1) Tính:
a) \(\sqrt{4.36}\)
b) (\(\sqrt{8}\) - 3\(\sqrt{2}\)) . \(\sqrt{2}\)
c)\(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}\)
d) \(\dfrac{2}{\sqrt{5}+2}\) + \(\dfrac{2}{\sqrt{5}-2}\)
1) Tìm ĐK của x để các căn thức sau có nghĩa:
a) \(\sqrt{x-2}\) b) \(\sqrt{2-3x}\)
2) Tính:
a) (\(\sqrt{8}-3\sqrt{2}\) ). \(\sqrt{2}\) b)\(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}\)
c) \(\sqrt{4.36}\) d) \(\sqrt{\dfrac{25}{81}.\dfrac{16}{49}}\)
3) Rút gọn:
a) \(\sqrt{19+\sqrt{136}}-\sqrt{19-\sqrt{136}}\) b) \(\sqrt[3]{27}+\sqrt[3]{-64}+2.\sqrt[3]{125}\)
4) Tìm x, biết:
\(\sqrt{4x+20}-2\sqrt{x+5}+\sqrt{9x+45}=6\)
5) Cho :
B = (\(\dfrac{1}{x+2\sqrt{x}}-\dfrac{1}{\sqrt{x}+2}\)) : \(\dfrac{1-\sqrt{x}}{x+4\sqrt{x}+4}\) ( với x > 0; x khác 1)
a) Rút gọn B
b) Tìm x để B = \(\dfrac{5}{2}\)
\(1,\\ a,ĐK:x-2\ge0\Leftrightarrow x\ge2\\ b,ĐK:2-3x\ge0\Leftrightarrow x\le\dfrac{2}{3}\\ 2,\\ a,=\sqrt{16}-3\sqrt{4}=4-6=-2\\ b,=\dfrac{-\sqrt{7}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=-\sqrt{7}\\ c,=\sqrt{4}\cdot\sqrt{36}=2\cdot6=12\\ d,=\sqrt{\dfrac{25}{81}}\cdot\sqrt{\dfrac{16}{49}}=\dfrac{5}{9}\cdot\dfrac{4}{7}=\dfrac{20}{63}\\ 3,\\ a,=\sqrt{19+2\sqrt{34}}-\sqrt{19-2\sqrt{34}}\\ =\sqrt{\left(\sqrt{17}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{17}-\sqrt{2}\right)^2}=\sqrt{17}+\sqrt{2}-\sqrt{17}+\sqrt{2}=2\sqrt{2}\\ b,=3-4+2\cdot5=9\)
\(4,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-2\sqrt{x+5}+3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=2\\ \Leftrightarrow x+5=4\Leftrightarrow x=-1\left(tm\right)\\ 5,\\ a,B=\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}+2\right)^2}{1-\sqrt{x}}=\dfrac{\sqrt{x}+2}{\sqrt{x}}\\ b,B=\dfrac{5}{2}\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}}=\dfrac{5}{2}\\ \Leftrightarrow2\sqrt{x}+4=5\sqrt{x}\\ \Leftrightarrow3\sqrt{x}=4\Leftrightarrow\sqrt{x}=\dfrac{4}{3}\Leftrightarrow x=\dfrac{16}{9}\)
* Thực hiện phép tính:
a. \(\dfrac{\sqrt{7}-5}{2}-\dfrac{6-2\sqrt{7}}{4}+\dfrac{6}{\sqrt{7}-2}-\dfrac{5}{4+\sqrt{7}}\)
b. \(\dfrac{2}{\sqrt{6}-2}+\dfrac{2}{\sqrt{6}+2}+\dfrac{5}{\sqrt{6}}\)
c. \(\dfrac{1}{\sqrt{3}+\sqrt{2}-\sqrt{5}}-\dfrac{1}{\sqrt{3}+\sqrt{2}+\sqrt{5}}\)
\(a,=\dfrac{\sqrt{7}-5}{2}-\dfrac{3-\sqrt{7}}{2}+\dfrac{6\left(\sqrt{7}+2\right)}{3}-\dfrac{5\left(4-\sqrt{7}\right)}{9}\\ =\dfrac{\sqrt{7}-5-3+\sqrt{7}}{2}+2\sqrt{7}+4-\dfrac{20-5\sqrt{7}}{9}\\ =\dfrac{2\sqrt{7}-8}{2}+2\sqrt{7}+4-\dfrac{20-5\sqrt{7}}{9}\\ =\sqrt{7}-4+2\sqrt{7}+4-\dfrac{20-5\sqrt{7}}{9}\\ =\dfrac{27\sqrt{7}-20+5\sqrt{7}}{9}=\dfrac{32\sqrt{7}-20}{9}\)
\(b,=\dfrac{2\left(\sqrt{6}+2\right)}{2}+\dfrac{2\left(\sqrt{6}-2\right)}{2}+\dfrac{5\sqrt{6}}{6}\\ =\sqrt{6}+2+\sqrt{6}-2+\dfrac{5\sqrt{6}}{6}\\ =\dfrac{12\sqrt{6}+5\sqrt{6}}{6}=\dfrac{17\sqrt{6}}{6}\)
