bạn viết thế này khó nhìn quá

nhìn hơi đau mắt nhá bạn hoa mắt quá

b: \(A=\dfrac{2-1}{3\cdot2}=\dfrac{1}{6}\)

a) Ta có: \(P=\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\cdot\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)

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

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

\(=\dfrac{x\left[x^2-4x+4+4x\right]}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{x^2-x-2}{x^2}\)

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

\(=\dfrac{x+1}{2x}\)

b) Thay \(x=\dfrac{1}{2}\) vào P, ta được:

\(P=\dfrac{1}{2}+1=\dfrac{3}{2}\)

\(B=\left(3x-1\right)^2-\left(x+7\right)^2-2\left(2x-5\right)\left(2x+5\right)\)

\(=9x^2-6x+1-\left(x^2+14x+49\right)-2\left(4x^2-25\right)\)

\(=9x^2-6x+1-x^2-14x-49-8x^2+50\)

\(=-20x+2\)

Thay x=1/5 vào B, ta được:

\(B=-20\cdot\dfrac{1}{5}+2=-4+2=-2\)

em học lớp 7 nên không biết 

đúng cho em nhé

chịu

a: \(A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\)

\(=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\sqrt{x}-1\)

a) \(A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\)

                        Đk: \(x>0\) và \(x\ne1\)

\(\Rightarrow A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\)

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

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

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

b) Thay \(x=3+2\sqrt{2}\) vào A ta được:

  \(A=\sqrt{3+2\sqrt{2}}-1=\sqrt{\left(\sqrt{2}+1\right)^2}-1\)

      \(=\sqrt{2}+1-1=\sqrt{2}\)

(Vì \(\sqrt{2}+1>0\Rightarrow\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\))