a, \(B=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\)ĐK : \(x>0;x\ne1\)

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

b,Ta có  \(x=3+2\sqrt{2}=\left(\sqrt{2}+1\right)^2\)

\(\Rightarrow\sqrt{x}=\sqrt{2}+1\)

Vậy \(B=\sqrt{2}+1-1=\sqrt{2}\)

a) Ta có: \(B=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\)

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

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

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

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

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

\(b,\) Thay \(x=2015\) ta có:

\(B=4\)

a, B=3-2x+\(\sqrt{4x^2+4x+1}\) =3-2x+\(\sqrt{\left(2x+1\right)^2}\) =3-2x+2x+1=4 b; Ta co B=4=4+0x thay x=2015 =>B=4+0x=4+0.2015=4 vay x=2015=>B=4

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

\(=\dfrac{-4x^2-8x}{\left(x+2\right)}\cdot\dfrac{1}{2x-1}=\dfrac{-4x\left(x+2\right)}{\left(x+2\right)\left(2x-1\right)}=\dfrac{-4x}{2x-1}\)

b: |x|=3

=>x=3 hoặc x=-3

Khi x=3 thì \(B=\dfrac{-4\cdot3}{2\cdot3-1}=\dfrac{-12}{5}\)

Khi x=-3 thì \(B=\dfrac{-4\cdot\left(-3\right)}{2\cdot\left(-3\right)-1}=\dfrac{12}{-7}=\dfrac{-12}{7}\)

a: \(B=\dfrac{3x\left(2x-3\right)-4\left(2x+3\right)-4x^2+23x+12}{\left(2x-3\right)\left(2x+3\right)}\cdot\dfrac{2x+3}{x+3}\)

\(=\dfrac{6x^2-9x-8x-12-4x^2+23x+12}{2x-3}\cdot\dfrac{1}{x+3}\)

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

b: 2x^2+7x+3=0

=>(2x+3)(x+2)=0

=>x=-3/2(loại) hoặc x=-2(nhận)

Khi x=-2 thì \(A=\dfrac{2\cdot\left(-2\right)}{-2-3}=\dfrac{-4}{-7}=\dfrac{4}{7}\)

d: |B|<1

=>B>-1 và B<1

=>B+1>0 và B-1<0

=>\(\left\{{}\begin{matrix}\dfrac{2x+2x-3}{2x-3}>0\\\dfrac{2x-2x+3}{2x-3}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-3< 0\\\dfrac{4x-3}{2x-3}>0\end{matrix}\right.\Leftrightarrow x< \dfrac{3}{4}\)

a)\(A=2x+\sqrt{1-4x+4x^2}\)

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

\(A=2x+\left|1-2x\right|\)

\(A=2x+\left|2x-1\right|\)

\(A=4x-1\)

b)\(x=\frac{1}{4}\Leftrightarrow A=4.\frac{1}{4}-1\)

\(\Leftrightarrow A=-1\)

a) \(A=2x+\sqrt{1-4x+4x^2}\)

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

\(A=2x+|1-2x|\)

\(A=2x+|2x-1|\)

\(A=4x-1\)

b) \(x=\frac{1}{4}\Leftrightarrow A=4.\frac{1}{4}-1\)

\(\Leftrightarrow A=-1\)

Đề không hiển thị biểu thức B. Bạn xem lại.

giải giúp em với mấy anh chị

\(A=\dfrac{2}{x-1}\sqrt{\dfrac{\left(x-1\right)^2}{4x^2}}=\dfrac{2}{x-1}\left|\dfrac{x-1}{2x}\right|=\dfrac{\left|x-1\right|}{\left(x-1\right)\left|x\right|}\)

\(B=\left(x^2-4\right)\sqrt{\dfrac{9}{x^2-4x+4}}=\dfrac{3\left(x^2-4\right)}{\left|x-2\right|}\)

a) Ta có: \(A=\dfrac{2}{x-1}\cdot\sqrt{\dfrac{x^2-2x+1}{4x^2}}\)

\(=\dfrac{2}{x-1}\cdot\dfrac{x-1}{2x}\)

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

b) Ta có: \(\left(x^2-4\right)\cdot\sqrt{\dfrac{9}{x^2-4x+4}}\)

\(=\dfrac{\left(x-2\right)\left(x+2\right)\cdot3}{\left(x-2\right)^2}\)

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