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a) \(\sqrt{25}+\sqrt{9}-\sqrt{16}\) = \(\sqrt{5^2}+\sqrt{3^2}-\sqrt{4^2}\) = 5 + 3 - 4 = 4
b) \(\sqrt{0,16}+\sqrt{0,01}+\sqrt{0,25}\) = 0,4 + 0,1 + 0,5 = 1
c) \(\left(\sqrt{3^2}\right)-\left(\sqrt{2^2}\right)+\left(\sqrt{5^2}\right)\)
= 3 - 2 + 5 = 6
d) \(\sqrt{4}-\left(-\sqrt{3}\right)^2+\sqrt{49}\) = 2 - 3 + 7 = 6
e) \(\left(2\sqrt{2}\right)^2-\left(3\sqrt{3}\right)^2\)
= \(\left(\sqrt{8}\right)^2-\left(\sqrt{27}\right)^2\) = 8 - 27 = -19
f) \(\left(-2\sqrt{2}\right)^2+\left(3\sqrt{3}\right)^2\) = 8 + 27 = 35
c) \(\frac{5}{8}+\frac{13}{10}-9+25=\frac{717}{40}\)
d) \(\sqrt{0,2^2}=\left|0,2\right|=0,2\)
e) \(\sqrt{\left(-0.3\right)^2}=0,3\)
g) \(-\sqrt{\left(-1.3\right)^2}=-1,3\)
h) \(-0,7\sqrt{\left(-0,7\right)^2}=-0,49\)
1. ĐK \(\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
a. Ta có \(R=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right).\left(\frac{1}{\sqrt{x}+2}+\frac{4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}+2}{\sqrt{x}}.\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
b. Với \(x=4+2\sqrt{3}\Rightarrow R=\frac{\sqrt{4+2\sqrt{3}}+2}{\sqrt{4+2\sqrt{3}}\left(\sqrt{4+2\sqrt{3}}-2\right)}=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}+2}{\sqrt{\left(\sqrt{3}+1\right)^2}\left(\sqrt{\left(\sqrt{3}+1\right)^2}-2\right)}\)
\(=\frac{\sqrt{3}+1+2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{\sqrt{3}+3}{3-1}=\frac{\sqrt{3}+3}{2}\)
c. Để \(R>0\Rightarrow\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}>0\Rightarrow\sqrt{x}-2>0\Rightarrow x>4\)
Vậy \(x>4\)thì \(R>0\)
2. Ta có \(A=6+2\sqrt{2}=6+\sqrt{8};B=9=6+3=6+\sqrt{9}\)
Vì \(\sqrt{8}< \sqrt{9}\Rightarrow A< B\)
3. a. \(VT=\frac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}.\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\left(\sqrt{a}-\sqrt{b}\right).\left(\sqrt{a}+\sqrt{b}\right)=a-b=VP\left(đpcm\right)\)
b. Ta có \(VT=\left(2+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right).\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)
\(=\left(2+\sqrt{a}\right)\left(2-\sqrt{a}\right)=4-a=VP\left(đpcm\right)\)