Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(3\sqrt{3}=\sqrt{27}>\sqrt{12}\)
b) \(3\sqrt{5}=\sqrt{45}>\sqrt{27}\)
c) \(\dfrac{1}{3}\sqrt{51}=\sqrt{\dfrac{51}{9}}< \sqrt{\dfrac{54}{9}}=6=\sqrt{\dfrac{150}{25}}=\dfrac{1}{5}\sqrt{150}\)
d) \(\dfrac{1}{2}\sqrt{6}=\sqrt{\dfrac{6}{4}}=\sqrt{\dfrac{3}{2}}< \sqrt{\dfrac{36}{2}}=6\sqrt{\dfrac{1}{2}}\)
b) \(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
\(=\dfrac{\sqrt{2}\cdot\sqrt{12-3\sqrt{7}}-\sqrt{2}\cdot\sqrt{12+3\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{24-6\sqrt{7}}-\sqrt{24+6\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{21}\right)^2-2\cdot\sqrt{21}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{21}\right)^2+2\cdot\sqrt{21}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{21}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{21}+\sqrt{3}\right)^2}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}\)
\(=\dfrac{-2\sqrt{3}}{\sqrt{2}}\)
\(=-\sqrt{6}\)
c) \(\sqrt[3]{\dfrac{3}{4}}\cdot\sqrt[3]{\dfrac{9}{16}}\)
\(=\sqrt[3]{\dfrac{3\cdot9}{4\cdot16}}\)
\(=\sqrt[3]{\left(\dfrac{3}{4}\right)^3}\)
\(=\dfrac{3}{4}\)
d) \(\dfrac{\sqrt[3]{54}}{\sqrt[3]{-2}}\)
\(=\sqrt[3]{\dfrac{54}{-2}}\)
\(=\sqrt[3]{-27}\)
\(=\sqrt[3]{\left(-3\right)^3}\)
\(=-3\)
a: Sửa đề: \(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\)
\(=\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{3}\cdot\sqrt{6}}+\dfrac{\sqrt{3}-\sqrt{2}}{12}\)
\(=\dfrac{\sqrt{6}+1}{3\sqrt{2}}+\dfrac{\sqrt{3}-\sqrt{2}}{12}\)
\(=\dfrac{2\sqrt{2}\left(\sqrt{6}+1\right)+\sqrt{3}-\sqrt{2}}{12}\)
\(=\dfrac{4\sqrt{3}+2\sqrt{2}+\sqrt{3}-\sqrt{2}}{12}\)
\(=\dfrac{5\sqrt{3}+\sqrt{2}}{12}\)
e: \(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)
\(=\sqrt[3]{2\sqrt{2}+3\sqrt{2}+6+1}-\sqrt[3]{2\sqrt{2}-3\sqrt{2}+6-1}\)
\(=\sqrt[3]{\left(\sqrt{2}+1\right)^3}-\sqrt[3]{\left(\sqrt{2}-1\right)^3}\)
\(=\sqrt{2}+1-\left(\sqrt{2}-1\right)\)
\(=\sqrt{2}+1-\sqrt{2}+1=2\)
a) \(\left(-\dfrac{1}{3}\sqrt{63}\right)^2=\dfrac{1}{9}\cdot63=7\)
\(\left(-2\sqrt{2}\right)^2=8\)
mà 7<8
nên \(-\dfrac{1}{3}\sqrt{63}>-2\sqrt{2}\)
b) Ta có: \(\left(2\sqrt{55}\right)^2=4\cdot55=220\)
\(\left(\dfrac{3}{5}\sqrt{750}\right)=\dfrac{9}{25}\cdot750=270\)
mà 220<270
nên \(2\sqrt{55}< \dfrac{3}{5}\sqrt{750}\)
hay \(-2\sqrt{55}< -\dfrac{3}{5}\sqrt{750}\)
a)
Có:
\(2\sqrt{29}=\sqrt{4.29}=\sqrt{116}\\ 3\sqrt{13}=\sqrt{9.13}=\sqrt{117}\)
Vì \(\sqrt{117}>\sqrt{116}\) nên \(3\sqrt{13}>2\sqrt{29}\)
b)
Có:
\(\dfrac{5}{4}\sqrt{2}=\sqrt{\dfrac{25}{16}.2}=\sqrt{\dfrac{25}{8}}\)
\(\dfrac{3}{2}\sqrt{\dfrac{3}{2}}=\sqrt{\dfrac{9}{4}.\dfrac{3}{2}}=\sqrt{\dfrac{27}{8}}\)
Do \(\sqrt{\dfrac{27}{8}}>\sqrt{\dfrac{25}{8}}\) nên \(\dfrac{3}{2}\sqrt{\dfrac{3}{2}}>\dfrac{5}{4}\sqrt{2}\)
c)
Có:
\(5\sqrt{2}=\sqrt{25.2}=\sqrt{50}\)
\(4\sqrt{3}=\sqrt{16.3}=\sqrt{48}\)
Vì \(\sqrt{50}>\sqrt{48}\) nên \(5\sqrt{2}>4\sqrt{3}\)
d)
Có:
\(\dfrac{5}{2}\sqrt{\dfrac{1}{6}}=\sqrt{\dfrac{25}{4}.\dfrac{1}{6}}=\sqrt{\dfrac{25}{24}}\)
\(6\sqrt{\dfrac{1}{37}}=\sqrt{36.\dfrac{1}{37}}=\sqrt{\dfrac{36}{37}}\)
lại có: \(\dfrac{25}{24}>\dfrac{36}{37}\)
\(\Rightarrow\dfrac{5}{2}\sqrt{\dfrac{1}{6}}>6\sqrt{\dfrac{1}{37}}\)
\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)
\(ĐK:x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)
\(\Leftrightarrow4x^2-9=4x+12\)
