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Xét \(\sqrt{2}.A=\sqrt{\dfrac{4+2\sqrt{3}}{2}}-\sqrt{\dfrac{4-2\sqrt{3}}{2}}\)
= \(\sqrt{\dfrac{\left(1+\sqrt{3}\right)^2}{2}}-\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{2}}\)
= \(\dfrac{1+\sqrt{3}}{\sqrt{2}}-\dfrac{\sqrt{3}-1}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}\)
<=> A = 1
A=\(\sqrt{2}\), cái kết quả này bấm máy tính là ra được, quan trọng là phải làm thế nào để ra
a) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\)
\(=\left[-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\cdot\left(\sqrt{2}-\sqrt{5}\right)\)
\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)
\(=-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)
\(=-\left(2-5\right)\)
\(=-\left(-3\right)\)
\(=3\)
b) Ta có:
\(x^2-x\sqrt{3}+1\)
\(=x^2-2\cdot\dfrac{\sqrt{3}}{2}\cdot x+\left(\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
\(=\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
Mà: \(\left(x-\dfrac{\sqrt{3}}{2}\right)^2\ge0\forall x\) nên
\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x\)
Dấu "=" xảy ra:
\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{\sqrt{3}}{2}\)
Vậy: GTNN của biểu thức là \(\dfrac{1}{4}\) tại \(x=\dfrac{\sqrt{3}}{2}\)
a)
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\\ =\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}^2-\sqrt{5}^2\right)\\ =-\left(2-5\right)\\ =-\left(-3\right)\\ =3\)
\(A=\frac{\sqrt{3}-1}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{\sqrt{3}+1}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}=\frac{\sqrt{3}-1}{1+\sqrt{\frac{2+\sqrt{3}}{2}}}+\frac{\sqrt{3}+1}{1-\sqrt{\frac{2-\sqrt{3}}{2}}}\)
\(=\frac{\sqrt{3}-1}{1+\frac{\sqrt{4+2\sqrt{3}}}{2}}+\frac{\sqrt{3}+1}{1-\frac{\sqrt{4-2\sqrt{3}}}{2}}=\frac{\sqrt{3}-1}{1+\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2}}+\frac{\sqrt{3}+1}{1-\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{2}}\)
\(=\frac{\sqrt{3}-1}{\frac{3+\sqrt{3}}{2}}+\frac{\sqrt{3}+1}{\frac{3-\sqrt{3}}{2}}=\frac{2\left(\sqrt{3}-1\right)}{\sqrt{3}\left(\sqrt{3}+1\right)}+\frac{2\left(\sqrt{3}+1\right)}{\sqrt{3}\left(\sqrt{3}-1\right)}\)
\(=\frac{2}{\sqrt{3}}\left(\frac{4-2\sqrt{3}+4+2\sqrt{3}}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\right)=\frac{2}{\sqrt{3}}.\frac{8}{2}=\frac{8}{\sqrt{3}}=\frac{8\sqrt{3}}{3}\)
1,
\(A=\left(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{a+2}{a-2}\left(đk:a\ne0;1;2;a\ge0\right)\)
\(=\frac{\left(a\sqrt{a}-1\right)\left(a+\sqrt{a}\right)-\left(a\sqrt{a}+1\right)\left(a-\sqrt{a}\right)}{a^2-a}.\frac{a-2}{a+2}\)
\(=\frac{a^2\sqrt{a}+a^2-a-\sqrt{a}-\left(a^2\sqrt{a}-a^2+a-\sqrt{a}\right)}{a\left(a-1\right)}.\frac{a-2}{a+2}\)
\(=\frac{2a\left(a-1\right)\left(a-2\right)}{a\left(a-1\right)\left(a+2\right)}=\frac{2\left(a-2\right)}{a+2}\)
Để \(A=1\)\(=>\frac{2a-4}{a+2}=1< =>2a-4-a-2=0< =>a=6\)
2,
a, Điều kiện xác định của phương trình là \(x\ne4;x\ge0\)
b, Ta có : \(B=\frac{2\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}-\frac{1}{\sqrt{x}+2}\)
\(=\frac{2\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{x-4}-\frac{\sqrt{x}-2}{x-4}\)
\(=\frac{2\sqrt{x}+2+2}{x-4}=\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{2}{\sqrt{x}-2}\)
