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a: \(\Leftrightarrow\dfrac{2x-3}{x-1}=4\)
=>4x-4=2x-3
=>2x=1
hay x=1/2
b: \(\Leftrightarrow\sqrt{\dfrac{2x-3}{x-1}}=2\)
=>(2x-3)=4x-4
=>4x-4=2x-3
=>2x=1
hay x=1/2(nhận)
c: \(\Leftrightarrow\sqrt{2x+3}\left(\sqrt{2x-3}-2\right)=0\)
=>2x+3=0 hoặc 2x-3=4
=>x=-3/2 hoặc x=7/2
e: \(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
=>căn (x-5)=2
=>x-5=4
hay x=9
a: \(=\left(-\sqrt{5}-\sqrt{7}\right)\cdot\left(\sqrt{7}-\sqrt{5}\right)\)
\(=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
=-2
b: \(=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
\(=\dfrac{\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}-1+\sqrt{3}+1}{\sqrt{2}}=\sqrt{6}\)
c: \(=\dfrac{\sqrt{10}\left(\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}-\sqrt{5}}-2-\sqrt{10}+3\sqrt{7}+2\)
\(=\sqrt{10}-\sqrt{10}+3\sqrt{7}=3\sqrt{7}\)
bạn nên tự nghiên cứu rồi giải đi chứ bạn đưa 1 loạt thế thì ai rảnh mà giải, với lại cứ bài gì không biết chưa chịu suy nghĩ đã hỏi rồi thì tiến bộ sao được, đúng không
\(1.\text{ }\dfrac{1}{\sqrt{k}-\sqrt{k+1}}=\dfrac{\left(\sqrt{k}+\sqrt{k+1}\right)}{\left(\sqrt{k}+\sqrt{k+1}\right)\left(\sqrt{k}-\sqrt{k+1}\right)}\\ =-\left(\sqrt{k}+\sqrt{k+1}\right)\\ \Rightarrow\dfrac{1}{\sqrt{1}-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}-...-\dfrac{1}{\sqrt{8}-\sqrt{9}}\\ =-\left(\sqrt{1}+\sqrt{2}\right)+\left(\sqrt{2}+\sqrt{3}\right)-\left(\sqrt{3}+\sqrt{4}\right)+...+\left(\sqrt{8}+\sqrt{9}\right)\\ =-\sqrt{1}-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+...+\sqrt{8}+\sqrt{9}\\ \\ =\sqrt{9}-\sqrt{1}=2\)
\(2.\text{ }\dfrac{1}{\left(k+1\right)\sqrt{k}+\sqrt{k+1}k}=\dfrac{1}{\sqrt{k\left(k+1\right)}\left(\sqrt{k+1}+\sqrt{k}\right)}\\ =\dfrac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}\left(\sqrt{k+1}+\sqrt{k}\right)\left(\sqrt{k+1}-\sqrt{k}\right)}\\ =\dfrac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}\left(k+1-k\right)}=\dfrac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}}\\ =\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\\ \Rightarrow\text{ }\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{7\sqrt{6}+6\sqrt{7}}\\ =\text{ }\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{6}}-\dfrac{1}{\sqrt{7}}\\ =\text{ }\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{7}}\\ \text{ }1-\dfrac{1}{\sqrt{7}}\)
1.\(\dfrac{1}{\sqrt{1}-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}-\dfrac{1}{\sqrt{4}-\sqrt{5}}+\dfrac{1}{\sqrt{5}-\sqrt{6}}-\dfrac{1}{\sqrt{6}-\sqrt{7}}+\dfrac{1}{\sqrt{7}-\sqrt{8}}-\dfrac{1}{\sqrt{8}-\sqrt{9}}=\dfrac{1+\sqrt{2}}{1-2}-\dfrac{\sqrt{2}+\sqrt{3}}{2-3}+\dfrac{\sqrt{3}+\sqrt{4}}{3-4}-\dfrac{\sqrt{4}+\sqrt{5}}{4-5}+\dfrac{\sqrt{5}+\sqrt{6}}{5-6}-\dfrac{\sqrt{6}+\sqrt{7}}{6-7}+\dfrac{\sqrt{7}+\sqrt{8}}{7-8}-\dfrac{\sqrt{8}+\sqrt{9}}{8-9}=-1-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+\sqrt{4}+\sqrt{5}-\sqrt{5}-\sqrt{6}+\sqrt{6}+\sqrt{7}-\sqrt{7}-\sqrt{8}+\sqrt{8}+\sqrt{9}=\sqrt{9}-1=3-1=2\)
