Ta có : \(\sqrt{25}=5;-\sqrt{25}=-5;\sqrt{\left(-5\right)^2}=\sqrt{25}=5\)
Theo mẫu trên, hãy tính :
a) \(\sqrt{36}\)
b) \(-\sqrt{16}\)
c) \(\sqrt{\dfrac{9}{25}}\)
d) \(\sqrt{3^2}\)
e) \(\sqrt{\left(-3\right)^2}\)
ta co :\(\sqrt{25=5};-\sqrt{25=-5;\sqrt{\left(-5\right)^2}=\sqrt{25=5}}\)
a) \(\sqrt{36}\)
b) -\(\sqrt{16}\)
c) \(\sqrt{\frac{9}{25}}\)
d)\(\sqrt{3^2}\)
e)\(\sqrt{\left(-3\right)^2}\)
a) \(\sqrt{36}=6\)
b)\(-\sqrt{16}=-4\)
c)\(\sqrt{\frac{9}{25}}=\frac{3}{5}\)
d)\(\sqrt{3^2}=\sqrt{9}=3\)
e)\(\sqrt{\left(-3\right)^2}=\sqrt{9}=3\)
a) \(\dfrac{2}{5}\sqrt{25}\) -\(\dfrac{1}{2}\sqrt{4}\) b)0,5\(\sqrt{0,09}\) +5\(\sqrt{0,81}\) c)\(\dfrac{2}{5}\sqrt{\dfrac{25}{36}}\) -\(\dfrac{5}{2}\sqrt{\dfrac{4}{25}}\)
d)-2\(\sqrt{\dfrac{-36}{-16}}\) + 5\(\sqrt{\dfrac{-81}{-25}}\)
`#3107.101107`
a)
`2/5 \sqrt{25} - 1/2 \sqrt{4}`
`= 2/5 * \sqrt{5^2} - 1/2 * \sqrt{2^2}`
`= 2/5*5 - 1/2*2`
`= 2 - 1`
`= 1`
b)
`0,5*\sqrt{0,09} + 5*\sqrt{0,81}`
`= 0,5*\sqrt{(0,3)^2} + 5*\sqrt{(0,9)^2}`
`= 0,5*0,3 + 5*0,9`
`= 0,15 + 4,5`
`= 4,65`
c)
`2/5\sqrt{25/36} - 5/2\sqrt{4/25}`
`= 2/5*\sqrt{(5^2)/(6^2)} - 5/2*\sqrt{(2^2)/(5^2)}`
`= 2/5*5/6 - 5/2*2/5`
`= 1/3 - 1`
`= -2/3`
d)
`-2 \sqrt{(-36)/(-16)} + 5 \sqrt{(-81)/(-25)}`
`= -2*\sqrt{36/16} + 5*\sqrt{81/25}`
`= -2*\sqrt{(6^2)/(4^2)} + 5*\sqrt{(9^2)/(5^2)}`
`= -2*6/4 + 5*9/5`
`= -3 + 9`
`= 6`
rút gọn biểu thức
a) \(\left(\sqrt{7}-\sqrt{2}\right).\left(\sqrt{9+2\sqrt{14}}\right)\)
b) \(\sqrt{\sqrt{13}-\sqrt{3-\sqrt{13}}-4\sqrt{3}}\)
c) \(\sqrt{80-\sqrt{321-16\sqrt{5}}-\sqrt{226-80\sqrt{5}-\sqrt{89-25\sqrt{5}}}}\)
d) \(\dfrac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}-\dfrac{6\sqrt{2}-4}{3-\sqrt{2}}\)
e) \(\dfrac{\sqrt{6-\sqrt{11}}}{\sqrt{22}-\sqrt{2}}+\dfrac{6}{\sqrt{2}}-\dfrac{3}{\sqrt{2}+1}\)
f) \(\dfrac{\sqrt{2}}{2\sqrt{2}+\sqrt{3}+\sqrt{5}}+\dfrac{\sqrt{2}}{2\sqrt{2}-\sqrt{3}-\sqrt{5}}\)
g) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
a) Ta có: \(\left(\sqrt{7}-\sqrt{2}\right)\cdot\sqrt{9+2\sqrt{14}}\)
\(=\left(\sqrt{7}-\sqrt{2}\right)\cdot\left(\sqrt{7}+\sqrt{2}\right)\)
=7-2
=5
d) Ta có: \(\dfrac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}-\dfrac{6\sqrt{2}-4}{3-\sqrt{2}}\)
\(=2\sqrt{2}-\sqrt{7}+5\sqrt{7}-\dfrac{2\sqrt{2}\left(3-\sqrt{2}\right)}{3-\sqrt{2}}\)
\(=2\sqrt{2}+4\sqrt{7}-2\sqrt{2}\)
\(=4\sqrt{7}\)
Ta co \(\sqrt{25}\)=5 ; -\(\sqrt{25}\)= -5 ; \(\sqrt{\left(-5\right)^2}\)= \(\sqrt{25}\)=5
Theo mau tren , hay tinh :
a) \(\sqrt{36}\)
b) - \(\sqrt{16}\)
c) \(\sqrt{\frac{9}{25}}\)
d) \(\sqrt{3^2}\)
e) \(\sqrt{\left(-3\right)^2}\)
Tính
a) \(2\sqrt{\dfrac{25}{16}}-3\sqrt{\dfrac{49}{36}}+4\sqrt{\dfrac{81}{64}}\)
