Bài 2:

\(a,\dfrac{x^2-4y^2}{x^2-4xy+4y^2}=\dfrac{\left(x-2y\right)\left(x+2y\right)}{\left(x-2y\right)^2}=\dfrac{x+2y}{x-2y}\)

\(b,\dfrac{x}{x-1}+\dfrac{3}{x+1}-\dfrac{6x-4}{x^2-1}=\dfrac{x\left(x+1\right)+3\left(x-1\right)-\left(6x-4\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2+x+3x-3-6x+4}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\)

\(c,\left(\dfrac{3}{x^2-3x}+\dfrac{x}{9-3x}\right).\dfrac{x}{x+3}=\left(\dfrac{3}{x\left(x-3\right)}-\dfrac{x}{3\left(x-3\right)}\right).\dfrac{x}{x+3}=\dfrac{9-x^2}{x\left(x-3\right)}.\dfrac{x}{x+3}=\dfrac{\left(3-x\right)\left(3+x\right)}{-x\left(3-x\right)}.\dfrac{x}{x+3}=-1\)

\(d,\dfrac{6+x}{3+x}-\dfrac{3x}{3x+x^2}-2=\dfrac{6+x}{3+x}-\dfrac{3}{3+x}-2=\dfrac{6+x-3}{3+x}-2=\dfrac{3+x}{3+x}-2=1-2=-1\)

Bài 1

\(x^3-2x^2+x=0\\ \Leftrightarrow x\left(x^2-2x+1\right)=0\\ \Leftrightarrow x\left(x-1\right)^2=0\)

\(\Leftrightarrow x=0\)       hoặc \(\left(x-1\right)^2=0\\ \Leftrightarrow x-1=0\\ \Leftrightarrow x=1\)

\(\left(x+2\right)^2=\left(x+2\right)\left(x-2\right)\\ \Leftrightarrow\left(x+2\right)^2-\left(x+2\right)\left(x-2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x+2-x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)4=0\\ \Leftrightarrow4x+8=0\\ \Leftrightarrow4x=-8\\ \Leftrightarrow x=-\dfrac{8}{4}\\ \Leftrightarrow x=-2\)

d. 3(x - 5) - x(x - 5)
⇔ (x - 5)(3 - x)
⇔ x - 5 = 0 hoặc 3 - x = 0

⇔ x       =5              - x = - 3

⇔ x       =5                x  = 3

4.2:

a: x^2-x+1=x^2-x+1/4+3/4

=(x-1/2)^2+3/4>=3/4>0 với mọi x

=>x^2-x+1 ko có nghiệm

b: 3x-x^2-4

=-(x^2-3x+4)

=-(x^2-3x+9/4+7/4)

=-(x-3/2)^2-7/4<=-7/4<0 với mọi x

=>3x-x^2-4 ko có nghiệm

5:

a: x^2+y^2=25

x^2-y^2=7

=>x^2=(25+7)/2=16 và y^2=16-7=9

x^4+y^4=(x^2)^2+(y^2)^2

=16^2+9^2

=256+81

=337

b: x^2+y^2=(x+y)^2-2xy

=1^2-2*(-6)

=1+12=13

x^3+y^3=(x+y)^3-3xy(x+y)

=1^3-3*1*(-6)

=1+18=19

em chịu_{em thua}

cái gì đây?

của bạn nè.Mik lớp 5 nhưng vẫn phải học thuộc hết 

1.2 với \(x\ge0,x\in Z\)

A=\(\dfrac{2\sqrt{x}+7}{\sqrt{x}+2}=2+\dfrac{3}{\sqrt{x}+2}\in Z< =>\sqrt{x}+2\inƯ\left(3\right)=\left(\pm1;\pm3\right)\)

*\(\sqrt{x}+2=1=>\sqrt{x}=-1\)(vô lí)

*\(\sqrt{x}+2=-1=>\sqrt{x}=-3\)(vô lí
*\(\sqrt{x}+2=3=>x=1\)(TM)

*\(\sqrt{x}+2=-3=\sqrt{x}=-5\)(vô lí)

vậy x=1 thì A\(\in Z\)

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