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

Answer:

Bài 1:

\(x^2-xy+5x-5y\)

\(=\left(x^2-xy\right)+\left(5x-5y\right)\)

\(=x\left(x-y\right)+5\left(x-y\right)\)

\(=\left(x+5\right)\left(x-y\right)\)

\(x^2-4x+4\)

\(=x^2-2x.2+2^2\)

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

Bài 2:

\(\left(x-3\right)^2+x\left(2x+7\right)\)

\(=x^2-6x++2x^2+7x\)

\(=3x^2+x+9\)

\(\frac{x+1}{x-2}+\frac{8-7x}{x^2-2x}+\frac{2}{x}\left(ĐK:x\ne0;x\ne2\right)\)

\(=\frac{x+1}{x-2}+\frac{8-7x}{x\left(x-2\right)}+\frac{2}{x}\)

\(=\frac{x\left(x+1\right)+8-7x+2\left(x-2\right)}{x\left(x-2\right)}\)

\(=\frac{x^2+x+8-7x+2x-4}{x\left(x-2\right)}\)

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

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

Bài 3:

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

\(\Rightarrow3x-x^2+x^2+10x+25=4\)

\(\Rightarrow x^2-x^2+3x+10x=4-25\)

\(\Rightarrow13x=-21\)

\(\Rightarrow x=\frac{-21}{13}\)

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

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

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

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

Bài 4:

Số tiền mua trà sữa khi chưa giảm giá

\(20000.5=100000\) đồng

Số tiền được giảm

\(100000.10:100=10000\) đồng

Vậy An cần trả

\(100000-10000=90000\) đồng