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b: \(=3x^{n-2+n+2}-3x^{n-2}y^{n+2}+3x^{n-2}y^{n+2}-y^{n+2+n-2}\)
\(=3x^{2n}-y^{2n}\)
c: \(=a^3+ab^2+ac^2-a^2b-abc-a^2c+a^2b+b^3+bc^2-ab^2-b^2c-abc+a^2c+b^2c+c^3-abc-bc^2-ca^2\)
\(=a^3+b^3+c^3-3acb\)
a: \(4x^2\left(3x^{n+1}-2x^n\right)\)
\(=4x^2\cdot3x^{n+1}-4x^2\cdot2x^n\)
\(=12x^{n+3}-8x^{n+2}\)
b: \(2\left(x^{2n}+2x^ny^n+y^{2n}\right)-y^n\left(4x^n+2y^n\right)\)
\(=2x^{2n}+4x^ny^n+2y^{2n}-4x^ny^n-2y^{2n}\)
\(=2x^{2n}\)
c: \(=\left(x^{3n}-y^{3n}\right)\left(x^{3n}+y^{3n}\right)\)
\(=x^{6n}-y^{6n}\)
d: \(=4^n\cdot4-3\cdot4^n=4^n\)
a: 4x2(3xn+1−2xn)4x2(3xn+1−2xn)
=4x2⋅3xn+1−4x2⋅2xn=4x2⋅3xn+1−4x2⋅2xn
=12xn+3−8xn+2=12xn+3−8xn+2
b: 2(x2n+2xnyn+y2n)−yn(4xn+2yn)2(x2n+2xnyn+y2n)−yn(4xn+2yn)
=2x2n+4xnyn+2y2n−4xnyn−2y2n=2x2n+4xnyn+2y2n−4xnyn−2y2n
=2x2n=2x2n
c: =(x3n−y3n)(x3n+y3n)=(x3n−y3n)(x3n+y3n)
=x6n−y6n=x6n−y6n
d: =4n⋅4−3⋅4n=4n
a) x^2yz - x^3y^3z + xyz^2
=xyz(x-x2y2+z)
b) 4x^3 + 24x^2 - 12xy^2
= 4(x3+6x2-3xy2)
c) x^2( m + n) - 3y^2 (m + n)
=(m+5)(x2-3y2)
d) 4x^2(x - y) + 9y^2( y - x)
= 4x2(x-y)-9y2(x-y)
= (x-y)(4x2-9y2)
e) x^2(a - b) + 2( b - a)
= x2(a-b)-2(a-b)
= (a-b)(x2-2)
g) 50x^2(x - y)^2 - 8y^2(y - x)^2
= 50x2(x-y)2+8y2(x-y)2
= 2(x-y)2(25x2+4y2)
f)10x^2( a - 2b)^2 - (x^2 + 2)(2b - a)^2
= 10x2(a-2b)2+(x2-2)(a-2b)2
= (a-2b)2(10+x2-2)
= (a-2b)2(8+x2)
h) 15am+nb - 45amb( m thuộc N*)
= 15am.15anb - 45amb
= 15amb(15an-3)
1)\(n^2\left(n-1\right)\left(n+1\right)-\left(n^2+2\right)\left(n^2-2\right)=n^2\left(n^2-1\right)-\left(n^4-4\right)=n^4-n^2-n^4+4\)
\(=-n^2+4\)
2)\(\left(y+3\right)\left(y-3\right)\left(y^2+9\right)-\left(y^2-4\right)\left(y^2+4\right)=\left(y^2-9\right)\left(y^2+9\right)-\left(y^4-16\right)\)
\(=y^4-81-y^4+16=-65\)
3)\(\left(x-2y+3\right)\left(x+2y-3\right)-\left(x-2y\right)\left(x+2y\right)=\left(x+3\right)^2-4y^2-\left(x^2-4y^2\right)\)
\(=x^2+6x+9-4y^2-x^2+4y^2=6x+9\)
4)\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
5)\(\left(a+b-c\right)^2=a^2+b^2+c^2+2ab-2bc-2ac\)
6)\(\left(a-b-c\right)^2=a^2+b^2+c^2-2ab+2bc-2ac\)
Học tốt nha bạn !
