Áp dụng BĐT phụ \(a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Leftrightarrow\left(a-b\right)^2\ge0\)

\(A\ge\dfrac{1}{2}\left(x+y+\dfrac{1}{x}+\dfrac{1}{y}\right)^2\ge\dfrac{1}{2}\left(x+y+\dfrac{4}{x+y}\right)^2=\dfrac{1}{2}\left(1+\dfrac{4}{1}\right)^2=\dfrac{25}{2}\)

Dấu "=" \(x=y=\dfrac{1}{2}\)

e: \(\left(a^2-1\right)\left(a^2+a+1\right)\left(a^2-a+1\right)\)

\(=\left(a^3-1\right)\left(a^3+1\right)\)

\(=a^6-1\)

b: Ta có: \(\left(1+x+x^2\right)\left(1-x\right)\left(1+x\right)\left(1-x+x^2\right)\)

\(=\left(1-x^3\right)\left(1+x^3\right)\)

\(=1-x^6\)

c: \(\left(a+1\right)\left(a+2\right)\left(a^2+4\right)\left(a-1\right)\left(a^2+1\right)\left(a-2\right)\)

\(=\left(a+1\right)\left(a-1\right)\left(a^2+1\right)\left(a+2\right)\left(a-2\right)\left(a^2+4\right)\)

\(=\left(a^2-1\right)\left(a^2+1\right)\left(a^2-4\right)\left(a^2+4\right)\)

\(=\left(a^4-1\right)\left(a^4-16\right)\)

\(=a^8-17a^4+16\)

d: \(\left(a^3+3\right)\left(a^6-3a^3+9\right)\)

\(=\left(a^3\right)^3+3^3\)

\(=a^9+27\)

a: \(A=\dfrac{1}{2a-1}\cdot\sqrt{5a^2}\cdot\left|2a-1\right|\)

\(=\dfrac{2a-1}{2a-1}\cdot a\sqrt{5}=a\sqrt{5}\)(do a>1/2)

b: \(A=\dfrac{\sqrt{x-1-2\sqrt{x-1}+1}}{\sqrt{x-1}-1}+\dfrac{\sqrt{x-1+2\sqrt{x-1}+1}}{\sqrt{x-1}+1}\)

\(=\dfrac{\left|\sqrt{x-1}-1\right|}{\sqrt{x-1}-1}+\dfrac{\sqrt{x-1}+1}{\sqrt{x-1}+1}\)

\(=\dfrac{\sqrt{x-1}-1}{\sqrt{x-1}-1}+1=1+1=2\)

c:

\(=\dfrac{a+b}{b^2}\cdot\dfrac{ab^2}{a+b}=a\)

d: Sửa đề: \(A=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)

\(=\left(1+\sqrt{a}+a+\sqrt{a}\right)\cdot\left(\dfrac{1}{1+\sqrt{a}}\right)^2\)

\(=\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)^2}=1\)

e:

\(A=\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^4}}\)

\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y}-1}{\left(x-1\right)^2}=\dfrac{1}{x-1}\)

f:

\(A=\sqrt{\dfrac{m}{\left(1-x\right)^2}\cdot\dfrac{4m\left(1-2x+x^2\right)}{81}}\)

\(=\sqrt{\dfrac{m}{\left(x-1\right)^2}\cdot\dfrac{4m\left(x-1\right)^2}{81}}\)

\(=\sqrt{\dfrac{4m^2}{81}}=\dfrac{2m}{9}\)

1: =(8+a^3)(8-a^3)=64-a^6

2: =x^3-6x^2+12x-8-x(x^2-1)+6x^2-18x

=x^3-6x-8-x^3+x

=-5x-8

3: =x^3+3x^2+3x+1-x^3+1-3x^2-3x

=2

1) \(3x\left(x-1\right)+5\left(x-1\right)\)

\(=\left(x-1\right)\left(3x+5\right)\)

2) \(4x(x-2y)-8y(2y-x)\)

\(=4x\left(x-2y\right)+8y\left(x-2y\right)\)

\(=\left(4x+8y\right)\left(x-2y\right)\)

\(=4\left(x+2y\right)\left(x-2y\right)\)

3) \(a^2\left(x-1\right)+b^2\left(1-x\right)\)

\(=a^2\left(x-1\right)-b^2\left(x-1\right)\)

\(=\left(a^2-b^2\right)\left(x-1\right)\)

\(=\left(a-b\right)\left(a+b\right)\left(x-1\right)\)

4) \(3x\left(x-a\right)+4a\left(a-x\right)\)

\(=3x\left(x-a\right)-4a\left(x-a\right)\)

\(=\left(x-a\right)\left(3x-4a\right)\)

5) \(5x\left(x-y\right)^2+10y^2\left(y-x\right)^2\)

\(=5x\left(x-y\right)^2+10y^2\left(x-y\right)^2\)

\(=\left(5x+10y^2\right)\left(x-y\right)^2\)

\(=5\left(x+2y^2\right)\left(x-y\right)^2\)

6) \(3x\left(x-3\right)^2+9\left(3-x\right)^2\)

\(=3x\left(x-3\right)^2+9\left(x-3\right)^2\)

\(=\left(3x+9\right)\left(x-3\right)^2\)

\(=3\left(x+3\right)\left(x-3\right)^2\)

7) \(x\left(m-a\right)^2-y\left(a-m\right)^2\)

\(=x\left(a-m\right)^2-y\left(a-m\right)^2\)

\(=\left(x-y\right)\left(a-m\right)^2\)

8) \(6y^2\left(x-1\right)^2+9y\left(1-x\right)^2\)

\(=6y^2\left(x-1\right)^2+9y\left(x-1\right)^2\)

\(=\left(6y^2+9x\right)\left(x-1\right)^2\)

\(=3\left(2y^2+3x\right)\left(x-1\right)^2\)

#Ayumu