Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
2. a) \(ĐKXĐ:x\ge\frac{1}{3}\)
\(\sqrt{3x-1}=4\)\(\Rightarrow\left(\sqrt{3x-1}\right)^2=4^2\)
\(\Leftrightarrow3x-1=16\)\(\Leftrightarrow3x=17\)\(\Leftrightarrow x=\frac{17}{3}\)( thỏa mãn ĐKXĐ )
Vậy \(x=\frac{17}{3}\)
b) \(ĐKXĐ:x\ge1\)
\(\sqrt{x-1}=x-1\)\(\Rightarrow\left(\sqrt{x-1}\right)^2=\left(x-1\right)^2\)
\(\Leftrightarrow x-1=x^2-2x+1\)\(\Leftrightarrow x^2-2x+1-x+1=0\)
\(\Leftrightarrow x^2-3x+2=0\)\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)( thỏa mãn ĐKXĐ )
Vậy \(x=1\)hoặc \(x=2\)
3. \(\sqrt{7-2\sqrt{6}}-\sqrt{10-4\sqrt{6}}=\sqrt{6-2\sqrt{6}+1}-\sqrt{6-4\sqrt{6}+4}\)
\(=\sqrt{\left(\sqrt{6}-1\right)^2}-\sqrt{\left(\sqrt{6}-2\right)^2}=\left|\sqrt{6}-1\right|-\left|\sqrt{6}-2\right|\)
Vì \(6>1\)\(\Leftrightarrow\sqrt{6}>\sqrt{1}=1\)\(\Rightarrow\sqrt{6}-1>0\)
\(6>4\)\(\Rightarrow\sqrt{6}>\sqrt{4}=2\)\(\Rightarrow\sqrt{6}-2>0\)
\(\Rightarrow\left|\sqrt{6}-1\right|-\left|\sqrt{6}-2\right|=\left(\sqrt{6}-1\right)-\left(\sqrt{6}-2\right)\)
\(=\sqrt{6}-1-\sqrt{6}+2=1\)
hay \(\sqrt{7-2\sqrt{6}}-\sqrt{10-4\sqrt{6}}=1\)
2a) \(\sqrt{3x-1}=4\)( ĐKXĐ : \(x\ge\frac{1}{3}\))
Bình phương hai vế
\(\Leftrightarrow\left(\sqrt{3x-1}\right)^2=4^2\)
\(\Leftrightarrow3x-1=16\)
\(\Leftrightarrow3x=17\)
\(\Leftrightarrow x=\frac{17}{3}\)( tmđk )
Vậy phương trình có nghiệm duy nhất là x = 17/3
b) \(\sqrt{x-1}=x-1\)( ĐKXĐ : \(x\ge1\))
Bình phương hai vế
\(\Leftrightarrow\left(\sqrt{x-1}\right)^2=\left(x-1\right)^2\)
\(\Leftrightarrow x-1=x^2-2x+1\)
\(\Leftrightarrow x^2-2x+1-x+1=0\)
\(\Leftrightarrow x^2-3x+2=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}\left(tmđk\right)}\)
Vậy phương trình có hai nghiệm là x = 1 hoặc x = 2
3. \(\sqrt{7-2\sqrt{6}}-\sqrt{10-4\sqrt{6}}\)
\(=\sqrt{6-2\sqrt{6}+1}-\sqrt{6-4\sqrt{6}+4}\)
\(=\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot1+1^2}-\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot2+2^2}\)
\(=\sqrt{\left(\sqrt{6}-1\right)^2}-\sqrt{\left(\sqrt{6}-2\right)^2}\)
\(=\left|\sqrt{6}-1\right|-\left|\sqrt{6}-2\right|\)
\(=\sqrt{6}-1-\left(\sqrt{6}-2\right)\)
\(=\sqrt{6}-1-\sqrt{6}+2\)
\(=1\)
a/ Bạn ghi nhầm đề rồi
c/ \(2\sqrt{18\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5\sqrt{48}}\)
\(=2\sqrt{18}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-3\sqrt{5}.\sqrt{\sqrt{48}}\)
\(=2.3\sqrt{2}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-3\sqrt{5}.\sqrt{4\sqrt{3}}\)
\(=2.3\sqrt{2}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-6\sqrt{5}.\sqrt{\sqrt{3}}\)
\(=2\sqrt{\sqrt{3}}\left(3\sqrt{2}-\sqrt{5}-3\sqrt{5}\right)\)
\(=2\sqrt{\sqrt{3}}\left(3\sqrt{2}-4\sqrt{5}\right)\)\(=2\sqrt{2\sqrt{3}}\left(3-2\sqrt{10}\right)\)
