=\(8\cdot\frac{10^n-1}{9}\)
đặt 10^n=X => A+B+C+7=(4*x^2-4+2*10*x-2+8x-8+63)/9=(4x^2+28x+49)/9
=> A+B+C+7=\(\frac{\left(2x+7\right)^2}{3^2}\)
2) = 4mn((m^2-1)-(n^2-1))=4mn(m+1)(m-1)-4mn(n-1)(n+1)
mà m,n nguyên => m-1,m,m+1 và n-1,n,n+1 là 3 số nguyên liên tiếp nên chia hết cho 6
do đó 4mn(m^2-n^2) chia hết 6*4=24
a)
\(\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\)
\(\Leftrightarrow\dfrac{2-x}{2002}+1=\dfrac{1-x}{2003}+1+\dfrac{-x}{2004}+1\)
\(\Leftrightarrow\dfrac{2004-x}{2002}-\dfrac{2004-x}{2003}-\dfrac{2004-x}{2004}=0\)
\(\Leftrightarrow\left(2004-x\right)\left(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\right)\)
\(\Leftrightarrow2004-x=0\) (vì \(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\ne0\))
\(\Leftrightarrow x=2004\)
S={2004}
a) C = c + d + 2 ( c − d ) 3 = ( 3 c − d ) 3 .
b) D = m − n ( n + p ) 3 = ( m − 2 n − p ) 3 .