Bài 1:
Ta có:
\(a+b+c=0\\ \Leftrightarrow a^3+b^3+c^3+3\left(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b+2abc\right)=0\\ \Leftrightarrow a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\\ \Leftrightarrow a^3+b^3+c^3-3abc=0\\ \Leftrightarrow a^3+b^3+c^3=3abc\left(dpcm\right)\)
Bài 1:
a) Ta có: \(a^2-b^2-2a+2b\)
\(=\left(a-b\right)\left(a+b\right)-2\left(a-b\right)\)
\(=\left(a-b\right)\left(a+b-2\right)\)
b) Ta có: \(3x-3y-5x\left(y-x\right)\)
\(=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(x-y\right)\left(3+5x\right)\)
c) Ta có: \(\left(x-y+4\right)^2-\left(2x+3y-1\right)^2\)
\(=\left(x-y+4-2x-3y+1\right)\left(x-y+4+2x+3y-1\right)\)
\(=\left(-x-2y+5\right)\left(3x+2y+3\right)\)
d) Ta có: \(16-x^2+4xy-4y^2\)
\(=16-\left(x^2-4xy+4y^2\right)\)
\(=16-\left(x-2y\right)^2\)
\(=\left(4-x+2y\right)\left(4+x-2y\right)\)
e) Ta có: \(\left(x+3\right)^3+\left(x-3\right)^3\)
\(=\left(x+3+x-3\right)\left[\left(x+3\right)^2-\left(x+3\right)\left(x-3\right)+\left(x-3\right)^2\right]\)
\(=2x\cdot\left(x^2+6x+9-x^2+9+x^2-6x+9\right)\)
\(=2x\cdot\left(x^2+27\right)\)
f) Ta có: \(x^4+x^3+2x^2+x+1\)
\(=\left(x^4+2x^2+1\right)+\left(x^3+x\right)\)
\(=\left(x^2+1\right)^2+x\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^2+1+x\right)\)
g) Ta có: \(9x^2-3xy+y-6x+1\)
\(=\left(9x^2-6x+1\right)-\left(3xy-y\right)\)
\(=\left(3x-1\right)^2-y\left(3x-1\right)\)
\(=\left(3x-1\right)\left(3x-1-y\right)\)
h) Ta có: \(x^3-4x^2+12x-27\)
\(=\left(x^3-27\right)-4x\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+3x+9\right)-4x\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+3x+9-4x\right)\)
\(=\left(x-3\right)\left(x^2-x+9\right)\)
Bài 2:
Ta có: \(x^3+x^2z+y^2z-xyz+y^3\)
\(=\left(x^3+y^3\right)+z\left(x^2-xy+y^2\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+z\left(x^2-xy+y^2\right)\)
\(=\left(x^2-xy+y^2\right)\left(x+y+z\right)\)
\(=0\cdot\left(x^2-xy+y^2\right)=0\)(đpcm)
T a c ó : x 2 + y 2 = a 2 + b 2 ⇔ x 2 - a 2 = b 2 - y 2 ⇔ x - a x + a = b - y b + y M à x + y = a + b ⇔ x - a = b - y n ê n t a c ó x - a x + a = x - a b + y ⇔ x - a x + a - x - a b + y = 0 ⇔ x - a x + a - b - y = 0 ⇔ x - a = 0 x + a - b - y = 0 ⇔ x = a x - y = b - a
+) Với x = a thay vào x + y = a + b ta có: a + y = a + b
Suy ra y = b
Do đó: x n + y n = a n + b n
+) Với x - y = b - a suy ra x = b - a + y thay vào x + y = a + b ta có:
b - a + y + y = a + b
2y = 2a
y = a
Suy ra x - a = b - a hay x = b
Do đó: x n + y n = b n + a n = a n + b n
Vậy x n + y n = a n + b n
Đáp án cần chọn là C