ý a sai sai bạn ạ

a,\(\sqrt{23-8\sqrt{7}}-\sqrt{7}=\sqrt{16-8\sqrt{7}+7}-\sqrt{7}=\sqrt{\left(4-\sqrt{7}\right)^2}-\sqrt{7}=\left|4-\sqrt{7}\right|-\sqrt{7}=4-\sqrt{7}-\sqrt{7}=4\)

câu b hình như thiếu đề thì phải bạn xem lại giúp mình

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-1}+1=2\\\sqrt{x-1}+1=-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-1}=1\Leftrightarrow x-1=1\Leftrightarrow x=2\\\sqrt{x-1}=-3\left(vl\right)\end{cases}}\)

Vậy phương trình có tập nghiệm  \(S=\left\{2\right\}\)

Ta có pt \(\Leftrightarrow\sqrt{a-2+4\sqrt{a-2}+4}+\sqrt{a-2-4\sqrt{a-2}+4}=2\)

<=> \(\left|\sqrt{a-2}+2\right|+\left|\sqrt{a-2}-2\right|=4\Leftrightarrow\left|\sqrt{a-2}+2\right|+\left|2-\sqrt{a-2}\right|=4\)

Áp dụng BĐT về giá trị tuyệt đối, ta có \(\left|\sqrt{a-2}+2\right|+\left|2-\sqrt{a-2}\right|\ge\left|\sqrt{a-2}+2+2-\sqrt{a-2}\right|=4\)

Dấu = xảy ra <=> \(2\ge\sqrt{a-2}\ge0\Leftrightarrow6\ge a\ge2\)

Vậy ...

^_^

\(\dfrac{\sqrt{a}-2}{a+2\sqrt{a}}+\dfrac{8}{a-4}\)

\(=\dfrac{\sqrt{a}-2}{\sqrt{a}\left(\sqrt{a}+2\right)}+\dfrac{8}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)

\(=\dfrac{\left(\sqrt{a}-2\right)^2+8\sqrt{a}}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)\cdot\sqrt{a}}\)

\(=\dfrac{\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)\cdot\sqrt{a}}=\dfrac{\sqrt{a}+2}{\sqrt{a}\left(\sqrt{a}-2\right)}\)

\(=\dfrac{\sqrt{a}+2}{a-2\sqrt{a}}\)

a,\(ĐK:a\ge1\)

\(\sqrt{a-1+2\sqrt{a-1}+1}+\sqrt{a-1-2\sqrt{a-1}+1}\)

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

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

Với \(\sqrt{a-1}\ge1\Leftrightarrow a\ge2\) thì \(\left|\sqrt{a-1}-1\right|=\sqrt{a-1}-1\)

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

Với \(0\le\sqrt{a-1}< 1\)\(\Leftrightarrow1\le a< 2\) thì 

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

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

Câu b tương tự:\(\sqrt{a+4\sqrt{a-2}+2}+\sqrt{a-4\sqrt{a-2}+2}\)

                         \(=\sqrt{a-2+4\sqrt{a-2}+4}+\sqrt{a-2-4\sqrt{a-2}+4}\)

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

                          \(=\left|\sqrt{a-2}+2\right|+\left|\sqrt{a-2}-2\right|\)

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

b)\(=\sqrt{a-2+4\sqrt{a-2}+4}+\sqrt{a-2-4\sqrt{a-2}+4}\)
\(=\sqrt{\left(\sqrt{a-2}+2\right)^2}+\sqrt{\left(\sqrt{a-2}-2\right)^2}=\sqrt{a-2}+2+\sqrt{a-2}-2=2\sqrt{a-2}\)

\(\sqrt{a+4\sqrt{a-2}+2}+\sqrt{a-4\sqrt{a-2}+2}=4\)

\(\Leftrightarrow\sqrt{a-2+4\sqrt{a-2}+4}+\sqrt{a-2-4\sqrt{a-2}+4}=4\)

\(\Leftrightarrow\sqrt{\left(\sqrt{a-2}+2\right)^2}+\sqrt{\left(\sqrt{a-2}-2\right)^2}=4\)

\(\Leftrightarrow\left|\sqrt{a-2}+2\right|+\left|\sqrt{a-2}-2\right|=4\)

Ta thấy :

\(VT=\left|\sqrt{a-2}+2\right|+\left|2-\sqrt{a-2}\right|\ge\left|\sqrt{a-2}+2+2-\sqrt{a-2}\right|=4\)

\(\Rightarrow VT\ge4\)

Dấu "=" xảy ra khi \(\left(\sqrt{a-2}+2\right)\left(2-\sqrt{a-2}\right)\ge0\Rightarrow a\le4\)

Kém theo ĐKXĐ ta tìm đc \(2\le a\le4\)

ai giúp mk với, các CTV các god đâu oy, mk cần gấp lắm

Phải có ĐK là \(a\le2\le6\) bạn nhé

Ta có

\(\sqrt{a+4\sqrt{a-2}+2}+\sqrt{a-4\sqrt{a-2}+2}\)

\(=\sqrt{a-2+4\sqrt{a-2}+4}+\sqrt{a-2-4\sqrt{a-2}+4}\)

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

\(=\sqrt{a-2}+2+\left|\sqrt{a-2}-2\right|\)

\(=\sqrt{a-2}+2+2-\sqrt{a-2}=4\)