Câu 1:
\(1,=18\sqrt{3}-10\sqrt{3}-5\sqrt{3}=3\sqrt{3}\\ 2,=\dfrac{2\sqrt{7}\left(3+\sqrt{7}\right)}{2}-\dfrac{\sqrt{7}\left(\sqrt{7}-2\right)}{\sqrt{7}-2}=3\sqrt{7}+7-7+2\sqrt{7}=5\sqrt{7}\\ 3,=10\sqrt{3}-15\sqrt{3}-8\sqrt{3}+16\sqrt{3}=3\sqrt{3}\\ 4,=\dfrac{7\left(\sqrt{10}+\sqrt{3}\right)}{7}-\dfrac{\sqrt{10}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{5}-\sqrt{2}}-2\sqrt{3}\\ =\sqrt{10}+\sqrt{3}-\sqrt{10}-2\sqrt{3}=-\sqrt{3}\\ 5,=2\sqrt{3}+6\sqrt{3}-5\sqrt{3}=3\sqrt{3}\\ 6,=\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}-\sqrt{\dfrac{\left(\sqrt{5}+2\right)^2}{1}}=\sqrt{5}-\sqrt{5}-2=-2\)
\(7,=2\sqrt{3}+12\sqrt{3}-6\sqrt{3}-2\sqrt{3}=6\sqrt{3}\\ 8,=\dfrac{\sqrt{3}\left(\sqrt{5}-2\right)}{2-\sqrt{5}}+2\sqrt{3}+\dfrac{13\left(\sqrt{3}+4\right)}{-13}=-\sqrt{3}+2\sqrt{3}-\sqrt{3}-4=-4\\ 9,=8\sqrt{3}+2\sqrt{3}-5\sqrt{3}-9\sqrt{3}=-4\sqrt{3}\\ 10,=\dfrac{\sqrt{6}\left(\sqrt{6}-1\right)}{\sqrt{6}-1}-3\sqrt{6}-\dfrac{4\left(2+\sqrt{6}\right)}{-2}=\sqrt{6}-3\sqrt{6}+4+2\sqrt{6}=4\)
Bài 1:
\(1,=18\sqrt{3}-10\sqrt{3}-5\sqrt{3}=3\sqrt{3}\)
\(2,=\dfrac{2\sqrt{7}\left(3+\sqrt{7}\right)}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}-\dfrac{\sqrt{7}\left(\sqrt{7}-2\right)}{\sqrt{7}-2}\)
\(=3\sqrt{7}+7-\sqrt{7}=2\sqrt{7}+7\)
\(3,=10\sqrt{3}-15\sqrt{3}-8\sqrt{3}+16\sqrt{3}=3\sqrt{3}\)
\(4,\dfrac{7\left(\sqrt{10}+\sqrt{3}\right)}{\left(\sqrt{10}+\sqrt{3}\right)\left(\sqrt{10}-\sqrt{3}\right)}-\dfrac{\sqrt{10}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{5}-\sqrt{2}}-\dfrac{6\sqrt{3}}{3}\)
\(=\sqrt{10}+\sqrt{3}-\sqrt{10}-3\sqrt{3}=-2\sqrt{3}\)
\(5,=2\sqrt{3}+6\sqrt{3}-5\sqrt{5}=3\sqrt{3}\)
\(6,\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}-\sqrt{\dfrac{\left(\sqrt{5}+2\right)^2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}\)
\(=\sqrt{5}-\sqrt{9+4\sqrt{5}}=\sqrt{5}-\left|\sqrt{5}+2\right|=\sqrt{5}-\sqrt{5}-2=-2\)
\(7,=14\sqrt{3}+12\sqrt{3}-6\sqrt{3}-2\sqrt{3}=16\sqrt{3}\)
\(8,=-\sqrt{3}+2\sqrt{3}-4-\sqrt{3}=-4\)
\(9,=8\sqrt{3}+2\sqrt{3}-5\sqrt{3}-6\sqrt{3}=-\sqrt{3}\)
\(10,=\sqrt{6}-3\sqrt{6}-\dfrac{\sqrt{6}}{3}=\dfrac{-7\sqrt{6}}{3}\)
Bài 2:
\(a,ĐKXĐ:x\ge2\)
\(< =>5\sqrt{x-2}-2\sqrt{x-2}-15=0\)
\(< =>3\sqrt{x-2}=15\)
\(< =>\sqrt{x-2}=5\)
\(< =>x-2=25\)
\(< =>x=27\left(TM\right)\)
Vậy pt có nghiệm là x=27
\(b,ĐKXĐ:x\ge2\)
\(< =>5\sqrt{x-2}=\sqrt{x-2}+16\)
\(< =>4\sqrt{x-2}=16\)
\(< =>\sqrt{x-2}=4\)
\(< =>x-2=16
\)
\(< =>x=18\left(TM\right)\)
Vậy pt có nghiệm là x=18
\(c,ĐKXĐ:x\ge1\)
\(< =>2\sqrt{\left(x-1\right)^2}=4\)
\(< =>\left|x-1\right|=2\)
\(< =>\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.< =>\left[{}\begin{matrix}x=3\left(TM\right)\\x=-1\left(Loại\right)\end{matrix}\right.\)
Vậy pt có nghiệm là x=3
\(d,ĐKXĐ:x\ge2
\)
\(< =>2\sqrt{x-2}+6\sqrt{x-2}=8+4\sqrt{x-2}\)
\(< =>4\sqrt{x-2}=8\)
\(< =>\sqrt{x-2}=2\)
\(< =>x-2=4< =>x=6\left(TM\right)\)
Vậy pt có nghiệm là x=6
e, ĐKXĐ: X≥\(\dfrac{1}{6}\)
\(13-\sqrt{\left(6x-1\right)^2}=8\)
\(< =>\left|6x-1\right|=5\)
\(< =>\left[{}\begin{matrix}6x-1=5\\6x-1=-5\end{matrix}\right.< =>\left[{}\begin{matrix}x=1\left(TM\right)\\x=\dfrac{-2}{3}\left(Loại\right)\end{matrix}\right.\)
Vậy pt có nghiệm là x=1
\(f,ĐKXĐ:x\ge5\)
\(< =>2\sqrt{x-5}-\sqrt{x-5}=\sqrt{1-x}\)
\(< =>x-5=1-x\)
\(< =>2x=6\)
\(< =>x=3\)(TM)
Vậy pt có nghiệm là x=3
