Answer:

Bài 2:

\(4x^2-x=0\)

\(\Rightarrow x\left(4x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\4x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{4}\end{cases}}\)

\(x\left(x-1\right)-x+1=0\)

\(\Rightarrow x\left(x-1\right)-\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x-1\right)=0\)

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

\(\Rightarrow x=1\)

Bài 3:

a) Tại \(x=\frac{3}{2}\) thay vào A ta được

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

\(A=\frac{2.\frac{3}{2}+2}{2.\frac{3}{2}-1}=2\)

b) \(B=\frac{1-2x}{2x}+\frac{2x}{2x-1}+\frac{1}{2x-4x^2}\)

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

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

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

\(=\frac{2\left(2x-1\right)}{\left(2x-1\right)2x}\)

\(=\frac{1}{x}\)

(x+3)^2 -x^2+9=0 

(x+3)^2-(x^2-9)=0

(x+3)^2-(x-3)(x+3)=0

(x+3)(x+3-x+3)=0

6(x+3)=0

suy ra x+3=0 

x=-3     k

Answer:

\(\left(x+3\right)^2-x^2+9=0\)

\(\Rightarrow\left(x+3\right)\left(x+3\right)-x^2+9=0\)

\(\Rightarrow x^2+3x+3x+9-x^2+9=0\)

\(\Rightarrow6x+18=0\)

\(\Rightarrow6x=-18\)

\(\Rightarrow x=-3\)

bài ni dể quá nên em ko biết

hihi

xl e mới lp 6 thui