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a) \(\left(x+y\right)^5-x-y=\left(x+y\right)^5-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^4-1\right]\)
= \(\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\) #áp dụng hàng đẳng thức#
c) \(x^9-x^7-x^6-x^5+x^4+x^3+x^2+1\)nhóm vào là đc
b) \(\left(x^2+y^2\right)^3+\left(z^2-x^2\right)^3+\left(y^2+z^2\right)^3\)
=\(\left(y^2+x^2\right)\left[\left(x^2+y^2\right)^2-\left(x^2+y^2\right)\left(z^2-x^2\right)+\left(z^2-x^2\right)^2\right]+\left(y^2+z^2\right)^3\)
= \(\left(y^2+z^2\right)\left[x^4+y^4+2x^2y^2-x^2z^2+x^4-y^2z^2+x^2y^2+z^4+x^4-2x^2z^2+y^4+z^4+2y^2z^2\right]\)
=\(=\left(y^2+z^2\right)\left(2x^4+2y^4+2z^4+3x^2y^2-3x^2z^2+y^2z^2\right)\)
a) \(x^7+x^5+x^4+x^3+x^2+1\)
\(=\left(x^7+x^4\right)+\left(x^5+x^2\right)+\left(x^3+1\right)\)
\(=x^4\left(x^3+1\right)+x^2\left(x^3+1\right)+\left(x^3+1\right)\)
\(=\left(x^3+1\right)\left(x^4+x^2+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)\left(x^4+x^2+1\right)\)
a) \(100x^2-\left(x^2+25\right)^2=\left(10x\right)^2-\left(x^2+25\right)^2=\left(10x-x^2-25\right)\left(10x+x^2+25\right)\)
\(=-\left(x-5\right)^2\left(x+5\right)^2\)
b) \(\left(x-y+5\right)^2-2\left(x-y+5\right)+1=\left(x-y+5-1\right)^2=\left(x-y+4\right)^2\)
c) \(\left(x^2+4y^2-5\right)^2-16\left(x^2+y^2+2xy+1\right)\)
Có lẽ bạn ghi sai đề rồi nha.
a) \(A=5\left(x-y\right)+ax-ay=\left(a+5\right)\left(x-y\right)\)
b) \(B=a\left(x+y\right)-4x-4y=\left(x+y\right)\left(a-4\right)\)
c) \(C=xz+yz-5\left(x+y\right)=\left(x+y\right)\left(z-5\right)\)
d) \(D=a\left(x-y\right)+bx-by=\left(a+b\right)\left(x-y\right)\)
e) \(E=x\left(x+y\right)-5x-5y=\left(x-5\right)\left(x+y\right)\)
f) \(F=x^2-x-y^2-y=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\)
g) \(G=x^2-xy+x-y=x\left(x-y\right)+x-y=\left(x+1\right)\left(x-y\right)\)
A = 5(x - y) + ax - ay = 5(x - y) + a(x - y) = (a + 5)(x - y)
B = a(x + y) - 4x - 4y = a(x + y) - 4(x + y) = (a - 4)(x + y)
C = xz + yz - 5(x + y) = z(x + y) - 5(x + y) = (z - 5)(x + y)
D = a(x - y) + bx - by = a(x - y) + b(x - y) = (a + b)(x - y)
E = x(x + y) - 5x - 5y = x(x + y) - 5(x + y) = (x - 5)(x + y)
F = x2 - x - y2 - y = (x2 - y2) - (x + y) = (x2 - xy + xy - y2) - (x + y) = [x(x - y) + y(x - y)] - (x + y) = (x - y)(x + y) - (x + y) = (x + y)(x - y - 1)
G = x2 - xy + x - y = x(x - y) + (x - y) = (x + 1)(x - y)
a,(x+y)(2a-4)
b,(x+y)(a-b)
c,a(b+a)(x-5)
d,2a(a+2)(x+y)
**** cho mk nha
a) \(x^3-5x^2+8x-4\)
\(=x^3-2x^2-3x^2+6x+2x-4\)
\(=x^2\left(x-2\right)-3x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-3x+2\right)\)
\(=\left(x-2\right)\left(x^2-x-2x+2\right)\)
\(=\left(x-2\right)\left[x\left(x-1\right)-2\left(x-1\right)\right]\)
\(=\left(x-2\right)\left(x-1\right)\left(x-2\right)\)
b) \(A=10x^2-15x+8x-12+7\)
\(A=5x\left(2x-3\right)+4\left(2x-3\right)+7\)
\(A=\left(2x-3\right)\left(5x+4\right)+7\)
Dễ thấy \(\left(2x-3\right)\left(5x+4\right)⋮\left(2x-3\right)=B\)
Vậy để \(A⋮B\)thì \(7⋮\left(2x-3\right)\)
\(\Rightarrow2x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
\(\Rightarrow x\in\left\{2;1;5;-2\right\}\)
Vậy.......
Ta có
2 a 2 x – 5 b y – 5 a 2 y + 2 b x = 2 a 2 x – 5 a 2 y + 2 b x – 5 b y = a 2 2 x – 5 y + b 2 x – 5 y = a 2 + b 2 x – 5 y
Đáp án cần chọn là: D