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Ta có: \(\frac{-3}{4\sqrt{3}-7}-\frac{3}{4\sqrt{3}+7}+\sqrt{4-2\sqrt{3}}\)
\(=\frac{-3\left(4\sqrt{3}+7\right)-3\left(4\sqrt{3}-7\right)}{\left(4\sqrt{3}\right)^2-7^2}+\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\)
\(=\frac{-12\sqrt{3}-21-12\sqrt{3}+21}{48-49}+\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\frac{-24\sqrt{3}}{-1}+\left|\sqrt{3}-1\right|\)
\(=24\sqrt{3}+\sqrt{3}-1\)(vì \(\sqrt{3}>1\))
\(=25\sqrt{3}-1\)
\(=\frac{-3\left(4\sqrt{3}+7\right)-3\left(4\sqrt{3}-7\right)}{\left(4\sqrt{3}\right)^2-49}+\sqrt{4-2\sqrt{3}}\)
Tiếp tục nhé
a,
\(\frac{\sqrt{6}\left(\sqrt{3}-1\right)}{\sqrt{3}\left(\sqrt{3}-1\right)}+\sqrt{\frac{\left(2-\sqrt{2}\right)^2}{\left(2+\sqrt{2}\right).\left(2-\sqrt{2}\right)}}\)
=\(\sqrt{2}+\frac{2-\sqrt{2}}{\sqrt{2}}\)
=\(\sqrt{2}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}}\)
=\(\sqrt{2}+\sqrt{2}-1\)
=\(2\sqrt{2}-1\)
còn tiếp
b=,\(\frac{6\sqrt{3}}{3}-\frac{\sqrt{3}\left(1-\sqrt{3}\right)}{\sqrt{3}}-\frac{\sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{2}-\sqrt{3}}\)
=\(6-1+\sqrt{3}-\sqrt{6}\)
=\(5+\sqrt{3}+\sqrt{6}\)
\(\frac{3}{\sqrt{7}-1}+\frac{3}{\sqrt{7}+1}=\frac{3\left[\sqrt{7}+1+\sqrt{7}-1\right]}{\left(\sqrt{7}+1\right)\left(\sqrt{7}-1\right)}=\frac{6\sqrt{7}}{6}=\sqrt{7}\)
\(\frac{3}{\sqrt{X}-1}-\frac{2}{\sqrt{X}+1}+\frac{X-7}{X-1}=\frac{3\left(\sqrt{X}+1\right)-2\left(\sqrt{X}-1\right)+X-7}{\left(\sqrt{X}+1\right)\left(\sqrt{X}-1\right)}=\frac{X+\sqrt{X}-2}{\left(\sqrt{X}+1\right)\left(\sqrt{X}-1\right)}=\frac{\sqrt{X}+2}{\sqrt{X}+1}\)
TÍNH GIÁ TRỊ BIỂU THỨC:
\(\frac{3}{\sqrt{7}-1}\) + \(\frac{3}{\sqrt{7}+1}\)= \(\frac{3\left(\sqrt{7}+1\right)+3\left(\sqrt{7}-1\right)}{\left(\sqrt{7}-1\right)\left(\sqrt{7}+1\right)}\)= \(\frac{3\sqrt{7}+3+3\sqrt{7}-3}{6}\)=\(\frac{6\sqrt{7}}{6}\)=\(\sqrt{7}\)
RÚT GỌN BIỂU THỨC:
\(\frac{3}{\sqrt{X}-1}\)-\(\frac{2}{\sqrt{X}+1}\)+\(\frac{X-7}{X-1}\)
= \(\frac{3\left(\sqrt{X}+1\right)}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)-\(\frac{2\left(\sqrt{X}-1\right)}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)+\(\frac{X-7}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)
= \(\frac{3\sqrt{X}+3-2\sqrt{X}+2+X-7}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)
= \(\frac{X+\sqrt{X}-2}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)
= \(\frac{\left(\sqrt{X}+1\right)\left(\sqrt{X}-2\right)}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)
= \(\frac{\sqrt{X}-2}{\sqrt{X}-1}\)
CHÚC EM HỌC TỐT!
