\(=\sqrt{\left(\sqrt{3}+2\right)^2}-\sqrt{\left(2\sqrt{3}-1\right)^2}\)

\(=\sqrt{3}+2-2\sqrt{3}+1\)

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

=\(\sqrt{3}-1+2-\) \(\sqrt{3}=1\)

b.=\(\frac{2+\sqrt{3}-2+\sqrt{3}}{2^2-3}=2\sqrt{3}\)

\(hpt\Leftrightarrow\hept{\begin{cases}6(x^3-y^3)=6(8x+2y)\\x^2-3y^2=6\end{cases}}\)

Suy ra \(6(x^3-y^3)=(8x+2y)(x^2-3y^2)\)

\(\Leftrightarrow x(x-3y)(x+4y)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=3y;x=-4y\end{cases}}\)

Thay vào giải tiếp nhé !!

Gõ đề cẩn thận hơn nhé !!

\(\frac{\sqrt{x^3+x}}{x^2-1}+\sqrt{1-x}-\sqrt{1+x}=0\)

ĐK:\(-1\le x\le1\)

\(pt\Leftrightarrow\frac{\sqrt{x^3+x}}{x^2-1}+\left(\sqrt{1-x}-1\right)-\left(\sqrt{1+x}-1\right)=0\)

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

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

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

Suy ra x=0

\(\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}=\sqrt{2+3+5+2\sqrt{2.3}+2\sqrt{2.5}+2\sqrt{3.5}}\)

\(=\sqrt{\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)^2}=\sqrt{2}+\sqrt{3}+\sqrt{5}\)

3) Đặt b+c=x;c+a=y;a+b=z.

=>a=(y+z-x)/2 ; b=(x+z-y)/2 ; c=(x+y-z)/2

BĐT cần CM <=> \(\frac{y+z-x}{2x}+\frac{x+z-y}{2y}+\frac{x+y-z}{2z}\ge\frac{3}{2}\)

VT=\(\frac{1}{2}\left(\frac{y}{x}+\frac{z}{x}-1+\frac{x}{y}+\frac{z}{y}-1+\frac{x}{z}+\frac{y}{z}-1\right)\)

\(=\frac{1}{2}\left[\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)+\left(\frac{x}{z}+\frac{z}{x}\right)-3\right]\)

\(\ge\frac{1}{2}\left(2+2+2-3\right)=\frac{3}{2}\)(Cauchy)

Dấu''='' tự giải ra nhá

Bài 4 

dễ chứng minh \(\left(a+b\right)^2\ge4ab;\left(b+c\right)^2\ge4bc;\left(a+c\right)^2\ge4ac\)

\(\Rightarrow\left(a+b\right)^2\left(b+c\right)^2\left(a+c\right)^2\ge64a^2b^2c^2\)

rồi khai căn ra \(\Rightarrow\)dpcm. 

đấu " = " xảy ra \(\Leftrightarrow\)\(a=b=c\)