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\(\sqrt{x^2-2x+4}+\sqrt{x^2+5}=9-2x\left(đk:x\le\dfrac{9}{2}\right)\)
\(\Leftrightarrow x^2-2x+4+x^2+5+2\sqrt{\left(x^2-2x+4\right)\left(x^2+5\right)}=81-36x+4x^2\)
\(\Leftrightarrow2\sqrt{\left(x^2-2x+4\right)\left(x^2+5\right)}=2x^2-34x+72\)
\(\Leftrightarrow4\left(x^2-2x+4\right)\left(x^2+5\right)=4x^4+1156x^2+5184-136x^3+288x^2-4896x\)
\(\Leftrightarrow4x^4-8x^3+36x^2-40x+80=4x^4-136x^3+1444x^2-4896x+5184\)
\(\Leftrightarrow128x^3-1408x^2+4856x-5104=0\)
\(\Leftrightarrow128x^2\left(x-2\right)-1152x\left(x-2\right)+2552\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(128x^2-1152x+2552\right)=0\)
\(\Leftrightarrow x=2\left(tm\right)\)(do \(128x^2-1152x+2552>0\))
\(=\dfrac{\sqrt{5}-2}{5-4}-\sqrt{\left(\sqrt{5}+2\right)^2}=\sqrt{5}-2-\sqrt{5}-2=-4\)
\(\sqrt{4-\sqrt{9+4\sqrt{2}}}=\sqrt{4-\sqrt{1+2.2.\sqrt{2}+\left(2\sqrt{2}\right)^2}}\)
\(=\sqrt{4-\sqrt{\left(1+2\sqrt{2}\right)^2}}\)
\(=\sqrt{4-\left|1+2\sqrt{2}\right|}=\sqrt{4-1-2\sqrt{2}}=\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{1-2.\sqrt{2}.1+\left(\sqrt{2}\right)^2}=\sqrt{\left(1-\sqrt{2}\right)^2}=\left|1-\sqrt{2}\right|=\sqrt{2}-1\)
Vậy ....
Lời giải:
a.
\(=\sqrt{5+2.2\sqrt{5}+2^2}-\sqrt{5-2.2\sqrt{5}+2^2}\)
$=\sqrt{(\sqrt{5}+2)^2}-\sqrt{(\sqrt{5}-2)^2}$
$=|\sqrt{5}+2|-|\sqrt{5}-2|=(\sqrt{5}+2)-(\sqrt{5}-2)=4$
b.
$=\sqrt{3-2.3\sqrt{3}+3^2}+\sqrt{3+2.3.\sqrt{3}+3^2}$
$=\sqrt{(\sqrt{3}-3)^2}+\sqrt{(\sqrt{3}+3)^2}$
$=|\sqrt{3}-3|+|\sqrt{3}+3|$
$=(3-\sqrt{3})+(\sqrt{3}+3)=6$
c.
$=\sqrt{2+2.3\sqrt{2}+3^2}-\sqrt{2-2.3\sqrt{2}+3^2}$
$=\sqrt{(\sqrt{2}+3)^2}-\sqrt{(\sqrt{2}-3)^2}$
$=|\sqrt{2}+3|-|\sqrt{2}-3|$
$=(\sqrt{2}+3)-(3-\sqrt{2})=2\sqrt{2}$
a) \(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)
\(=2\sqrt{5}+2+\sqrt{5}-2\)
\(=3\sqrt{5}\)
b) \(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\)
\(=3-2\sqrt{2}+2\sqrt{2}-1\)
=2
c) \(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\)
\(=2-\sqrt{2}+3\sqrt{2}-2\)
\(=2\sqrt{2}\)
a)
\(M=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{4+4\sqrt{5}+5}-\sqrt{4-4\sqrt{5}+5}\)
\(=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(=\left|2+\sqrt{5}\right|-\left|2-\sqrt{5}\right|\)
\(=2+\sqrt{5}-\left(\sqrt{5}-2\right)\) (vì \(2+2\sqrt{5}>0;2-\sqrt{5}< 0\) )
\(=2+\sqrt{5}-\sqrt{5}+2\\ =4\)
b)
\(N=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\left|\sqrt{7}-1\right|-\left|\sqrt{7}+1\right|\)
\(=\sqrt{7}-1-\left(\sqrt{7}+1\right)\) (vì \(\sqrt{7}-1>0;\sqrt{7}+1>0\) )
\(=\sqrt{7}-1-\sqrt{7}-1\\ =-2\)
Ta có: \(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)
\(=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)
\(=-13\sqrt{3}\)
\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\\ =2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}=-13\sqrt{3}\)
a,\(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}=\sqrt{2^2+2\cdot2\cdot\left(2\sqrt{5}\right)+\left(2\sqrt{5}\right)^2}\) \(+\sqrt{\left(\sqrt{5}\right)^2-2\cdot2\sqrt{5}+2^2}=\sqrt{\left(2+2\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}\)=\(2+2\sqrt{5}+\sqrt{5}-2=3\sqrt{5}\)
b,\(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}=3-2\sqrt{2}+2\sqrt{2}+1=4\)
c,\(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-2\right)^2}=2-\sqrt{2}+3\sqrt{2}-2=2\sqrt{2}\)
\(2\sqrt{9-\sqrt{4}}\\ =2\sqrt{9-2}\\ =2\sqrt{7}\)