\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)
\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(ĐK:x\ge5\)
\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)
\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)
ĐK:x>=1
\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)
\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)
\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)
\(ĐK:x\ge3\)
\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)
\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)
\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}=0\) (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)
\(1,\\ a,=\sqrt{\left(3+\sqrt{7}\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}=3+\sqrt{7}-\sqrt{7}+1=4\\ b,K=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{\sqrt{3}-1}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\ c,=\sqrt{\left(6-2\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-4\right)^2}=6-2\sqrt{6}+2\sqrt{6}-4=2\\ e,=\sqrt{\left(2-\sqrt{2}\right)^2}-\left(\sqrt{6}-\sqrt{2}\right)=2-\sqrt{2}-\sqrt{6}+\sqrt{2}=2-\sqrt{6}\)
\(2,\\ a,A=\dfrac{x-3\sqrt{x}+3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+3}{x+9}\\ A=\dfrac{x+9}{\left(\sqrt{x}-3\right)\left(x+9\right)}=\dfrac{1}{\sqrt{x}-3}\\ b,x=4+2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{3}+1\\ \Leftrightarrow A=\dfrac{1}{\sqrt{3}+1-3}=\dfrac{1}{\sqrt{3}+2}=2-\sqrt{3}\)
1: \(\sqrt{3+\sqrt{5}}\cdot\sqrt{2}=\sqrt{6+2\sqrt{5}}=\sqrt{5}+1\)
3) \(\left(\sqrt{\dfrac{3}{4}}-\sqrt{3}+5\cdot\sqrt{\dfrac{4}{3}}\right)\cdot\sqrt{12}\)
\(=\left(\dfrac{\sqrt{3}}{2}-\dfrac{2\sqrt{3}}{2}+5\cdot\dfrac{2}{\sqrt{3}}\right)\cdot\sqrt{12}\)
\(=\dfrac{17\sqrt{3}}{6}\cdot2\sqrt{3}\)
\(=\dfrac{34\cdot3}{6}=\dfrac{102}{6}=17\)
a: \(=\dfrac{2\sqrt{7}-10-6+2\sqrt{7}}{4}+4+2\sqrt{7}-\dfrac{20}{9}+\dfrac{5}{9}\sqrt{7}\)
\(=\sqrt{7}-4+\dfrac{23}{9}\sqrt{7}+\dfrac{16}{9}\)
\(=\dfrac{32}{9}\sqrt{7}-\dfrac{20}{9}\)
b:\(=\dfrac{2\sqrt{6}+4+2\sqrt{6}-4}{2}+\dfrac{5}{6}\sqrt{6}\)
\(=2\sqrt{6}+\dfrac{5}{6}\sqrt{6}=\dfrac{17}{6}\sqrt{6}\)
c: \(=\dfrac{1}{3}\sqrt{3}+\dfrac{1}{6}\sqrt{2}+\dfrac{1}{\sqrt{3}}\cdot\sqrt{\dfrac{5-2\sqrt{6}}{12}}\)
\(=\dfrac{1}{3}\sqrt{3}+\dfrac{1}{6}\sqrt{2}+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\)
\(=\dfrac{2\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{6}=\dfrac{3\sqrt{3}}{6}=\dfrac{\sqrt{3}}{2}\)
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: \(=\dfrac{\sqrt{10}\left(\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}-\sqrt{5}}-2-\sqrt{10}+3\sqrt{7}+2\)
\(=\sqrt{10}-\sqrt{10}+3\sqrt{7}=3\sqrt{7}\)
a: \(4\sqrt{7}=\sqrt{4^2\cdot7}=\sqrt{112}\)
\(3\sqrt{13}=\sqrt{3^2\cdot13}=\sqrt{117}\)
mà 112<117
nên \(4\sqrt{7}< 3\sqrt{13}\)
b: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)
\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)
mà 108>64
nên \(3\sqrt{12}>2\sqrt{16}\)
c: \(\dfrac{1}{4}\sqrt{84}=\sqrt{\dfrac{1}{16}\cdot84}=\sqrt{\dfrac{21}{4}}\)
\(6\sqrt{\dfrac{1}{7}}=\sqrt{36\cdot\dfrac{1}{7}}=\sqrt{\dfrac{36}{7}}\)
mà \(\dfrac{21}{4}>\dfrac{36}{7}\)
nên \(\dfrac{1}{4}\sqrt{84}>6\sqrt{\dfrac{1}{7}}\)
d: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)
\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)
mà 108>64
nên \(3\sqrt{12}>2\sqrt{16}\)