\(e,=\dfrac{\left(3+\sqrt{2}\right)\left(2\sqrt{2}+1\right)}{7}-\sqrt{\dfrac{\left(\sqrt{2}+1\right)^2}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}}\\ =\dfrac{7\sqrt{2}+7}{7}-\dfrac{\sqrt{2}+1}{1}=\sqrt{2}+1-\sqrt{2}-1=0\)

\(f,=\sqrt{\dfrac{\left(2\sqrt{3}-3\right)^2}{\left(2\sqrt{3}-3\right)\left(2\sqrt{3}+3\right)}}\left(2+\sqrt{3}\right)\\ =\dfrac{\left(2\sqrt{3}-3\right)\left(2+\sqrt{3}\right)}{\sqrt{3}}\\ =\dfrac{\sqrt{3}}{\sqrt{3}}=1\)

\(h,=\sqrt{\dfrac{\left(3\sqrt{5}-1\right)\left(2\sqrt{5}-3\right)}{20-9}}\left(\sqrt{2}+\sqrt{10}\right)\\ =\sqrt{\dfrac{2\left(33-11\sqrt{5}\right)}{11}}\left(\sqrt{5}+1\right)\\ =\sqrt{\dfrac{22\left(3-\sqrt{5}\right)}{11}}\left(\sqrt{5}+1\right)\\ =\sqrt{6-2\sqrt{5}}\left(\sqrt{5}+1\right)=\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=4\)

a: \(A=\dfrac{x\left(x+2\right)}{\left(x-2\right)^2}:\dfrac{x^2-4+x+6-x^2}{x\left(x-2\right)}\)

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

c: A<0

=>x-2<0

=>x<2

d: B nguyên

=>x^2-4+4 chia hết cho x-2

=>x-2 thuộc {1;-1;2;-2;4;-4}

=>x thuộc {3;1;4;6}

c) \(=\left(5a-\dfrac{1}{3}\right)^2\)

d) \(=\left(y-\dfrac{1}{3}\right)^2\)

e) \(=\left(2x-y+1\right)^2\)

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

g) \(=\left(2xy^2-3\right)^2\)

\(c,=\left(5a-\dfrac{1}{3}\right)^2\\ d,=\left(y^4-\dfrac{1}{3}\right)^2\\ e,=\left(2x-y+1\right)^2\\ f,=\left(2x-4y\right)^2+4\left(x-2y\right)+1=\left(2x-4y+1\right)\\ g,=\left(2xy^2-3\right)^2\)

c: \(\Leftrightarrow\dfrac{x}{2012}-1+\dfrac{x+1}{2013}-1+\dfrac{x+2}{2014}-1+\dfrac{x+3}{2015}-1+\dfrac{x+4}{2016}-1=0\)

=>x-2012=0

hay x=2012

d: \(\Leftrightarrow\dfrac{x-90}{10}-1+\dfrac{x-76}{12}-2+\dfrac{x-58}{14}-3+\dfrac{x-36}{16}-4+\dfrac{x-15}{17}-5=0\)

=>x-100=0

hay x=100

b: =>(x+1)(x-1)-(x+3)(x-3)=2x^2+6x

=>2x^2+6x=x^2-1-x^2+9=8

=>2x^2+6x-8=0

=>x^2+3x-4=0

=>(x+4)(x-1)=0

=>x=-4 hoặc x=1(loại)

a: =>x^3+2x-2x(x^2+1)=0

=>x^3+2x-2x^3-2x=0

=>-x^3=0

=>x=0(nhận)

c: =>(x-2)(x+2)-(x+5)^2=x^2-8

=>x^2-4-x^2-10x-25=x^2-8

=>x^2-8=-10x-29

=>x^2+10x+21=0

=>(x+3)(x+7)=0

=>x=-3 hoặc x=-7

a 13+22-20-(-133)+8

=13+22-20+133-8

=133-13+22+8-20

=120-30-20

=70

Gfgdvcgvcgdvyvgdyfvfgdggfy₫6Fsvgxdgsgscgcvcgxvgc  dvgcxvgsfsgg#vgcg#gsv v#ggd*#gsvggdy6g6egy₫ỳ₫udgyfh6&h₫hfyvhvihfifojfhvufhfuifuf

a: \(\left(x-1.2\right)^2=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1.2=2\\x-1.2=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3.2\\x=-0.8\end{matrix}\right.\)

b: Ta có: \(\left(x+1\right)^3=-125\)

\(\Leftrightarrow x+1=-5\)

hay x=-6

c) 3^(4-x)=27

3^(4-x) = 3^3

4-x = 3

x = 1

d) \(​​\dfrac{5x+2}{6}\) +\(\dfrac{3-4x}{2}\) = 2-\(\dfrac{x+7}{3}\) 

=>5x+2+3(3-4x)=12-2(x+7)

5x+2+9-12x=12-2x-14

-5x=-13

x=\(\dfrac{13}{5}\) 

e) \(\dfrac{-20}{9}x +4=\dfrac{8}{3}x-40\)

=>-20x+36=24x-360

-44x=-396

x=9

f) 3x(2x-5)-4X+10=0

6X² -15X-4X+10=0

2x(3x-2)-5(3x-2)=0

(3x-2)(2x-5)=0

\(\left[\begin{array}{} Biểu thức (3x-2=0)\\ Biểu thức (2x-5=0) \end{array} \right.\)\(\left[\begin{array}{} (x=\dfrac{2}{3})\\ (x=\dfrac{5}{2}) \end{array} \right.\)

j) \(\dfrac{x-45}{55}+\dfrac{x-47}{53}=\dfrac{x-55}{45}+\dfrac{x-53}{47}\)

\(\dfrac{x-45}{55}-1+\dfrac{x-47}{53}-1=\dfrac{x-55}{45}-1+\dfrac{x-53}{47}-1\)

\(\dfrac{x-100}{55}+\dfrac{x-100}{53}=\dfrac{x-100}{45}+\dfrac{x-100}{47}\)

\(\dfrac{x-100}{55}+\dfrac{x-100}{53}-\dfrac{x-100}{45}-\dfrac{x-100}{47}=0\)

(x-100)(\(\dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}\))=0

=> x-100=0(\(\dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}\) >0)

=> x= 100