\(c,=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{5}-\sqrt{3}-\sqrt{2}+\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}\right)^2-5}\\ =\dfrac{2\sqrt{5}}{5+2\sqrt{6}-5}=\dfrac{2\sqrt{5}}{2\sqrt{6}}=\dfrac{\sqrt{30}}{6}\)
tính:
a) \(\sqrt{\dfrac{1}{8}}.\sqrt{2}.\sqrt{125}.\sqrt{\dfrac{1}{5}}\)
b)\(\sqrt{\sqrt{2}-1}.\sqrt{\sqrt{2}+1}\)
c) \(\sqrt{11-6\sqrt{2}}.\sqrt{11+6\sqrt{2}}\)
d) \(\sqrt{12-6\sqrt{3}}.\sqrt{\dfrac{1}{3-\sqrt{3}}}\)
e) \(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}\)
f) \(\dfrac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}\)
g) \(\left(\dfrac{1}{5-2\sqrt{6}}+\dfrac{2}{5+2\sqrt{6}}\right)\left(15+2\sqrt{6}\right)\)
a) \(\sqrt{\dfrac{1}{8}}\cdot\sqrt{2}\cdot\sqrt{125}\cdot\sqrt{\dfrac{1}{5}}\) = \(\sqrt{\dfrac{1}{8}\cdot2}.\sqrt{125\cdot\dfrac{1}{5}}=\sqrt{\dfrac{1}{4}}.\sqrt{25}=\dfrac{1}{2}\cdot5=2,5\)
b)\(\sqrt{\sqrt{2}-1}.\sqrt{\sqrt{2}+1}=\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\sqrt{2-1}=1\)
c) \(\sqrt{11-6\sqrt{2}}.\sqrt{11+6\sqrt{2}}=\sqrt{\left(11-6\sqrt{2}\right)\left(11+6\sqrt{2}\right)}=\sqrt{121-72}=\sqrt{49}=7\)
* Thực hiện phép tính:
a. \(2\sqrt{18}-9\sqrt{50}+3\sqrt{8}\)
b. \(\left(\sqrt{7}-\sqrt{3}\right)^2+7\sqrt{84}\)
c. \(\left(\dfrac{6-2\sqrt{2}}{3-\sqrt{2}}\dfrac{5}{\sqrt{5}}\right):\dfrac{1}{2-\sqrt{5}}\)
* Tìm x, biết:
a. \(\sqrt{\left(2x+3\right)^2}=8\)
b. \(\sqrt{9x}-7\sqrt{x}=8-6\sqrt{x}\)
c. \(\sqrt{9x-9}+1=13\)
bài 1:
a: Ta có: \(2\sqrt{18}-9\sqrt{50}+3\sqrt{8}\)
\(=6\sqrt{2}-45\sqrt{2}+6\sqrt{2}\)
\(=-33\sqrt{2}\)
b: Ta có: \(\left(\sqrt{7}-\sqrt{3}\right)^2+7\sqrt{84}\)
\(=10-2\sqrt{21}+14\sqrt{21}\)
\(=12\sqrt{21}+10\)
Bài 2:
a: Ta có: \(\sqrt{\left(2x+3\right)^2}=8\)
\(\Leftrightarrow\left|2x+3\right|=8\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=8\\2x+3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{11}{2}\end{matrix}\right.\)
b: Ta có: \(\sqrt{9x}-7\sqrt{x}=8-6\sqrt{x}\)
\(\Leftrightarrow4\sqrt{x}=8\)
hay x=4
c: Ta có: \(\sqrt{9x-9}+1=13\)
\(\Leftrightarrow3\sqrt{x-1}=12\)
\(\Leftrightarrow x-1=16\)
hay x=17
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: \(=\dfrac{5}{8}\cdot4\sqrt{3}-\dfrac{1}{33}\cdot11\sqrt{3}+\dfrac{3}{14}\cdot7\sqrt{3}-\dfrac{1}{4}\cdot8\sqrt{3}\)
\(=\dfrac{5}{2}\sqrt{3}-\dfrac{1}{3}\sqrt{3}+\dfrac{3}{2}\sqrt{3}-2\sqrt{3}=\dfrac{5}{3}\sqrt{3}\)
rút gọn
a.\(\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right)\div\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
b.\(\sqrt{2}+\dfrac{1}{\sqrt{5+2\sqrt{6}}}+\dfrac{2}{\sqrt{8+2\sqrt{15}}}\)
`a)((sqrt(14)-sqrt7)/(1-sqrt2)+(sqrt{15}-sqrt5)/(1-sqrt3)):1/(sqrt7-sqrt5)`