f/ \(\sqrt{2}.\sqrt{2+\sqrt{3}}-2\left(\sqrt{3}-1\right)=\sqrt{4+2\sqrt{3}}-2\left(\sqrt{3}-1\right)\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-2\left(\sqrt{3}-1\right)=\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)\)
\(=\sqrt{3}+1-2\sqrt{3}+2=3-\sqrt{3}=\sqrt{3}\left(\sqrt{3}-1\right)\)
g/ \(\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}-2\sqrt{3}+2007\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-2\sqrt{3}+2007\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}-2\sqrt{3}+2007\)
\(=2007\)
Bài 2:
\(D=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{120\sqrt{121}+121\sqrt{120}}\)
Với mọi \(n\inℕ^∗\)ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)
\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{[\left(n+1\right)\sqrt{n}]^2-\left(n\sqrt{n+1}\right)^2}\)
\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)^2-n^2\left(n+1\right)}\)
\(=\frac{\left(n+1\right)\sqrt{n}-n\left(\sqrt{n}+1\right)}{n\left(n+1\right)\left(n+1-n\right)}\)
\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}\)
\(=\frac{\left(n+1\right)\sqrt{n}}{n\left(n+1\right)}-\frac{n\sqrt{n+1}}{n\left(n+1\right)}\)
\(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
\(\Rightarrow D=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{4}}+....+\frac{1}{\sqrt{120}}-\frac{1}{\sqrt{121}}\)
\(=1-\frac{1}{\sqrt{121}}=\frac{10}{11}\)
Bài 1: chắc lại phải "liên hợp" gì đó rồi:V
\(\sqrt{2009}-\sqrt{2008}=\frac{1}{\sqrt{2009}+\sqrt{2008}}\)
\(\sqrt{2007}-\sqrt{2006}=\frac{1}{\sqrt{2007}+\sqrt{2006}}\)
Đó \(\sqrt{2009}+\sqrt{2008}>\sqrt{2007}+\sqrt{2006}\)
Nên \(\sqrt{2009}-\sqrt{2008}< \sqrt{2007}-\sqrt{2006}\)
Tổng quát ta có bài toán sau, với So sánh \(\sqrt{n}-\sqrt{n-1}\text{ và }\sqrt{n-2}-\sqrt{n-3}\)
Với \(n\ge3\). Lời giải xin mời các bạn:)
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=,\(\frac{6\sqrt{3}}{3}-\frac{\sqrt{3}\left(1-\sqrt{3}\right)}{\sqrt{3}}-\frac{\sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{2}-\sqrt{3}}\)
=\(6-1+\sqrt{3}-\sqrt{6}\)
=\(5+\sqrt{3}+\sqrt{6}\)
\(a,\frac{6}{4+\sqrt{4-2\sqrt{3}}}=\frac{6}{4+\sqrt{\sqrt{3}^2-2\sqrt{3}+\sqrt{1}^2}}\)
\(=\frac{6}{4+\sqrt{\left(\sqrt{3}-\sqrt{1}\right)^2}}=\frac{6}{4+|\sqrt{3}-1|}=\frac{6}{3+\sqrt{3}}\)
\(=\frac{6}{\sqrt{3}\left(\sqrt{3}+1\right)}=\frac{\sqrt{36}}{\sqrt{3}\left(\sqrt{3}+1\right)}=\frac{\sqrt{3}.\sqrt{12}}{\sqrt{3}\left(\sqrt{3}+1\right)}=\frac{\sqrt{12}}{\sqrt{3}+1}\)
\(d,\frac{1}{\sqrt{7-2\sqrt{10}}}+\frac{1}{\sqrt{7+2\sqrt{10}}}\)
\(=\frac{1}{\sqrt{\sqrt{5}^2-2.\sqrt{2}.\sqrt{5}+\sqrt{2}^2}}+\frac{1}{\sqrt{\sqrt{5}^2+2.\sqrt{2}.\sqrt{5}+\sqrt{2}^2}}\)
\(=\frac{1}{\sqrt{\left(\sqrt{5}-\sqrt{2}\right)}}+\frac{1}{\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}\)
