\(Q=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ Q=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}=\dfrac{\left(\sqrt{x}+1\right)^2}{x}\)

Q=√x+1√x(√x−1):√x−1+1(√x−1)(√x+1)Q=√x+1√x(√x−1)⋅(√x−1)(√x+1)√x=(√x+1)2x

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

trên bài mink đã ẩn đi bước quy đồng!!

\(\dfrac{18}{\left(x-3\right)\left(x^2-9\right)}-\dfrac{3}{x^2-6x+9}-\dfrac{x}{x^2-9}=\dfrac{18}{\left(x-3\right)\left(x+3\right)\left(x-3\right)}-\dfrac{3}{\left(x-3\right)^2}-\dfrac{x}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{18}{\left(x-3\right)^2\left(x+3\right)}-\dfrac{3}{\left(x-3\right)^2}-\dfrac{x}{\left(x-3\right)\left(x+3\right)}=\dfrac{18-3\left(x+3\right)-x\left(x-3\right)}{\left(x-3\right)^2\left(x+3\right)}\)

\(=\dfrac{18-3x-9-x^2+3x}{\left(x-3\right)^2\left(x+3\right)}=\dfrac{9-x^2}{\left(x-3\right)^2\left(x+3\right)}=\dfrac{-\left(x-3\right)\left(x+3\right)}{\left(x-3\right)^2\left(x+3\right)}=\dfrac{-1}{x-3}\)

Bạn ơi làm ý nào đéy ??

2 ý trên à ?????

a: Xét tứ giác ADHE có

góc ADH=góc AEH=góc DAE=90 độ

nên ADHE là hình chữ nhật

b: \(HD=\sqrt{10^2-8^2}=6\left(cm\right)\)

\(S_{ADHE}=6\cdot8=48\left(cm^2\right)\)

c: Để ADHE là hình vuông thì AH là phân giác của góc BAC

=>góc B=45 độ

\(\dfrac{x+9}{x^2-9}-\dfrac{3}{x^2+3x}\)

= \(\dfrac{x+9}{\left(x-3\right).\left(x+3\right)}-\dfrac{3}{x.\left(x+3\right)}\)

=\(\dfrac{\left(x+9\right).x}{\left(x-3\right).\left(x+3\right).x}-\dfrac{3.\left(x-3\right)}{x.\left(x+3\right).\left(x-3\right)}\)

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

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

=\(\dfrac{x^2-3x+18}{3\left(x-3\right)\left(x+3\right)}\)

\(\dfrac{x+9}{x^2-9}-\dfrac{3}{x^2+3x}\)

\(=\dfrac{x+9}{\left(x-3\right)\left(x+3\right)}-\dfrac{3}{x\left(x+3\right)}\)

\(=\dfrac{x\left(x+9\right)}{x\left(x-3\right)\left(x+3\right)}-\dfrac{3\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}\)

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

\(=\dfrac{x^2+6x+9}{x\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x+3\right)^2}{x\left(x-3\right)\left(x+3\right)}=\dfrac{x+3}{x\left(x-3\right)}\)

______________________________________________________
\(\dfrac{x+1}{2x+6}-\dfrac{x-6}{2x^2+6x}\)

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

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

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

a) \(\dfrac{x+9}{x^2-9}\)-\(\dfrac{3}{x^2+3x}\) = \(\dfrac{x+9}{\left(x-3\right)\left(x+3\right)}\)-\(\dfrac{3}{x\left(x+3\right)}\)

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

                                = \(\dfrac{x^2+6x+9}{x\left(x-3\right)\left(x+3\right)}\)

                                = \(\dfrac{\left(x+3\right)^2}{x\left(x-3\right)\left(x+3\right)}\)

                                = \(\dfrac{x+3}{x\left(x-3\right)}\)

Bài này trong sách là trên mạng có ấy c lên ytb shears cho nhanh