`=((sqrt7(sqrt2-1))/(1-sqrt2)+(sqrt5(sqrt3-1))/(1-sqrt3)).(sqrt7-sqrt5)`
`=(-sqrt7-sqrt5)*(sqrt7-sqrt5)`
`=-(sqrt7+sqrt5)(sqrt7+sqrt5)`
`=-(7-5)=-2`
`b)sqrt2+1/sqrt{5+2sqrt6}+2/sqrt{8+2sqrt{15}}`
`=sqrt2+1/sqrt{3+2sqrt{3}.sqrt2+2}+2/sqrt{5+2sqrt{5}.sqrt3+3}`
`=sqrt2+1/sqrt{(sqrt3+sqrt2)^2}+2/sqrt{(sqrt5+sqrt3)^2}`
`=sqrt2+1/(sqrt3+sqrt2)+2/(sqrt5+sqrt3)`
`=sqrt2+((sqrt3+sqrt2)(sqrt3-sqrt2))/(sqrt3+sqrt2)+((sqrt5+sqrt3)(sqrt5-sqrt3))/(sqrt5+sqrt3)`
`=sqrt2+sqrt3-sqrt2+sqrt5-sqrt3=sqrt5`
a) Ta có: \(\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
\(=\left(-\dfrac{\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\dfrac{\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
\(=-2\)
b) Ta có: \(\sqrt{2}+\dfrac{1}{\sqrt{5+2\sqrt{6}}}+\dfrac{2}{\sqrt{8+2\sqrt{15}}}\)
\(=\sqrt{2}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{2}{\sqrt{5}+\sqrt{3}}\)
\(=\sqrt{2}+\sqrt{3}-\sqrt{2}+\sqrt{5}-\sqrt{3}\)
\(=\sqrt{5}\)
rút gọn :
a)\(\left(\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}+\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
b) \(\sqrt{\dfrac{3\sqrt{3}-4}{2\sqrt{3}+1}}+\sqrt{\dfrac{\sqrt{3}+4}{5-2\sqrt{3}}}\)
c) \(\dfrac{2\sqrt{5}-5\sqrt{2}}{\sqrt{2}-\sqrt{5}}+\dfrac{6}{2-\sqrt{10}}+\sqrt{67+12\sqrt{7}}\)
d) \(\left(\dfrac{\sqrt{5}}{\sqrt{2}+1}+\dfrac{14}{2\sqrt{2}-1}-\dfrac{6}{2-\sqrt{2}}\right).\sqrt{17-12\sqrt{2}}\)
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: \(=\dfrac{\sqrt{10}\left(\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}-\sqrt{5}}-2-\sqrt{10}+3\sqrt{7}+2\)
\(=\sqrt{10}-\sqrt{10}+3\sqrt{7}=3\sqrt{7}\)
Rút gọn:
a)\(\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}+\dfrac{6}{2-\sqrt{10}}\)
b)\(\dfrac{6}{\sqrt{5}-1}+\dfrac{7}{1-\sqrt{3}}-\dfrac{2}{\sqrt{3}-\sqrt{5}}\)
c)\(\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right)\div\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
d)\(\sqrt{2}+\dfrac{1}{\sqrt{5+2\sqrt{6}}}+\dfrac{2}{\sqrt{8+2\sqrt{15}}}\)
e)\(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\times\left(\sqrt{6}+11\right)\)
Lm nhanh giúp mk nhé, mk đang cần gấp!
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B1: thực hiện phép tính
a )\(\dfrac{\sqrt{6}-\sqrt{15}}{\sqrt{35}-\sqrt{14}}\)
b ) \(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\)
c )\(\dfrac{\sqrt{3-\sqrt{5}.}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
d ) \(\dfrac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{1}{\sqrt{2}-\sqrt{2+\sqrt{3}}}\)
B2:chúng minh vế phải bằng vế trái
a) \(\dfrac{21+8\sqrt{5}}{4+\sqrt{5}}.\sqrt{9-4\sqrt{5}}=\sqrt{5}-2\)
b) \(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}=-2\sqrt{3}\)