Áp dụng quy tắc chia hai căn bậc hai, hãy tính :
a) \(\dfrac{\sqrt{2300}}{\sqrt{23}}\)
b) \(\dfrac{\sqrt{12,5}}{\sqrt{0,5}}\)
c) \(\dfrac{\sqrt{192}}{\sqrt{12}}\)
d) \(\dfrac{\sqrt{6}}{\sqrt{150}}\)
e,\(\sqrt{\dfrac{9}{169}}\)
f,\(\sqrt{1\dfrac{9}{16}}\)
g,\(\dfrac{\sqrt{2300}}{\sqrt{23}}\)
h,\(\dfrac{\sqrt{12,5}}{\sqrt{0,5}}\)
\(e,\sqrt{\dfrac{9}{169}}=\dfrac{\sqrt{9}}{\sqrt{169}}=\dfrac{\sqrt{3^2}}{\sqrt{13^2}}=\dfrac{3}{13}\)
\(f,\sqrt{1\dfrac{9}{16}}=\sqrt{\dfrac{25}{16}}=\dfrac{\sqrt{25}}{\sqrt{16}}=\dfrac{\sqrt{5^2}}{\sqrt{4^2}}=\dfrac{5}{4}\)
\(g,\dfrac{\sqrt{2300}}{\sqrt{23}}=\sqrt{\dfrac{2300}{23}}=\sqrt{100}=\sqrt{10^2}=10\)
\(h,\dfrac{\sqrt{12,5}}{\sqrt{0,5}}=\sqrt{\dfrac{12,5}{0,5}}=\sqrt{25}=\sqrt{5^2}=5\)
e, \(\sqrt{\dfrac{9}{169}}\)
\(=\sqrt{\dfrac{3^2}{13^2}}\)
\(=\dfrac{3}{13}\)
f, \(\sqrt{1\dfrac{9}{16}}\)
\(=\sqrt{\dfrac{25}{16}}\)
\(=\sqrt{\dfrac{5^2}{4^2}}\)
\(=\dfrac{5}{4}\)
g, \(\dfrac{\sqrt{2300}}{\sqrt{23}}\)
\(=\dfrac{10\sqrt{23}}{\sqrt{23}}\)
\(=10\)
h, \(\dfrac{\sqrt{12,5}}{\sqrt{0,5}}\)
\(=\dfrac{\dfrac{5\sqrt{2}}{2}}{\dfrac{\sqrt{2}}{2}}\)
\(=\dfrac{\dfrac{5\sqrt{2}}{2}\cdot2}{\sqrt{2}}\)
\(=\dfrac{5\sqrt{2}}{\sqrt{2}}=5\)
1. Áp dụng quy tắc khai phương một thương, hãy tính:
a, \(\sqrt{\dfrac{36}{121}}\) b, \(\sqrt{\dfrac{9}{16}:\dfrac{25}{36}}\) c, \(\sqrt{0,0169}\)
d,\(\dfrac{\sqrt{15}}{\sqrt{735}}\) e, \(\sqrt{\dfrac{81}{8}:\sqrt{3\dfrac{1}{8}}}\) g, \(\dfrac{\sqrt{12,5}}{\sqrt{0,5}}\)
2. Tính:
a,\(\sqrt{\dfrac{25}{144}}\) b,\(\sqrt{2\dfrac{7}{81}}\) c,\(\sqrt{\dfrac{2,25}{16}}\) d, \(\sqrt{\dfrac{1,21}{0,49}}\)
3. Áp dụng quy tắc chia hai căn bậc hai, hãy tính:
a, \(\sqrt{18}:\sqrt{2}\) b, \(\sqrt{45}:\sqrt{80}\)
c, (\(\sqrt{20}-\sqrt{45}+\sqrt{5}\) ) : \(\sqrt{5}\) d, \(\dfrac{\sqrt{8^2}}{\sqrt{4^5.2^3}}\)
