ta có B=3-2x+2x+1
=>B=4
b)với x=2015 thì B=4

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

Nếu \(x\ge-\frac{1}{2}\Rightarrow B=3-2x+2x+1=4\)

Nếu \(x< -\frac{1}{2}\Rightarrow B=3-2x-2x-1=4-4x\)

b, x = 2015 tức là \(x>-\frac{1}{2}\)

Vậy với x = 2015 thì B = 4

\(\sqrt{1+4x+4x^2}=\sqrt{\left(2x+1\right)^2}=\left|2x+1\right|\)

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

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

\(=\sqrt{x}-1+\sqrt{x}\)(vì x>0)

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

Vậy....

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

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

Nếu \(x\ge1\) thì \(A=\sqrt{x}-1+\sqrt{x}=2\sqrt{x}-1\)

Nếu 0 < x < 1 thì \(A=1-\sqrt{x}+\sqrt{x}=1\)

b, \(x=2\frac{1}{4}\Rightarrow x=\frac{9}{4}>1\)

Vậy \(x=2\frac{1}{4}\) thì 

\(A=2\sqrt{x}-1=2.\sqrt{\frac{9}{4}}-1=2.\frac{3}{2}-1=2\)

\(\sqrt{8-2\sqrt{15}}\)

\(=\sqrt{5-2\sqrt{5.3}+3}\)

\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)

\(=\sqrt{5}-\sqrt{3}\)

\(\sqrt{x^2-2x+1}=x^2-1\)\(,DKXD:x\ge-1\)

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

\(\Leftrightarrow x-1=\left(x-1\right)\left(x+1\right)\)

\(\Leftrightarrow\left(x-1\right):\left(x-1\right)=x+1\)

\(\Leftrightarrow x+1=1\)

\(\Leftrightarrow x=0\)(thỏa mãn ĐKXĐ)

Vậy.......

ĐK: \(x^2\ge1\)

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

\(\Leftrightarrow\left|x-1\right|=\left(x-1\right)\left(x+1\right)\)

Nếu \(x\ge1\Rightarrow x-1=\left(x-1\right)\left(x+1\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow x=1\left(x\ge1\right)\)

Nếu \(x< 1\Rightarrow1-x=\left(x-1\right)\left(x+1\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)+x-1=0\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\Leftrightarrow x=-2\left(x\le-1\right)\)

Vậy \(x\in\left\{1;-2\right\}\)

\(TXD:\hept{\begin{cases}x\supseteq1\\-\sqrt{\frac{1}{2}}\subseteq x\subseteq\sqrt{\frac{1}{2}}\end{cases}}\)

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

<=> \(1-2x^2=x^2-2x+1\)

\(3x^2-2x=0\)

\(x\left(3x-2\right)=0\orbr{\begin{cases}x=0\left(KTM\right)\\x=\frac{2}{3}\left(TM\right)\end{cases}}\)

\(Vậy\)\(x=\frac{2}{3}\)\(là\)\(nghiệm\)\(phương\)\(trình\).

\(\sqrt{x^2-4}-x+2=0\)                   ĐK : x \(\ge\)2 ; x \(\le\)- 2

\(\Leftrightarrow\sqrt{x^2-4}=x-2\)

\(\Leftrightarrow(\sqrt{x^2-4})^2=(x-2)^2\)

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

\(\Leftrightarrow4x=4+4\)

\(\Leftrightarrow4x=8\)

\(\Leftrightarrow x=2\) ( thỏa mãn ĐK )