Ta có: 1<2

nên \(1-\sqrt{2}< 2-\sqrt{2}\)

\(\Leftrightarrow f\left(1-\sqrt{2}\right)>f\left(2-\sqrt{2}\right)\)(Vì hàm số y=f(x)=-x+4 nghịch biến trên R nên nếu \(x_1< x_2\) thì \(f\left(x_1\right)>f\left(x_2\right)\))

Ta có \(1-\sqrt{2}< 2-\sqrt{2}\) \(\Rightarrow-\left(1-\sqrt{2}\right)>-\left(2-\sqrt{2}\right)\) \(\Rightarrow-\left(1-\sqrt{2}\right)+4>-\left(2-\sqrt{2}\right)+4\) Mà \(f\left(1-\sqrt{2}\right)=-\left(1-\sqrt{2}\right)+4,f\left(2-\sqrt{2}\right)=-\left(2-\sqrt{2}\right)+4\)

\(\Rightarrow f\left(1-\sqrt{2}\right)>f\left(2-\sqrt{2}\right)\)

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

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

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

\(=\sqrt{5}+\sqrt{5}\)

\(=2\sqrt{5}\)

b) \(\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(\sqrt{2}-5\right)^2}\)

\(=\left|\sqrt{2}-1\right|-\left|\sqrt{2}-5\right|\)

\(=\sqrt{2}-1-\left(5-\sqrt{2}\right)\)

\(=\sqrt{2}-1-5+\sqrt{2}\)

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

\(\sqrt{3}-\frac{5}{2}>\sqrt{3}-4\text{ vì }-\frac{5}{2}>-4\)

\(\Rightarrow2.\left(\sqrt{3}-\frac{5}{2}\right)>\sqrt{3}-4\)

\(\Rightarrow2.\sqrt{3}-5>\sqrt{3}-4\)

b) vì \(\sqrt{5}-\sqrt{12}< 0\), ta có: 

 \(5\sqrt{5}-2\sqrt{3}=4\sqrt{5}+\sqrt{5}-\sqrt{12}< 4\sqrt{5}< 4\sqrt{5}+6\) 

Vậy \(5\sqrt{5}-2\sqrt{3}< 6+4\sqrt{5}\)

c)\(\sqrt{2}-\sqrt{6}=\sqrt{2}.\left(\sqrt{1}-\sqrt{3}\right)>\left(1-\sqrt{3}\right)\)

Vậy \(\sqrt{2}-\sqrt{6}>1-\sqrt{3}\)

Tính $P=x^5+2x^4+3x^3+4x^2+5x+6$ - Đại số - Diễn đàn Toán học

Ta có \(\left\{{}\begin{matrix}\left(2\sqrt{3}\right)^2=12\\\left(3\sqrt{2}\right)^2=18\end{matrix}\right.\) \(\Rightarrow2\sqrt{3}< 3\sqrt{2}\)

- Nếu \(m+1>0\Rightarrow m>-1\Rightarrow f\left(x\right)\) đồng biến \(\Rightarrow f\left(2\sqrt{3}\right)< f\left(3\sqrt{2}\right)\)

- Nếu \(m+1< 0\Rightarrow m< -1\Rightarrow f\left(x\right)\) nghịch biến \(\Rightarrow f\left(2\sqrt{3}\right)>f\left(3\sqrt{2}\right)\)

- Nếu \(m=-1\Rightarrow f\left(2\sqrt{3}\right)=f\left(3\sqrt{2}\right)=-2\)