4. Khẳng định nào sau đây là đúng?
A. \(\sqrt{\dfrac{3}{\left(-5\right)^2}}=-\dfrac{\sqrt{3}}{5}\) B. \(\left(\sqrt{\dfrac{-3}{-5}}\right)^2=\dfrac{3}{5}\)
5. Tính.
a, \(\sqrt{2\dfrac{7}{81}}:\dfrac{\sqrt{6}}{\sqrt{150}}\) b, \(\left(\sqrt{12}+\sqrt{27}-\sqrt{3}\right):\sqrt{3}\)
c, \(\left(\sqrt{\dfrac{1}{5}-\sqrt{\dfrac{9}{5}}+\sqrt{5}}\right):\sqrt{5}\) d, \(\sqrt{\dfrac{2+\sqrt{3}}{\sqrt{2}}}\)
6. So sánh
a, So sánh \(\sqrt{144-49}\) và \(\sqrt{144}-\sqrt{49}\);
b, Chứng minh rằng , với hai số a,b thỏa mãn a> b> 0 thì \(\sqrt{a}-\sqrt{b}< \sqrt{a-b}\)
1
a,\(\sqrt{\dfrac{36}{121}}=\sqrt{\dfrac{6^2}{11^2}}=\dfrac{6}{11}\)
\(\sqrt{\dfrac{9}{16}:\dfrac{25}{36}}=\sqrt{\dfrac{81}{100}}=\sqrt{\dfrac{9^2}{10^2}}=\dfrac{9}{10}\)
Bài 2:
a: \(\sqrt{\dfrac{25}{144}}=\dfrac{5}{12}\)
b: \(\sqrt{2+\dfrac{7}{81}}=\sqrt{\dfrac{169}{81}}=\dfrac{13}{9}\)
c: \(\sqrt{\dfrac{2.25}{16}}=\dfrac{1.5}{4}=\dfrac{3}{8}\)
d: \(\sqrt{\dfrac{1.21}{0.49}}=\sqrt{\dfrac{121}{49}}=\dfrac{11}{7}\)
Bài3:
a: \(=\sqrt{\dfrac{18}{2}}=\sqrt{9}=3\)
b: \(=\sqrt{\dfrac{45}{80}}=\sqrt{\dfrac{9}{16}}=\dfrac{3}{4}\)
c: \(=\dfrac{2\sqrt{5}-3\sqrt{5}+\sqrt{5}}{\sqrt{5}}=0\)
d: \(=\sqrt{\dfrac{2^6}{2^{10}\cdot2^3}}=\sqrt{\dfrac{1}{2^7}}=\dfrac{1}{8\sqrt{2}}=\dfrac{\sqrt{2}}{16}\)
Tìm x để các căn bậc hai sau có nghĩa
a) \(\sqrt{\dfrac{15+3x^2}{-6}}\) b) \(\sqrt{\dfrac{-81}{-12-x^2}}\)
c) \(\sqrt{\dfrac{31\left(x^2+21\right)}{3}}\) d) \(\sqrt{\dfrac{-12}{11+x^2}}\)
e) \(\sqrt{\dfrac{21}{-x^2-17}}\)
a: ĐKXĐ: 3x^2+15/-6>=0
=>3x^2+15<=0(vô lý)
b: ĐKXĐ: -81/-x^2-12>=0
=>-x^2-12<0
=>-x^2<12
=>x^2>-12(luôn đúng)
c: ĐKXĐ: 31(x^2+21)/3>=0
=>x^2+21>=0(luôn đúng)
d: ĐKXĐ: -12/x^2+11>=0
=>x^2+11<0(vô lý)
e: ĐKXĐ: 21/-x^2-17>=0
=>-x^2-17>0
=>x^2+17<0(vô lý)
Các số sau đây có căn bậc hai không?
a) A = \(\left(1-\dfrac{\sqrt{3}-1}{2}\right):\left(\dfrac{\sqrt{3}-1}{2}+2\right)\)
b) B = \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right):\dfrac{1}{\sqrt{5}-\sqrt{2}}\)
a) \(A=\left(1-\dfrac{\sqrt{3}-1}{2}\right):\left(\dfrac{\sqrt{3}-1}{2}+2\right)\)
\(=\left(\dfrac{2}{2}-\dfrac{\sqrt{3}-1}{2}\right):\left(\dfrac{\sqrt{3}-1}{2}+\dfrac{4}{2}\right)\)
\(=\dfrac{2-\left(\sqrt{3}-1\right)}{2}:\dfrac{\left(\sqrt{3}-1\right)+4}{2}\)
\(=\dfrac{3-\sqrt{3}}{2}.\dfrac{2}{\sqrt{3}+3}\)
\(=\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}\left(1+\sqrt{3}\right)}\)
\(=\dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\dfrac{\left(\sqrt{3}-1\right)^2}{2}\)
Vì \(\left\{{}\begin{matrix}\left(\sqrt{3}-1\right)^2>0\\2>0\end{matrix}\right.\) \(\Rightarrow\dfrac{\left(\sqrt{3}-1\right)^2}{2}>0\) hay A>0
=> A có căn bậc 2
Vậy......
