Ta có: M= \(\frac{1+2x}{1+\sqrt{1+2x}}+\frac{1-2x}{1-\sqrt{1-2x}}\)= \(\frac{\left(1+2x\right)\left(1-\sqrt{1+2x}\right)+\left(1-2x\right)\left(1+\sqrt{1+2x}\right)}{1-\left(1-2x\right)}\)=\(\frac{1-\sqrt{1+2x}+2x-2x\sqrt{1+2x}+1+\sqrt{1+2x}-2x-2x\sqrt{1+2x}}{2x}\)

=\(\frac{2}{2x}=\frac{1}{x}\)

Với x=\(\frac{\sqrt{3}}{4}\)=> M=\(\frac{4}{\sqrt{3}}\)

bài này dài phết @@

M = \(\dfrac{4}{\sqrt{3}}\)

Ta có \(\sqrt{\left(1+2x\right)^2}\)= 1 + 2x   (1)

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

(1) +(2) = 2

Có \(\sqrt{1+2x}.\sqrt{1-2x}\)= \(\sqrt{1-4x^2}=\frac{1}{2}\)    (3)

Từ (1),(2),(3) \(\Rightarrow\)\(\left(\sqrt{1+2x}+\sqrt{1-2x}\right)^2\)= 3 \(\Rightarrow\)\(\sqrt{1+2x}+\sqrt{1-2x}\)=\(\sqrt{3}\)    (4)

                            \(\left(\sqrt{1+2x}-\sqrt{1-2x}\right)^2\)= 1 \(\Rightarrow\) \(\sqrt{1+2x}-\sqrt{1-2x}\)= 1             (5)

Có M= \(\frac{\left(1+2x\right).\left(1-\sqrt{1-2x}\right)+\left(1-2x\right).\left(1+\sqrt{1+2x}\right)}{\left(1+\sqrt{1+2x}\right).\left(1-\sqrt{1-2x}\right)}\)

Xét TS= \(1-\sqrt{1-2x}+2x-2x.\sqrt{1-2x}+1+\sqrt{1+2x}-2x-2x.\sqrt{1+2x}\)

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

Thay (4), (5) và x vào TS ta có TS= \(2+1-2.\frac{\sqrt{3}}{4}.\sqrt{3}=\frac{3}{2}\)          (6)

Xét MS=\(1-\sqrt{1-2x}+\sqrt{1+2x}-\sqrt{1-4x^2}\)

Thay (5) và x vào MS ta có MS= \(1+1-\frac{1}{2}\)=\(\frac{3}{2}\)                                   (7)

Từ (6),(7) ta có giá trị của M= 1

Ta có: \(Q=\frac{1-2x}{1-\sqrt{1-2x}}+\frac{1-2x}{1-\sqrt{1-2x}}\)

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

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

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

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

Thế \(x=\frac{\sqrt{3}}{4}\) ta được: \(Q=\left(\frac{1-2.\frac{\sqrt{3}}{4}}{\sqrt{\frac{\sqrt{3}}{4}}}\right)^2=0,04145188433\)

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

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

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

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

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

Thế \(x=\frac{\sqrt{3}}{4}\) ta được: \(\frac{1+2.\frac{\sqrt{3}}{4}}{2+\sqrt{\frac{\sqrt{3}}{4}}}+\frac{1-2.\frac{\sqrt{3}}{4}}{\sqrt{\frac{\sqrt{3}}{4}}}=1\)

P/s: Nếu làm chưa chuẩn, mong mọi người sửa chữa giúp em chứ đừng tk sai ạ. Em cảm ơn

Đặt \(\hept{\begin{cases}\sqrt{1+2x}=a\\\sqrt{1-2x}=b\end{cases}}\)

\(\Rightarrow a^2+b^2=2;ab=\frac{1}{2};a-b=1\)

Ta có:

\(B=\frac{a^2}{1+a}+\frac{b^2}{1-b}=\frac{a^2-a^2b+b^2+b^2a}{1-ab+a-b}\)

\(=\frac{a^2+b^2-ab\left(a-b\right)}{1-ab+\left(a-b\right)}=\frac{2-\frac{1}{2}.1}{1-\frac{1}{2}+1}=1\)

Mấy cái khác bỏ qua đi tập trung chỗ e sai thôi nha. Theo em suy ra thì

\(\sqrt{1+2x}=\sqrt{1^2+2.x.1+\left(\sqrt{x}\right)^2}\)

\(\Leftrightarrow1+2x=1+2x+x\)

\(\Leftrightarrow x=0\)(sai)

Thứ 2:

\(\frac{1+2\frac{\sqrt{3}}{4}}{2+\sqrt{\frac{\sqrt{3}}{4}}}+\frac{1-2\frac{\sqrt{3}}{4}}{\sqrt{\frac{\sqrt{3}}{4}}}\ne1\)