c, Với \(x=3+2\sqrt{3}\)thì \(B=\frac{2}{3-2+2\sqrt{3}}=\frac{2}{1+2\sqrt{3}}\)
\(\frac{\sqrt{2-\sqrt{3}}}{2}:\left(\frac{\sqrt{2+\sqrt{3}}}{2}-\frac{2}{\sqrt{6}}+\frac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}\right).\)
\(=\frac{2\sqrt{2-\sqrt{3}}}{4}:\left(\frac{2\sqrt{2+\sqrt{3}}}{4}-\frac{2}{\sqrt{6}}+\frac{2\sqrt{2+\sqrt{3}}}{4\sqrt{3}}\right)\)
\(=\frac{\sqrt{4-2\sqrt{3}}}{4}:\left(\frac{\sqrt{4+2\sqrt{3}}}{4}-\frac{2}{\sqrt{6}}+\frac{\sqrt{4+2\sqrt{3}}}{4\sqrt{3}}\right)\)
\(=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{4}:\left[\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{4}-\frac{2}{\sqrt{6}}+\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{4\sqrt{3}}\right]\)
\(=\frac{\sqrt{3}-1}{4}:\left[\frac{\sqrt{6}\left(\sqrt{3}+1\right)}{4\sqrt{6}}-\frac{2.4}{4\sqrt{6}}+\frac{\sqrt{2}\left(\sqrt{3}+1\right)}{4\sqrt{6}}\right]\)
\(=\frac{\sqrt{3}-1}{4}:\frac{\sqrt{18}+\sqrt{6}-8+\sqrt{6}+\sqrt{2}}{4\sqrt{6}}\)
\(=\frac{\sqrt{3}-1}{4}.\frac{4\sqrt{6}}{\sqrt{2}\left(\sqrt{9}+2\sqrt{3}+1\right)}\)
\(=\frac{\sqrt{6}\left(\sqrt{3}-1\right)}{\sqrt{2}\left(\sqrt{3}+1\right)^2}=\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)^2}\)............
\(\sqrt{\frac{2}{2-\sqrt{3}}}-\sqrt{\frac{2}{2+\sqrt{3}}}\)
\(=\sqrt{\frac{2\left(2+\sqrt{3}\right)}{4-3}}-\sqrt{\frac{2\left(2-\sqrt{3}\right)}{4-3}}\)
\(=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left|\sqrt{3}+1\right|-\left|\sqrt{3}-1\right|\)
\(=\sqrt{3}+1-\sqrt{3}+1\)( \(\sqrt{3}+1>0\) và \(\sqrt{3}-1>0\) )
\(=2\)
\(\)
\(\sqrt{2\left(2+\sqrt{3}\right)}-\sqrt{2\left(2-\sqrt{3}\right)}\))
\(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)
\(\sqrt{\left(1+\sqrt{3}\right)^2}-\sqrt{\left(1-\sqrt{3}\right)^2}\)
\(1+\sqrt{3}-\sqrt{3}+1\)
\(2\)
a) \(\frac{2+\sqrt{3}}{2-\sqrt{3}}=\frac{\left(2+\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\frac{\left(2+\sqrt{3}\right)^2}{4-3}\)
\(=\left(2+\sqrt{3}\right)^2=7+4\sqrt{3}\)
\(\frac{5+2\sqrt{6}}{5-2\sqrt{6}}=\frac{\left(5+2\sqrt{6}\right)^2}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}=\frac{\left(5+2\sqrt{6}\right)^2}{25-24}\)
\(=\left(5+2\sqrt{6}\right)^2=49+20\sqrt{6}\)
b) \(\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{\left(\sqrt{3}-1\right)^2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{3-2\sqrt{3}+1}{3-1}\)
\(=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}\)
c) \(\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}=\frac{\left(2+\sqrt{3}\right)^2+\left(2-\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=\frac{4+4\sqrt{3}+3+4-4\sqrt{3}+3}{4-3}=14\)
d) \(\frac{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}-\frac{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}+\sqrt{2-\sqrt{3}}}}\)
\(=\frac{\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)^2-\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)^2}{\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)}\)
\(=\frac{2+\sqrt{3}+2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+2-\sqrt{3}-\left(2+\sqrt{3}-2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+2-\sqrt{3}\right)}{2+\sqrt{3}-\left(2-\sqrt{3}\right)}\)
\(=\frac{4\sqrt{4-3}}{2\sqrt{3}}=\frac{4}{2\sqrt{3}}=\frac{2}{\sqrt{3}}\)