\(VT=x+2\sqrt{2x-4}\)

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

\(=\left(\sqrt{x-2}+\sqrt{2}\right)^2=VP\left(\text{đ}pcm\right)\)

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

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

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

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

a) có \(\sqrt{x^2+2x+5}=\sqrt{x^2+2x+1+4}=\sqrt{\left(x+1\right)^2+4}\)Vì \(\left(x+1\right)^2\ge0\forall x\in R\rightarrow\left(x+1\right)^2+4\ge0+4=4\forall x\in R\)

\(\Rightarrow\sqrt{x^2+2x+5}\ge\sqrt{0+4}=\sqrt{4}=2\) (đpcm)

Dấu "=" xảy ra khi và chỉ khi \(x=-1.\)

b) \(x>\sqrt{x}\Leftrightarrow x^2>x\Leftrightarrow x^2-x>0\)

\(\Leftrightarrow x\left(x-1\right)\ge0\)

Vì \(x>1\rightarrow x>0;x-1>0\)

\(\Rightarrow x\left(x-1\right)>0\) với mọi \(x>1\)

hay \(x>\sqrt{x}\) (đpcm)

Chúc bạn học tốt!

a)

√(x^2+2x+5)>2

<=>x^2+2x+5>4

<=>x^+2x+1>0

(x+1)^2 > 0 =>dpcm

b)

x>1<=>x^2>x

x(x-1)>0

luon dung

Q=\(\frac{\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}}{\sqrt{x+\sqrt{2x-1}-\sqrt{x-\sqrt{2x-1}}}}\)(x\(\ge2\))

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

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

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

vậy \(Q=\sqrt{2x-1}với\)\(x\ge2\)

~ happy new year~

Câu 2b đề là tìm x chứ nhỉ???

b) \(\sqrt{x^2-4}+\sqrt{x-2}=0\)

Ta có: \(\left\{{}\begin{matrix}\sqrt{x^2-4}\ge0\\\sqrt{x-2}\ge0\end{matrix}\right.\)

=> Dấu = xảy ra <=> \(\left\{{}\begin{matrix}\sqrt{x^2-4}=0\\\sqrt{x-2}=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x^2-4=0\\x-2=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=\pm2\\x=2\end{matrix}\right.\) <=> x = 2

Vậy x = 2

bài 2 câu b) đề sai rồi bạn

còn bài 1 câu b) mình cảm thấy sai sai