b)\(B=\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right):\dfrac{1}{\sqrt{5}-\sqrt{2}}\)
\(=\left(\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)\left(1+\sqrt{3}\right)}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}-\sqrt{5}\right):\dfrac{\sqrt{5}+\sqrt{2}}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}\)
\(=\left(\dfrac{\sqrt{2}\left(3-1\right)}{1-3}-\sqrt{5}\right).\dfrac{5-2}{\sqrt{5}+\sqrt{2}}\)
\(=\left(-\sqrt{2}-\sqrt{5}\right).\dfrac{3}{\sqrt{5}+\sqrt{2}}\)
\(=-\left(\sqrt{2}+\sqrt{5}\right).\dfrac{3}{\sqrt{5}+\sqrt{2}}\)
\(=-3\)
Vì -3 < 0 hay B < 0
=> B không có căn bậc 2
Vậy.....
Áp dụng quy tắc khai phương một thương, hãy tính :
a) \(\sqrt{\dfrac{9}{169}}\)
b) \(\sqrt{\dfrac{25}{144}}\)
c) \(\sqrt{1\dfrac{9}{16}}\)
d) \(\sqrt{2\dfrac{7}{81}}\)
Áp dụng quy tắc khai phương một thương, hãy tính :
9169" id="MathJax-Element-1-Frame" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; font-size:18px; font-style:normal; font-weight:normal; letter-spacing:normal; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; text-align:left; text-indent:0px; text-transform:none; white-space:nowrap; word-spacing:normal; word-wrap:normal" tabindex="0">9169−−−−√ = \(\sqrt{\dfrac{3^2}{13^2}}\) = \(\left|\dfrac{3}{13}\right|\) = \(\dfrac{3}{13}\)
25144" id="MathJax-Element-2-Frame" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; font-size:18px; font-style:normal; font-weight:normal; letter-spacing:normal; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; text-align:left; text-indent:0px; text-transform:none; white-space:nowrap; word-spacing:normal; word-wrap:normal" tabindex="0">25144−−−−√ = \(\sqrt{\dfrac{5^2}{12^2}}\) = \(\left|\dfrac{5}{12}\right|\) = \(\dfrac{5}{12}\)
916" id="MathJax-Element-3-Frame" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; font-size:18px; font-style:normal; font-weight:normal; letter-spacing:normal; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; text-align:left; text-indent:0px; text-transform:none; white-space:nowrap; word-spacing:normal; word-wrap:normal" tabindex="0">1916−−−−√ = \(\sqrt{\dfrac{25}{16}}\) = \(\sqrt{\dfrac{5^2}{4^2}}\) = \(\left|\dfrac{5}{4}\right|\) = \(\dfrac{5}{4}\)
781" id="MathJax-Element-4-Frame" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-table; float:none; font-size:18px; font-style:normal; font-weight:normal; letter-spacing:normal; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; text-align:left; text-indent:0px; text-transform:none; white-space:nowrap; word-spacing:normal; word-wrap:normal" tabindex="0">2781−−−−√ = \(\sqrt{\dfrac{169}{81}}\) = \(\sqrt{\dfrac{13^2}{9^2}}\) = \(\left|\dfrac{13}{9}\right|\) = \(\dfrac{13}{9}\)
Tính:
a. \(5\sqrt{2}-2\sqrt{48}+6\sqrt{75}-\sqrt{108}\)
b.\(2\sqrt{147}-\dfrac{3}{32}\sqrt{192}+\dfrac{4}{18}\sqrt{243}-\dfrac{1}{10}\sqrt{300}\)
c. \(-\dfrac{1}{2}\sqrt{108}+\dfrac{1}{15}\sqrt{75}-\dfrac{1}{22}\sqrt{363}+\sqrt{12}\)