\(=\frac{1}{\sqrt{5}-\sqrt{2}}+\frac{1}{\sqrt{5}+\sqrt{2}}=\frac{\sqrt{5}+\sqrt{2}+\sqrt{5}-\sqrt{2}}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}\)
\(=\frac{2\sqrt{5}}{\sqrt{5}^2-\sqrt{2}^2}=\frac{\sqrt{5.4}}{5-2}=\frac{\sqrt{20}}{3}\)
a) \(\frac{3}{\sqrt{5}}=\frac{3\sqrt{5}}{\sqrt{5}.\sqrt{5}}=\frac{3\sqrt{5}}{5}\)
\(\frac{2\sqrt{3}}{\sqrt{2}}=\frac{2\sqrt{3}.\sqrt{2}}{\sqrt{2}.\sqrt{2}}=\frac{2\sqrt{6}}{2}=\sqrt{6}\)
\(\frac{a}{\sqrt{b}}=\frac{a\sqrt{b}}{\sqrt{b}.\sqrt{b}}=\frac{a\sqrt{b}}{b}\)
\(\frac{x+1}{\sqrt{x^2-1}}=\frac{\left(x+1\right)\left(\sqrt{x^2-1}\right)}{\left(\sqrt{x^2-1}\right)\left(\sqrt{x^2-1}\right)}\) = \(\frac{\left(\sqrt{x^2-1}\right)\left(x+1\right)}{x^2-1}\)
b/ \(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{n}+\sqrt{n+1}}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{n+1}-\sqrt{n}\)
\(=\sqrt{n+1}-1\)
Câu a quy đồng từ từ từ phải qua trái là ra
a) \(\sqrt{75}-\sqrt{5\frac{1}{3}}+\frac{9}{2}\sqrt{2\frac{2}{3}}+2\sqrt{27}\)
\(=\sqrt{75}-\sqrt{\frac{16}{3}}+\frac{9}{2}\sqrt{\frac{8}{3}}+2\sqrt{27}\)
\(=5\sqrt{3}-\frac{4}{\sqrt{3}}+3\sqrt{6}+6\sqrt{3}\)
\(=-\frac{4}{\sqrt{3}}+5\sqrt{3}+3\sqrt{6}+6\sqrt{3}\)
\(=-\frac{4}{\sqrt{3}}+11\sqrt{3}+3\sqrt{6}\)
\(=-\frac{4\sqrt{3}}{3}+11\sqrt{3}+3\sqrt{6}\)
b) \(\sqrt{48}-\sqrt{5\frac{1}{3}}+2\sqrt{75}-5\sqrt{1\frac{1}{3}}\)
\(=\sqrt{48}-\sqrt{\frac{16}{3}}+2\sqrt{75}-5\sqrt{\frac{4}{3}}\)
\(=4\sqrt{3}-\frac{4}{\sqrt{3}}+10\sqrt{3}-\frac{10}{\sqrt{3}}\)
\(=-\frac{4}{\sqrt{3}}-\frac{10}{\sqrt{3}}+4\sqrt{3}+10\sqrt{3}\)
\(=-\frac{14\sqrt{3}}{3}+4\sqrt{3}+10\sqrt{3}\)
\(=-\frac{14\sqrt{3}}{3}+14\sqrt{3}\)
c)\(\left(\sqrt{15}+2\sqrt{3}\right)^2+12\sqrt{5}\)
\(=27+12\sqrt{5}+12\sqrt{5}\)
\(=27+24\sqrt{5}\)
d)\(\left(\sqrt{6}+2\right)\left(\sqrt{3}-\sqrt{2}\right)\)
\(=\sqrt{6}+2-\sqrt{3}-\sqrt{2}\)
e) \(\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4\)
\(=4+2\sqrt{3}-2\sqrt{3}+4\)
= 8
f) \(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}\)
\(=\frac{7-4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}+\frac{7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
\(=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
\(=\frac{14}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
= 14
a) \(2\sqrt{2}.\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}=9\)
\(=2\sqrt{2}.\left(\sqrt{3}-2\right)+9+4\sqrt{2}-2\sqrt{6}\)
\(=2\sqrt{6}-4\sqrt{2}+9+4\sqrt{2}-2\sqrt{6}\)
= 9 (đpcm)
b) \(\sqrt{\sqrt{2}+1}-\sqrt{\sqrt{2}-1}=\sqrt{2\left(\sqrt{2}-1\right)}\)
\(=\sqrt{\sqrt{2}+1}-\sqrt{\sqrt{2}-1}=\sqrt{2^{\frac{1}{2}}\left(\sqrt{2}-1\right)}\)
\(=\sqrt{2\left(\sqrt{2}-1\right)}\) (đpcm)