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\(\dfrac{a+2b-ab-2b^2}{1-b}=\dfrac{\left(a+2b\right)-b\left(a+2b\right)}{1-b}=\dfrac{\left(1-b\right)\left(a+2b\right)}{1-b}=a+2b\)
\(A=\left(\frac{4x}{x^2-4}+\frac{2x-4}{x+2}\right).\frac{x+2}{2x}+\frac{2}{2-x}\\=\left(\frac{4x}{x^2-4}+\frac{\left(2x-4\right)\left(x-2\right)}{x^2-4}\right)\frac{x+2}{2x}+\frac{2}{2-x}=\left(\frac{4x}{x^2-4}+\frac{2x^2-4x-4x+8}{x^2-4}\right) \frac{x+2}{2x}+\frac{2}{2-x}\)
\(=\left(\frac{4x+2x^2-8x+8}{x^2-4}\right).\frac{x+2}{2x}+\frac{2}{2-x}\\ =\frac{2x\left(x+2\right)-8\left(x-1\right)}{x^2-4}.\frac{x+2}{2x}+\frac{2}{2-x}\)
a) \(a^4+b^4\)
\(=\left(a^2\right)^2+\left(b^2\right)^2\)
\(=\left(a^2-b^2\right).\left(a^2+b^2\right)\)
b) Tương tự
c) \(a^5+b^5\)
\(=\left(\sqrt{a}^5\right)^2+\left(\sqrt{b}^5\right)^2\)
\(=\left(\sqrt{a}^5+\sqrt{b}^5\right).\left(\sqrt{a}^5-\sqrt{b}^5\right)\)
a: \(=a\left(y^2-2yz+z^2\right)\)
\(=a\left(y-z\right)^2\)
b: \(=\left(x^2+6xy+9y^2\right)-16\)
=(x+3y)^2-16
=(x+3y+4)(x+3y-4)
c: \(=7\left(a-b\right)+\left(a-b\right)\left(a+b\right)\)
=(a-b)(7+a+b)
d: \(36x^4-13x^2\)
=x^2*36x^2-x^2*13
=x^2(36x^2-13)
f: x^2-2xy+y^2-49
=(x-y)^2-49
=(x-y-7)(x-y+7)
e: 2x^3-18x
=2x(x^2-9)
=2x(x-3)(x+3)
g: 2x+2y-x^2-xy
=2(x+y)-x(x+y)
=(x+y)(2-x)
h: (x^2+3)^2+16
=x^4+6x^2+25
=x^4+10x^2+25-4x^2
=(x^2+5)^2-4x^2
=(x^2-2x+5)(x^2+2x+5)
a) \(=6a-3+15-5a=a+12\)
b) \(=25x-12x+4+35-14x=-x+39\)
d) \(=2ab+8a^2-b^2-4ab+2ab-6a^2=2a^2-b^2\)
e) \(=x+x^2-x^3+x^4-x^5+1+x-x^2+x^3-x^4=-x^5+2x+1\)
f) \(=6y^3-3y^2+y-y+y^2-y^3-y^2+y=5y^3-3y^2+y\)
a) 3( 2a -1) +5( 3-a)
= 3. 2a -3.1 +5. 3- 5.a
= 6a -3+ 15-5a
=(6a -5a )+ (-3+ 15)
b) 25x - 4(3x - 1) +7(5 - 2x)
= 25x -4.3x + 4.1 + 7.5 - 7.2
=25x - 12x + 4 +35 - 14x
= (25x-12x-14x)+(4+35)
= -x=39
c) -12x3 -x1-2x-18x2
= -36x-x-2x-36x
= -75x
d) (2a-b)(b+4a)+2a(b-3a)
= 2ab+2a4a-bb-b4a+2ab-2a3b
= 2ab+8a2-b2-4ab+2ab-6a2
=(2ab-4ab+2ab)+(8a2-6a2)-b2
= 2a2-b2
e) (x+1)(2+x-x2+x3-x4)
= (x+1)(2-2x)
= x2-x2x+1.2-1.2x
=(2x-2x)-2x2+2
= -2x2+2
\(=\left(a^2-b^2\right)\left(a^4+a^2b^2+b^4\right)\\ =a^6+a^4b^2+a^2b^4-a^4b^2-a^2b^4-b^6\\ =a^6-b^6\)
Cảm ơn b