`a)|x|-4/5=0`

`=>|x|=4/5`

`=>x=+-4/5`

`b)|x+5,5|=-4`

Mà: `|x+5,5|>=0` với mọi x 

`-4<0=>` không có x thỏa

`d)7,5-3|5-2x|=-4,5`

`=>3|5-2x|=7,5+4,5`

`=>3|5-2x|=12`

`=>|5-2x|=4`

`=>5-2x=4` hoặc `5-2x=-4`

`=>2x=1` hoặc `2x=9`

`=>x=1/2` hoặc `x=9/2`

`a)x^3-3/4x^2y+3/4xy^2-1/8y^3`

`=x^3-3*x^2*1/2y+3*x*(1/2y)^2-(1/2y)^3`

`=(x-1/2y)^3`

`b)(x-y)^3+6(x-y)^2+12(x-y)+8`

`=(x-y)^3+3*(x-y)^2*2+3*(x-y)*2^2+2^3`

`=[(x-y)+2]^3`

`=(x-y+2)^3`

Ta có:

`(202^6*15^6+201^5*15^5)`

`=15^5*(202^6*15+201^5)`

Ta có: `15^5` là số lẻ 

`=>` Để phép tính chia hết cho 2 thì `202^6*15+201^5` là số chẵn 

Mà: `201^5` là số lẻ 

`=>202^6*15` phải là số lẻ 

Nhưng `202^6` chẵn `=>202^6*15` chẵn 

`=>202^6*15^6+201^5*15^5` chia hết cho 2 là vô lý 

`a)2x^2+7x-4`

`=(2x^2-x)+(8x-4)`

`=2x^2-x+8x-4`

`=x(2x-1)+4(2x-1)`

`=(2x-1)(x+4)`

`b)5x^2+17x+6`

`=5x^2+2x+15x+6`

`=(5x^2+2x)+(15x+6)`

`=x(5x+2)+3(5x+2)`

`=(x+3)(2x+5)`

\(A=\dfrac{x+\sqrt{x}-6}{x-9}+\dfrac{x-7\sqrt{x}+19}{x+\sqrt{x}-12}-\dfrac{x-5\sqrt{x}}{x+4\sqrt{x}}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)+3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{x-7\sqrt{x}+19}{\sqrt{x}\left(\sqrt{x}-3\right)+4\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\sqrt{x}\left(\sqrt{x}+4\right)}\\ =\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{x-7\sqrt{x}+19}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}-5}{\sqrt{x}+4}\\ =\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)}+\dfrac{x-7\sqrt{x}+19}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)}-\dfrac{\left(\sqrt{x}-5\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)}\\ =\dfrac{x-2\sqrt{x}+4\sqrt{x}-8+x-7\sqrt{x}+19-x+5\sqrt{x}+3\sqrt{x}-15}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)}\\ =\dfrac{x+3\sqrt{x}-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}-3}\)

\(1,A=\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}-\dfrac{10\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}-\dfrac{5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\\ =\dfrac{x+5\sqrt{x}-10\sqrt{x}-5\sqrt{x}+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\\ =\dfrac{x-10\sqrt{x}+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\\ =\dfrac{\left(\sqrt{x}-5\right)^2}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ =\dfrac{\sqrt{x}-5}{\sqrt{x}+5}\)

\(2,A=\dfrac{\sqrt{9}-5}{\sqrt{9}+5}=\dfrac{3-5}{3+5}=\dfrac{-2}{8}=-\dfrac{1}{4}\)

`(x+1)(5x-125)=0`

`<=>x+1=0` hoặc `5x-125=0`

`<=>x=-1` hoặc `5x=125`

`<=>x=-1` hoặc `x=125/5`

`<=>x=-1` hoặc `x=25`

Vậy: `S={-1;25}`

\(1,P=\dfrac{\sqrt{x} }{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{3}{\sqrt{x}+3}\) 

\(2,P=\dfrac{1}{3}\Rightarrow\dfrac{3}{\sqrt{x}+3}=\dfrac{1}{3}\Rightarrow\sqrt{x}=6\Rightarrow x=36\left(tm\right)\)

3, Ta có:

\(\sqrt{x}\ge0\Rightarrow\sqrt{x}+3\ge3\Rightarrow P=\dfrac{3}{\sqrt{x}+3}\le1\)

Dấu "=" xảy ra: `x=0` (tm)

Vậy: ...

`a)(x-1/2)^3`

`=x^3-3*x^2*1/2+3*x*(1/2)^2-(1/2)^3`

`=x^3-3/2x^2+3/4x-1/8`

`b)(x-2y)^3`

`=x^3-3*x^2*2y+3*x*(2y)^2-(2y)^3`

`=x^3-6x^2+12xy^2-8y^3`

`c)(x+y^2/2)^3`

`=x^3+3*x^2*y^2/2+3*x*(y^2/2)^2+(y^2/2)^3`

`=x^3+3/2x^2y^2+3/4xy^2+y^6/8` 

`A=3+3^2+3^3+...+3^100`

`3A=3(3+3^2+...+3^100)`

`3A=3^2+3^3+...+3^101`

`3A-A=(3^2+3^3+...+3^101)-(3+3^2+...+3^100)`

`2A=3^2+3^3+...+3^101-3-3^2-...-3^100`

`2A=3^101-3`

`A=(3^101-3)/2`