\(x+y+z\le2xyz+\sqrt{2}\)

\(\Leftrightarrow x+y+z-2xyz\le\sqrt{2}\)

\(\Leftrightarrow x\left(1-2yz\right)+y+z\le\sqrt{2}\)

Ta có: \(x\left(1-2yz\right)+y+z\le\sqrt{\left[x^2+\left(y+z\right)^2\right]\left[\left(1-2yz\right)^2+1\right]}\)( bđt bunhiacopxki )

Ta cần chứng minh \(\sqrt{\left[x^2+\left(y+z\right)^2\right]\left[\left(1-2yz\right)^2+1\right]}\le\sqrt{2}\)

\(\Leftrightarrow\left[x^2+y^2+z^2+2yz\right]\left[1-4yz+4y^2z^2+1\right]\le2\)

\(\Leftrightarrow\left(1+2yz\right)\left(2-4yz+4y^2z^2\right)\le2\)

\(\Leftrightarrow2-4yz+4y^2z^2+4yz-8y^2z^2+8y^3z^3\le2\)

\(\Leftrightarrow8y^3z^3-4y^2z^2\le0\)

\(\Leftrightarrow4y^2z^2\left(2yz-1\right)\le0\)(1) 

Ta có: \(1=x^2+y^2+z^2\ge y^2+z^2\ge2yz\)

\(\Rightarrow yz\le\frac{1}{2}\)

\(\Rightarrow\left(1\right)\)đúng

Vậy đẳng thức đc chứng minh

ok bạn

a)\(-5\sqrt{80}+4\sqrt{45}-2\sqrt{245}\)

\(=-20\sqrt{5}+12\sqrt{5}-14\sqrt{5}\)

\(=\left(-20+12-14\right)\sqrt{5}=-22\sqrt{5}\)

b)\(\sqrt{12-6\sqrt{3}}-\sqrt{21-12\sqrt{3}}\)

\(=\sqrt{9-6\sqrt{3}+3}-\sqrt{12-6\sqrt{12}+9}\)

\(=\sqrt{\left(3-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{12}-3\right)^2}\)

\(=\left|3-\sqrt{3}\right|-\left|\sqrt{12}-3\right|\)

\(=3-\sqrt{3}-\sqrt{12}+3\)(do \(3>\sqrt{3};\sqrt{12}>3\))

\(=6-\sqrt{12}-\sqrt{3}\)

\(=6-2\sqrt{3}-\sqrt{3}=6-3\sqrt{3}\)

c)\(\sqrt{7-\sqrt{40}}-\sqrt{7+\sqrt{40}}\)

\(=\sqrt{5-2\sqrt{5}.\sqrt{2}+2}-\sqrt{5+2\sqrt{5}.\sqrt{2}+2}\)

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

\(=\left|\sqrt{5}-\sqrt{2}\right|-\left|\sqrt{5}+\sqrt{2}\right|\)

\(=\sqrt{5}-\sqrt{2}-\sqrt{5}-\sqrt{2}\)(do \(\sqrt{5}>\sqrt{2}\))

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

ĐK: \(x< 0\)hoặc \(x>\frac{1}{3}\).

Đặt \(t=\sqrt{\frac{3x-1}{x}}>0\)

Phương trình ban đầu tương đương với:

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

\(\Leftrightarrow2t^3-t^2-1=0\)

\(\Leftrightarrow2t^3-2t^2+t^2-1=0\)

\(\Leftrightarrow\left(t-1\right)\left(2t^2+t+1\right)=0\)

\(\Leftrightarrow t-1=0\)(vì \(2t^2+t+1>0\))

\(\Leftrightarrow t=1\)

\(\Leftrightarrow\frac{3x-1}{x}=1\)

\(\Rightarrow3x-1=x\)

\(\Leftrightarrow x=\frac{1}{2}\)(thỏa mãn) 

Mình mới thử chương trình lớp 9 nên chưa hiểu nhiều lắm. Cảm ơn nhé!

\(\sqrt{12-3\sqrt{15}}=\frac{1}{\sqrt{2}}\sqrt{24-6\sqrt{15}}=\frac{1}{\sqrt{2}}\sqrt{15-2.3.\sqrt{15}+9}\)

\(=\frac{1}{\sqrt{2}}\sqrt{\left(\sqrt{15}-3\right)^2}=\frac{1}{\sqrt{2}}\left|\sqrt{15}-3\right|=\frac{\sqrt{15}-3}{\sqrt{2}}\)

\(=\frac{\sqrt{30}-3\sqrt{2}}{2}\)