d. \(\dfrac{5}{8}\sqrt{48}-\dfrac{1}{33}\sqrt{363}+\dfrac{3}{14}\sqrt{147}-\dfrac{1}{4}\sqrt{192}\)
e. \(\dfrac{3}{2}\sqrt{12}+\dfrac{7}{5}\sqrt{75}-\dfrac{9}{10}\sqrt{300}+\dfrac{11}{6}\sqrt{108}\)
a: \(5\sqrt{2}-8\sqrt{3}+30\sqrt{3}-6\sqrt{3}=5\sqrt{2}+16\sqrt{3}\)
b: \(=14\sqrt{3}-\dfrac{3}{32}\cdot8\sqrt{3}+\dfrac{4}{18}\cdot9\sqrt{3}-\dfrac{1}{10}\cdot10\sqrt{3}\)
\(=14\sqrt{3}-\dfrac{3}{4}\sqrt{3}+2\sqrt{3}-1\sqrt{3}=\dfrac{57}{4}\sqrt{3}\)
c: \(=\dfrac{-1}{2}\cdot6\sqrt{3}+\dfrac{1}{15}\cdot5\sqrt{3}-\dfrac{1}{22}\cdot11\sqrt{3}+2\sqrt{3}\)
\(=-3\sqrt{3}+\dfrac{1}{3}\sqrt{3}-\dfrac{1}{2}\sqrt{3}+2\sqrt{3}=-\dfrac{7}{6}\sqrt{3}\)
d: \(=\dfrac{5}{8}\cdot4\sqrt{3}-\dfrac{1}{33}\cdot11\sqrt{3}+\dfrac{3}{14}\cdot7\sqrt{3}-\dfrac{1}{4}\cdot8\sqrt{3}\)
\(=\dfrac{5}{2}\sqrt{3}-\dfrac{1}{3}\sqrt{3}+\dfrac{3}{2}\sqrt{3}-2\sqrt{3}=\dfrac{5}{3}\sqrt{3}\)
Áp dụng quy tắc chia hai căn bậc hai , hãy tính :
\(\frac{\sqrt{8^2}}{\sqrt{4^{5.}}\sqrt{2^3}}\)
Rút gọn các biểu thức sau:
a) \(\dfrac{2}{5}\sqrt{75}-0,5\sqrt{48}+\sqrt{300}-\dfrac{2}{3}\sqrt{12}\)
b) \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}\)
c) \(3\sqrt{2}-2\sqrt{3}+2\sqrt{3}+3\sqrt{2}\)
d) \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
e) \(\dfrac{\sqrt{a}-\sqrt{b}^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}\) với a > 0, b > 0
a, \(\dfrac{2}{5}\sqrt{75}-0,5\sqrt{48}+\sqrt{300}-\dfrac{2}{3}\sqrt{12}=2\sqrt{3}-2\sqrt{3}+10\sqrt{3}-\dfrac{4}{3}\sqrt{3}=\dfrac{26}{3}\sqrt{3}\)
b, \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}=\dfrac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}+\dfrac{3}{\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}\)
\(=\dfrac{\sqrt{6}}{2}+\dfrac{\sqrt{3}}{\sqrt{3}+\sqrt{2}}\)
\(=\dfrac{\sqrt{6}}{2}+\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)\)
\(=\dfrac{\sqrt{6}}{2}+3-\sqrt{6}=\dfrac{6-\sqrt{6}}{2}\)
c, \(3\sqrt{2}-2\sqrt{3}+2\sqrt{3}+3\sqrt{2}=6\sqrt{2}\)
d, \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{\left(\sqrt{6}-3\right)^2}+\sqrt{\left(2\sqrt{6}+3\right)^2}\)
\(=-\sqrt{6}+3+2\sqrt{6}+3=\sqrt{6}+6\)
e, Ghi đúng đề.
\(\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}=\dfrac{a+b-2\sqrt{ab}+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}=2\sqrt{b}\)
THỰC HIỆN PHÉP TÍNH
1,\(\sqrt{1\dfrac{9}{16}}\)
2,\(\dfrac{\sqrt{12,5}}{0,5}\)
3,\(\sqrt{\dfrac{25}{64}}\)
4,\(\dfrac{\sqrt{230}}{\sqrt{2,3}}\)
5,\((\sqrt{\dfrac{2}{3}}+\sqrt{\dfrac{50}{3}}-\sqrt{24}).\sqrt{6}\)
1) \(\sqrt{1\dfrac{9}{16}}=\sqrt{\dfrac{25}{16}}=\dfrac{5}{4}\)
2) \(\dfrac{\sqrt{12.5}}{0.5}=\sqrt{\dfrac{12.5}{0.25}}=5\sqrt{2}\)
3) \(\sqrt{\dfrac{25}{64}}=\dfrac{5}{8}\)
4) \(\dfrac{\sqrt{230}}{\sqrt{2.3}}=\sqrt{\dfrac{230}{2.3}}=\sqrt{100}=10\)
5) \(\left(\sqrt{\dfrac{2}{3}}+\sqrt{\dfrac{50}{3}}-\sqrt{24}\right)\cdot\sqrt{6}\)
\(=\left(\dfrac{\sqrt{2}}{\sqrt{3}}+\dfrac{5\sqrt{2}}{\sqrt{3}}-2\sqrt{6}\right)\cdot\sqrt{6}\)
\(=\left(\dfrac{6\sqrt{2}}{\sqrt{3}}-2\sqrt{6}\right)\cdot\sqrt{6}\)
\(=0\cdot\sqrt{6}=0\)