\(A=\frac{\left(x-1\right)-5\sqrt{x-1}+6}{\sqrt{x-1}\cdot\left(\sqrt{x-1}-3\right)}=\frac{\left(\sqrt{x-1}-2\right)\cdot\left(\sqrt{x-1}-3\right)}{\sqrt{x-1}\cdot\left(\sqrt{x-1}-3\right)}\)    Đk x\(\ne\) 1;10

\(A=\frac{\sqrt{x-1}-2}{\sqrt{x-1}}=1-\frac{2}{\sqrt{x-1}}\) 

\(\frac{x^3-x^2-x-2}{x^5-3x^4+4x^3-5x^2+3x-2}\)

\(=\frac{x^3-2x^2+x^2-2x+x-2}{x^5-2x^4-x^4+2x^3+2x^3-4x^2-x^2+2x+x-2}\)

\(=\frac{\left(x^3-2x^2\right)+\left(x^2-2x\right)+\left(x-2\right)}{\left(x^5-2x^4\right)-\left(x^4-2x^3\right)+\left(2x^3-4x^2\right)-\left(x^2-2x\right)+\left(x-2\right)}\)

\(=\frac{x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)}{x^4\left(x-2\right)-x^3\left(x-2\right)+2x^2\left(x-2\right)-x\left(x-2\right)+\left(x-2\right)}\)

\(=\frac{\left(x-2\right)\left(x^2+x+1\right)}{\left(x-2\right)\left(x^4-x^3+2x^2-x+1\right)}=\frac{x^2+x+1}{x^4-x^3+2x^2-x+1}\)

a, x.(x-y) +y.(x+y)

=x²-xy+xy+y²

=x²+y²

b, (x²-5).(2x+3)-2x.(x-3)

=2x³+3x²-10x-15-2x²+6x

=2x³-x²-4x-15

c, 8-5x.(x+2) +4 .( x-2) . (x+1) +2.( x+2)+ 2.(x-2)+10

=8-5x²-10x+4.(x²+x-2x-2)+2x+4+2x-4+10

=18-6x-5x²+4x²+4x-8x-8

=10-10x-x²

a)  x.(x-y) +y.(x+y)

= x^2-xy+xy+y^2

=x^2+y^2

b) (x^2-5).(2x+3)-2x(x-3)

= 2x^3+3x^2-10x-15-2x^2+6x

=2x^3+x^2-4x-15

Bài 1: 

a: \(P=\left(\dfrac{x-2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+2}{\left(x+1\right)^2}\right)\cdot\dfrac{\left(x-1\right)^2\cdot\left(x+1\right)^2}{4}\)

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

\(=\dfrac{-2x}{1}\cdot\dfrac{x-1}{4}=-\dfrac{x\left(x-1\right)}{2}\)

b: Để \(\dfrac{P-4}{5}=x\) thì P-4=5x

=>P=5x+4

\(\Leftrightarrow-\dfrac{x\left(x-1\right)}{2}=5x+4\)

=>-x²+x=10x+8

=>x²-x=-10x-8

=>x²+9x+8=0

=>x=-8(nhận) hoặc x=-1(loại)

Bài 1:

\(a,\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)\)

\(=x^6-3x^4+3x^2-1-x^6+1\)

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

\(b,\left(x^4-3x^2+9\right)\left(x^2+3\right)-\left(3+x^2\right)^3\)

\(=x^6+27-27-27x^2-9x^4-x^6\)

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

Bài 5:

\(A=x^2-2x+1\)

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

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

Với mọi giá trị của x ta có:

\(\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2-2\ge-2\)

Vậy Min A = -2

Để A = -2 thì \(x-1=0\Rightarrow x=1\)

b, \(B=4x^2+4x+5\)

\(=\left(4x^2+4x+1\right)+4\)

\(=\left(2x+1\right)^2+4\)

Với mọi giá trị của x ta có:

\(\left(2x+1\right)^2\ge0\Rightarrow\left(2x+1\right)^2+4\ge4\)

Vậy Min B = 4

Để B = 4 thì \(2x+1=0\Rightarrow2x=-1\Rightarrow x=-\dfrac{1}{2}\)

c, \(C=2x-x^2-4\)

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

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

Với mọi giá trị của x ta có:

\(\left(x-1\right)^2\ge0\Rightarrow-\left(x-1\right)^2\le0\Rightarrow-\left(x-1\right)^2-3\le-3\)Vậy Max C = -3

để C = -3 thì \(x-1=0\Rightarrow x=1\)

\(a\text{)}.\:\left(x^2+2\right)^2-\left(x+2\right)\left(x-2\right)\left(x^2+4\right)\\ =x^4+4x^2+4-\left(x^2-4\right)\left(x^2+4\right)\\ =x^4+4x^2+4-x^4+16\\ =4x^2+20\)

\(b\text{)}.\:\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\\ =\left(x+1+x-1\right)\left(x+1-x+1\right)-3\left(x^2-1\right)\\ =4x-3x^2+3\)

a.\(=\dfrac{8}{15}\times\dfrac{1}{2}=\dfrac{4}{15}\)

b.\(=\dfrac{1}{3}\times\dfrac{1}{2}-\dfrac{1}{5}\times\dfrac{1}{2}=\dfrac{1}{6}-\dfrac{1}{10}=\dfrac{1}{15}\)

a.(1/3+1/5) x 1/2

=8/15 x 1/2=8/30=4/15

b)(1/3-1/5) x 1/2=2/15 x 1/2=1/15