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a) x5 : xn
= x5-n
=> 5\(\ge\)n => n = 1;2;3;4;5
b) x2n:x5
= x2n-5
=> 2n - 5 \(\ge\) 0
=> 2n \(\ge\) 5
con c) chép thiếu đúng ko
d) xn+2y3 : x5y3
= (xn+2 : x5)(y3:y3)
= xn+2 : x5
= xn+2-5 = xn-3
=> n - 3\(\ge\)0
=> n \(\ge\) 3
e) x3n+1:x7
= x3n+1-7
= x3n-6
=> 3n -6 \(\ge\) 0
=> 3n \(\ge\) 6
=> n \(\ge\) 2
f) xnyn+3 : x6y10
= xn-6yn+3-10
= xn-6yn-7
=> n \(\ge\) 7
tik cho mik đi nha
a, n= 1;2;3;4
b, n= bội của 5
tương tự nha! làm sao cho số mũ chia hết cho nha là được.
a)Để xn+2.yn+1 chia hết x5.y6 thì
\(\Leftrightarrow\hept{\begin{cases}n+2\ge5\\n+1\ge6\end{cases}\Leftrightarrow\hept{\begin{cases}n\ge3\\n\ge5\end{cases}\Leftrightarrow}n\ge3}\)
Vậy n=0;1;2;3(vì n thuộc N)
a: \(=n^3+2n^2+3n^2+6n-n-2-n^3+5\)
\(=5n^2+5n+3⋮̸5\)
b:\(=6n^2+30n+n+5-6n^2+3n-10n+5\)
\(=24n+10=2\left(12n+5\right)⋮2\)
d: \(=4x^2y^2-2x^2y+2xy^2-xy-4x^2y^2+xy\)
\(=-2\left(x^2y-xy^2\right)⋮2\)
b) x8 +7x4+16
= x8+8x4-x4+16
= (x8+8x4+16) - x4
=(x4+4)2-x4
= (x4+4+x2)(x4+4-x2)
c) x5+x-1
= x5 - x4+x3+x4-x3+x2-x2+x-1
= x3(x2-x+1) + x2(x2-x+1) - (x2-x+1)
= (x2-x+1)(x3+x2 -1)
d)x7+x2+1
=x7-x+x2 +x+1
= x (x6-1) + (x2+x+1)
= x(x3-1)(x3+1) + (x2+x+1)
= x(x3+1)(x-1)(x2+x+1)+(x2+x+1)
= (x2+x+1)[x(x3+1)(x-1) +1]
= (x2+x+1)(x5-x4+x2-x+1)
= x (x-1)(x2+x+1)
e) x5+x4+1
= x5+x4+x3 - x3+1
= x3(x2+x+1) - (x-1)(x2+x+1)
= (x2+x+1)(x3-x+1)
f) x8+x+1
= x8-x2+x2+x+1
= x2(x6-1)+(x2+x+1)
= x2(x3-1)(x3+1) +(x2+x+1)
= (x5+x2)(x-1)(x2+x+1) +(x2+x+1)
= (x2+x+1)(x6-x5+x3-x2+1)
Bài làm :
\(a,\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)+10\)
\(=8x+16-5x^2-10x+\left(4x-8\right)\left(x+1\right)+2\left(x^2-2^2\right)+10\)
\(=8x+16-5x^2-10x+4x^2+4x-8x-8+2x^2-8+10\)
\(=\left(8x-10x+4x-8x\right)+\left(-5x^2+4x^2+2x^2\right)+\left(16-8-8+10\right)\)
\(=-6x+x^2+10\)
a)\(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)+10\)\(=8x+16-5x^2-2+4x-8x-8+2x-4x-4+10\)\(=\left(8x+4x-8x+2x-4x\right)+\left(16-2-8-4+10\right)+5x^2\)
\(=2x+12+5x^2\)
b)\(4\left(x-1\right)\left(x+5\right)-\left(x+2\right)\left(x+5\right)-3\left(x-1\right)\left(x+2\right)\)
\(=4x-4x-20-\left[x^2+5x+2x+10\right]-3\left[x^2+2x-1x-2\right]\)
\(=4x-4x-20-x^2-5x-2x-10-3x^2-6x+3x+6\)
\(=\left(4x-4x-5x-2x-6x+3x\right)+\left(-20-10+6\right)+\left(-x^2-3x^2\right)\)
\(=-10x-24-4x^2\)
c)\(\left(x^{2n}+x^ny^n+y^{2n}\right)\left(x^n-y^n\right)\left(x^{3n}+y^{3n}\right)\)
Xét tích \(\left(x^{2n}+x^ny^n+y^{2n}\right)\left(x^n-y^n\right)\Leftrightarrow\left(x^n\right)^3-\left(y^n\right)^3=x^{3n}-y^{3n}\)
Thay vào bt đã cho ta có \(\left(x^{3n}-y^{3n}\right)\left(x^{3n}+y^{3n}\right)\)
\(\Leftrightarrow\left(x^{3n}\right)^2-\left(y^{3n}\right)^2=x^{6n}-y^{6n}\)