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

\(\Rightarrow\left|A\right|=\left|\dfrac{\left|2x-1\right|}{2\left(2x-1\right)}\right|=\dfrac{\left|2x-1\right|}{2\left|2x-1\right|}=\dfrac{1}{2}\)

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

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

\(=\left[{}\begin{matrix}-\dfrac{\left(2x-1\right)}{2\left(2x-1\right)}=-\dfrac{1}{2}\left(x< \dfrac{1}{2}\right)\\\dfrac{2x-1}{2\left(2x-1\right)}=\dfrac{1}{2}\left(x\ge\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left|A\right|=0.5\)

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

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

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

\(\left|A\right|=\dfrac{2x-1}{2\left(2x-1\right)}\) \(\Rightarrow\left|A\right|=\dfrac{1}{2}=0,5\left(x\ne0,5\right)\)

a) A \(=\frac{x^2-4}{2}\cdot\sqrt{\frac{2^2}{\left(x-2\right)^2}}\) \(=\frac{x^2-4}{2}\cdot\left|\frac{2}{x-2}\right|\)

+ Với x < 2 ta có \(A=\frac{x^2-4}{2}\cdot\frac{2}{2-x}\)

\(A=\frac{\left(x+2\right)\left(x-2\right)}{2-x}=-\left(x+2\right)\)

+ Với x > 2 ta có : \(A=\frac{x^2-4}{2}\cdot\frac{2}{x-2}\)

\(A=\frac{\left(x-2\right)\left(x+2\right)}{x-2}=x+2\)

câu b và c tương tự

ấn vào đúng cho mk đi mk ân cho bạn ok

I not sure for this answer if have any trouble you can ask me

a)\(\sqrt{x^2-4x+5}\ge\forall x\)

\(\Leftrightarrow\sqrt{x^2-4x+4+1}\)

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

mà \(\sqrt{\left(x+1\right)^2}\ge0\forall x\)

nên \(\sqrt{\left(x+1\right)^2}+1>0\forall x\)

câu a chưa rõ đề, bắt chứng minh nhưng ko biết \(\ge\) cái j ms đc chứ ạ ?

Trả lời:

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

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

Đẳng thức đã cho trở thành:

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

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

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

\(=t+1+t-1\)

\(=2t\)

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

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

b, \(\sqrt{2x+\sqrt{4x-1}}+\sqrt{2x-\sqrt{4x-1}}=\sqrt{6}\)

Đặt \(\sqrt{4x-1}=t\)\(\Rightarrow2x=\frac{t^2+1}{2}\)

Đẳng thức đã cho trở thành:

\(VT=\sqrt{\frac{t^2+1}{2}+t}+\sqrt{\frac{t^2+1}{2}-t}\)

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

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

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

\(=\frac{2t}{\sqrt{2}}\)

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

14:

\(A=\sqrt{-4x^2+4x+7}\)

\(=\sqrt{-\left(4x^2-4x-7\right)}\)

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

\(=\sqrt{-\left(2x-1\right)^2+8}< =\sqrt{8}=2\sqrt{2}\)

Dấu = xảy ra khi 2x-1=0

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

13:

\(a+b+c>=\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\)

=>\(2a+2b+2c-2\sqrt{ab}-2\sqrt{bc}-2\sqrt{ac}>=0\)

=>\(\left(a-2\sqrt{ab}+b\right)+\left(b-2\sqrt{bc}+c\right)+\left(a-2\sqrt{ac}+c\right)>=0\)

=>\(\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt{b}-\sqrt{c}\right)^2+\left(\sqrt{a}-\sqrt{c}\right)^2>=0\)(luôn đúng)