2]\(\sqrt{3}\)+1+\(\sqrt{4-4\sqrt{3}+3}\)=\(\sqrt{3}+1+\sqrt{\left(2-\sqrt{3}\right)^2}=\sqrt{3}+1+2-\sqrt{3}=3\)
4\(\left(\dfrac{\sqrt{3}.\left(2+\sqrt{3}\right)+2.\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right).\left(2+\sqrt{3}\right)}\right)=\dfrac{\sqrt{3}.\left(2+\sqrt{3}\right)+2.\left(2-\sqrt{3}\right)}{1}\)
1: \(=2\sqrt{7}-12\sqrt{7}+15\sqrt{7}+27\sqrt{7}=32\sqrt{7}\)
3: \(=\sqrt{5}-2-\sqrt{14+6\sqrt{5}}\)
\(=\sqrt{5}-2-3-\sqrt{5}=-5\)
4: \(=2\sqrt{3}+3+4-2\sqrt{3}=7\)
5: \(=3-\sqrt{2}+3+\sqrt{2}+4-3=7\)
6: \(=\sqrt{\dfrac{6+2\sqrt{5}}{4}}+\sqrt{\dfrac{14-6\sqrt{5}}{4}}\)
\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}}{2}=\dfrac{4}{2}=2\)
8: \(=\sqrt{5}-1+\sqrt{\dfrac{\left(3-\sqrt{5}\right)^2}{4}}-\sqrt{\dfrac{\left(3+\sqrt{5}\right)^2}{4}}\)
\(=\sqrt{5}-1+\dfrac{3-\sqrt{5}}{2}-\dfrac{3+\sqrt{5}}{2}\)
\(=\dfrac{2\sqrt{5}-2+3-\sqrt{5}-3-\sqrt{5}}{2}=\dfrac{-2}{2}=-1\)
a: \(=\dfrac{1}{2}\cdot2\sqrt{3}+3\sqrt{3}-5\sqrt{3}=-\sqrt{3}\)
b: \(=2-\sqrt{3}-\sqrt{3}-1=1\)
c: \(=18\sqrt{3}-10\sqrt{3}-\dfrac{1}{2}\cdot10\sqrt{3}=3\sqrt{3}\)
d: \(=\sqrt{10}+\sqrt{3}-\sqrt{5}+\sqrt{2}-2\sqrt{3}=\sqrt{10}+\sqrt{2}-\sqrt{3}-\sqrt{5}\)
d) \(\dfrac{4}{\sqrt{7}-\sqrt{3}}+\dfrac{6}{3+\sqrt{3}}+\dfrac{\sqrt{7}-7}{\sqrt{7}-1}\)
\(=\dfrac{4\left(\sqrt{7}+\sqrt{3}\right)}{\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)}+\dfrac{6\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}-\dfrac{7-\sqrt{7}}{\sqrt{7}-1}\)
\(=\dfrac{4\left(\sqrt{7}+\sqrt{3}\right)}{4}+\dfrac{6\left(3-\sqrt{3}\right)}{6}-\dfrac{\sqrt{7}\left(\sqrt{7}-1\right)}{\sqrt{7}-1}\)
\(=\sqrt{7}+\sqrt{3}+3-\sqrt{3}-\sqrt{7}=3\)
b) \(\dfrac{\sqrt{5}-\sqrt{15}}{1-\sqrt{3}}-\sqrt{21+4\sqrt{5}}=\dfrac{\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}-\sqrt{20+2\sqrt{20}+1}\)
\(=\sqrt{5}-\sqrt{\left(\sqrt{20}+1\right)^2}=\sqrt{5}-\left(\sqrt{20}+1\right)=\sqrt{5}-2\sqrt{5}-1=-1-\sqrt{5}\)
1,
a,\(4\sqrt{\dfrac{9}{2}}+\sqrt{2}+\sqrt{\dfrac{1}{18}}=4\sqrt{\dfrac{18}{4}}+\sqrt{2}+\sqrt{\dfrac{1}{9.2}}=4\dfrac{\sqrt{18}}{2}+\sqrt{2}+\dfrac{1}{3}\sqrt{\dfrac{1}{2}}=2\sqrt{9.2}+\sqrt{2}+\dfrac{1}{3}\sqrt{\dfrac{2}{4}}=2.3\sqrt{2}+\sqrt{2}+\dfrac{\sqrt{2}}{6}=6\sqrt{2}+\sqrt{2}+\sqrt{2}\dfrac{1}{6}=\dfrac{43}{6}\sqrt{2}\) b,\(4\sqrt{20}-3\sqrt{125}+5\sqrt{45}-15\sqrt{\dfrac{1}{5}}=4\sqrt{4.5}-3\sqrt{25.5}+5\sqrt{9.5}-15\dfrac{\sqrt{5}}{5}=4.2\sqrt{5}-3.5\sqrt{5}+5.3\sqrt{5}-3\sqrt{5}=8\sqrt{5}-15\sqrt{5}+15\sqrt{5}-3\sqrt{5}=5\sqrt{5}\)
*) Giải phương trình :
\(\sqrt{4x-8}+5\sqrt{x-2}-\sqrt{9x-18}=20\) ( ĐKXĐ : x \(\ge\) 2 )
\(\Leftrightarrow\sqrt{4\left(x-2\right)}+5\sqrt{x-2}-\sqrt{9\left(x-2\right)}=20\)
\(\Leftrightarrow2\sqrt{x-2}+5\sqrt{x-2}-3\sqrt{x-2}=20\)
\(\Leftrightarrow4\sqrt{x-2}=20\)
\(\Leftrightarrow\sqrt{x-2}=5\)
\(\Leftrightarrow x-2=25\)
\(\Leftrightarrow x=27\) ( thỏa mãn điều kiện )
Vậy phương trình có nghiệm x = 27 .