thiếu đề rồi,đề còn phải có điều kiên a,b,c>=0 nữa,nếu a,b,c <0 thì sai đề chắc rồi

ĐK : \(a;b;c\ge0\)

Biến đổi tương đương

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

\(\Leftrightarrow a+b+c-\sqrt{ab}-\sqrt{bc}-\sqrt{ac}\ge0\)

\(\Leftrightarrow2a+2b+2c-2\sqrt{ab}-2\sqrt{bc}-2\sqrt{ac}\ge0\)

\(\Leftrightarrow\left(a-2\sqrt{ab}+b\right)+\left(b-2\sqrt{bc}+c\right)+\left(c-2\sqrt{ac}+a\right)\ge0\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt{b}-\sqrt{c}\right)^2+\left(\sqrt{c}-\sqrt{a}\right)^2\ge0\)(luôn đúng \(\forall a;b;c\ge0\))

Vậy \(a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\)

ko xếp dc

ĐKXĐ: \(2x-5\ge0\Leftrightarrow x\ge2,5\)

pt\(\Leftrightarrow\sqrt{2x+4-2.3\sqrt{2x-5}}+\sqrt{2x-4+2\sqrt{2x-5}}=4\)\(\Leftrightarrow\sqrt{2x-5-2.3\sqrt{2x-5}+9}+\sqrt{2x-5+2\sqrt{2x-5}+1}=4\)

\(\Leftrightarrow\sqrt{\left(\sqrt{2x-5}-3\right)^2}+\sqrt{\left(\sqrt{2x-5}+1\right)^2}=4\)

\(\Leftrightarrow\left|\sqrt{2x-5}-3\right|+\left|\sqrt{2x-5}+1\right|=4\)

\(\Leftrightarrow\left|3-\sqrt{2x-5}\right|+\left|\sqrt{2x-5}+1\right|=4\)

Có: \(VT=\left|3-\sqrt{2x-5}\right|+\left|\sqrt{2x-5}+1\right|\ge\left|3-\sqrt{2x-5}+\sqrt{2x-5}+1\right|=4=VP\)

Dấu "=" xảy ra khi \(\left(3-\sqrt{2x-5}\right)\left(\sqrt{2x-5}+1\right)\ge0\)

Mà \(\sqrt{2x-5}+1\ge0\Rightarrow3-\sqrt{2x-5}\ge0\Rightarrow\sqrt{2x-5}\le3\)

\(\Rightarrow0\le\sqrt{2x-5}\le3\)

\(\Leftrightarrow0\le2x-5\le9\)

\(\Leftrightarrow2,5\le x\le7\)(TM)

bình 2 vế

\(\Leftrightarrow\left(\sqrt{1-x}-\sqrt{2+x}\right)^2=1\Leftrightarrow1-x-2\sqrt{\left(1-x\right)\left(2+x\right)}+2+x=1\)

\(\Leftrightarrow3-2\sqrt{\left(1-x\right)\left(2+x\right)}=1\Leftrightarrow2\sqrt{\left(1-x\right)\left(2+x\right)}=2\)

\(\Leftrightarrow\sqrt{-x^2-x+2}=1\Leftrightarrow-x^2-x+2=1\Leftrightarrow-x^2-x+1=0\)

\(\Leftrightarrow-\left(x^2+2.\frac{1}{2}.x+\frac{1}{4}\right)+\frac{5}{4}=0\Leftrightarrow\left(x+\frac{1}{2}\right)^2=\frac{5}{4}\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=-\frac{\sqrt{5}}{2}\\x+\frac{1}{2}=\frac{\sqrt{5}}{2}\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-\sqrt{5}-1}{2}\\x=\frac{\sqrt{5}-1}{2}\end{cases}}\)