Ta có:
\(\left(\sqrt{5-3\sqrt{2}}+\sqrt{3\sqrt{2}-4}\right)^2=5-3\sqrt{2}+3\sqrt{2}-4+2\sqrt{5-3\sqrt{2}}\sqrt{3\sqrt{2}-4}\)
\(=1+2\sqrt{27\sqrt{2}-38}\)
Áp dụng vào bài toán t được
\(\dfrac{\sqrt{1+2\sqrt{27\sqrt{2}-38}}-\sqrt{5-3\sqrt{2}}}{\sqrt{3\sqrt{2}-4}}\)
\(=\dfrac{\sqrt{\left(\sqrt{5-3\sqrt{2}}+\sqrt{3\sqrt{2}-4}\right)^2}-\sqrt{5-3\sqrt{2}}}{\sqrt{3\sqrt{2}-4}}\)
\(=\dfrac{\sqrt{5-3\sqrt{2}}+\sqrt{3\sqrt{2}-4}-\sqrt{5-3\sqrt{2}}}{\sqrt{3\sqrt{2}-4}}=1\)
ĐKXĐ: \(x\ge\frac{3}{2}\)
PT (=) \(\sqrt{\left(\sqrt{2x-3}+1\right)^2}+\sqrt{\left(\sqrt{2x-3}+4\right)^2}=7\)
(=) \(\sqrt{2x-3}+1+\sqrt{2x-3}+4=7\)
(=) \(2\sqrt{2x-3}=2\) (=) \(\sqrt{2x-3}=1\)(=) 2x = 4 (=) x = 2 ( Thỏa mãn điều kiện )
Vậy x=2
\(1)\dfrac{{14}}{{\sqrt 7 }} = \dfrac{{14\sqrt 7 }}{{\sqrt 7 .\sqrt 7 }} = \dfrac{{14\sqrt 7 }}{7} = 2\sqrt 7 \\ 2)\dfrac{{\sqrt 3 }}{{\sqrt 2 }} = \dfrac{{\sqrt 3 .\sqrt 2 }}{{\sqrt 2 .\sqrt 2 }} = \dfrac{{\sqrt 6 }}{2}\\ 3)\dfrac{5}{{\sqrt {10} }} = \dfrac{{5\sqrt {10} }}{{\sqrt {10} .\sqrt {10} }} = \dfrac{{5\sqrt {10} }}{{10}} = \dfrac{{\sqrt {10} }}{2}\\ 4)\dfrac{3}{{2\sqrt 5 }} = \dfrac{{3.2\sqrt 5 }}{{2\sqrt 5 .2\sqrt 5 }} = \dfrac{{6\sqrt 5 }}{{20}} = \dfrac{{3\sqrt 5 }}{{10}}\\ 5)\dfrac{{7 + \sqrt 7 }}{{\sqrt 7 + 1}} = \dfrac{{\left( {7 + \sqrt 7 } \right)\left( {\sqrt 7 - 1} \right)}}{{\left( {\sqrt 7 + 1} \right)\left( {\sqrt 7 - 1} \right)}} = \dfrac{{6\sqrt 7 }}{6} = \sqrt 7 \\ 6)\dfrac{{\sqrt 2 - \sqrt 6 }}{{3\sqrt 3 - 3}} = \dfrac{{\left( {\sqrt 2 - \sqrt 6 } \right)\left( {3\sqrt 3 + 3} \right)}}{{\left( {3\sqrt 3 - 3} \right)\left( {3\sqrt 3 + 3} \right)}} = \dfrac{{ - 2\sqrt 2 }}{6} = \dfrac{{ - \sqrt 2 }}{3}\\ 7)\dfrac{{\sqrt 3 }}{{3 - \sqrt 3 }} = \dfrac{{\sqrt 3 \left( {3 + \sqrt 3 } \right)}}{{\left( {3 - \sqrt 3 } \right)\left( {3 + \sqrt 3 } \right)}} = \dfrac{{3\sqrt 3 + 3}}{6} = \dfrac{{3\left( {\sqrt 3 + 1} \right)}}{6} = \dfrac{{\sqrt 3 + 1}}{2}\\ 8)\dfrac{2}{{2 - \sqrt 3 }} = \dfrac{{2\left( {2 + \sqrt 3 } \right)}}{{\left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)}} = 4 + 2\sqrt 3 \\ 9)\dfrac{{\sqrt 3 + 2}}{{2 - \sqrt 3 }} = \dfrac{{\left( {\sqrt 3 + 2} \right)\left( {2 + \sqrt 3 } \right)}}{{\left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)}} = 7 + 4\sqrt 3 \\ 10)\dfrac{{3\sqrt 5 }}{{2\sqrt 5 - 1}} = \dfrac{{3\sqrt 5 \left( {2\sqrt 5 + 1} \right)}}{{\left( {2\sqrt 5 - 1} \right)\left( {2\sqrt 5 + 1} \right)}} = \dfrac{{30 + 3\sqrt 5 }}{{19}}\\ 11)\dfrac{1}{{\sqrt 3 }} = \dfrac{{1.\sqrt 3 }}{{\sqrt 3 .\sqrt 3 }} = \dfrac{{\sqrt 3 }}{3} \)