b) \(\left(3\sqrt{2}\right)^2-\left(4\sqrt{\dfrac{1}{2}}\right)^2+\dfrac{1}{16}.\left(\sqrt{\dfrac{3}{4}}\right)^2\)
c) \(\dfrac{2}{3}\sqrt{\dfrac{81}{16}}-\dfrac{3}{4}\sqrt{\dfrac{64}{9}}+\dfrac{7}{5}.\sqrt{\dfrac{25}{196}}\)
a: \(=2\cdot\dfrac{5}{4}-3\cdot\dfrac{7}{6}+4\cdot\dfrac{9}{8}=\dfrac{5}{2}-\dfrac{7}{2}+\dfrac{9}{2}=\dfrac{7}{2}\)
b: \(=18-16\cdot\dfrac{1}{2}+\dfrac{1}{16}\cdot\dfrac{3}{4}\)
=10+3/64
=643/64
c: \(=\dfrac{2}{3}\cdot\dfrac{9}{4}-\dfrac{3}{4}\cdot\dfrac{8}{3}+\dfrac{7}{5}\cdot\dfrac{5}{14}=\dfrac{3}{2}-2+\dfrac{1}{2}=2-2=0\)
bài 2
a) \(4\sqrt{28}+3\sqrt{63}-3\sqrt{11.2}-2\sqrt{175}\)
b) \(\sqrt{5}.\left(\sqrt{5}-3\sqrt{20}+2\sqrt{80}\right)\)
c) \(\left(\sqrt{\dfrac{16}{3}}-\sqrt{\dfrac{25}{3}}\right).\sqrt{3}\)
e) \(\left(\sqrt{\dfrac{32}{3}}-\sqrt{54}+\sqrt{\dfrac{50}{3}}\right).\sqrt{6}\)
f) \(\left(\sqrt{6}-2\right).\left(\sqrt{3}+\sqrt{2}\right)\)
a) Ta có: \(4\sqrt{28}+3\sqrt{63}-3\sqrt{112}-2\sqrt{175}\)
\(=8\sqrt{7}+9\sqrt{7}-12\sqrt{7}-10\sqrt{7}\)
\(=-5\sqrt{7}\)
b) Ta có: \(\sqrt{5}\left(\sqrt{5}-3\sqrt{20}+2\sqrt{80}\right)\)
\(=\sqrt{5}\left(\sqrt{5}-6\sqrt{5}+8\sqrt{5}\right)\)
\(=\sqrt{5}\cdot3\sqrt{5}=15\)
c) Ta có: \(\left(\sqrt{\dfrac{16}{3}}-\sqrt{\dfrac{25}{3}}\right)\cdot\sqrt{3}\)
\(=\dfrac{-1}{\sqrt{3}}\cdot\sqrt{3}\)
=-1
e) Ta có: \(\left(\sqrt{\dfrac{32}{3}}-\sqrt{54}+\sqrt{\dfrac{50}{3}}\right)\cdot\sqrt{6}\)
\(=\left(\dfrac{4\sqrt{2}}{\sqrt{3}}+\dfrac{5\sqrt{2}}{\sqrt{3}}-3\sqrt{6}\right)\cdot\sqrt{6}\)
\(=\dfrac{9\sqrt{12}}{\sqrt{3}}-18\)
\(=0\)
f) Ta có: \(\left(\sqrt{6}-2\right)\left(\sqrt{3}+\sqrt{2}\right)\)
\(=3\sqrt{2}+2\sqrt{3}-2\sqrt{2}-2\sqrt{2}\)
\(=\sqrt{2}\)
bài 1: tính
a) 3/4+(-5/2)+(-3/5)
b) \(\sqrt{\left(7\right)^2}+\sqrt{\dfrac{25}{16}-\dfrac{3}{2}}\)
c)\(\dfrac{1}{2}.\sqrt{100}-\sqrt{\dfrac{1}{16}+\left(\dfrac{1}{3}\right)^0}\)
a)\(\dfrac{3}{4}-\dfrac{5}{2}-\dfrac{3}{5}=\dfrac{15}{20}-\dfrac{50}{20}-\dfrac{12}{20}=-\dfrac{47}{20}\)
b) \(\sqrt{7^2}+\sqrt{\dfrac{25}{16}-\dfrac{3}{2}}=7+\sqrt{\dfrac{1}{16}}=7+\dfrac{1}{4}=\dfrac{29}{4}\)
c) \(\dfrac{1}{2}.\sqrt{100}-\sqrt{\dfrac{1}{16}+\left(\dfrac{1}{3}\right)^0}=\dfrac{1}{2}.10-\sqrt{\dfrac{1}{16}+1}=5-\sqrt{\dfrac{17}{16}}\)
Tính:
a) \(2\sqrt{\dfrac{27}{4}}-\sqrt{\dfrac{48}{9}}-\dfrac{2}{5}\sqrt{\dfrac{75}{16}}\)
b) \(\left(1+\dfrac{5-\sqrt{5}}{1-\sqrt{5}}\right)\left(\dfrac{5+\sqrt{5}}{1+\sqrt{5}}+1\right)\)
c) \(3\sqrt{\dfrac{9}{8}}-\sqrt{\dfrac{49}{2}}+\sqrt{\dfrac{25}{18}}\)
d) \(\dfrac{1}{\sqrt{3}-\sqrt{2}}+\dfrac{1}{\sqrt{3}+\sqrt{2}}\)
\(a,2\sqrt{\dfrac{27}{4}}-\sqrt{\dfrac{48}{9}}-\dfrac{2}{5}.\sqrt{\dfrac{75}{16}}\)
\(\Leftrightarrow2.\dfrac{\sqrt{27}}{2}-\sqrt{\dfrac{48}{3}}-\dfrac{2}{5}.\dfrac{\sqrt{75}}{4}\)
\(\Leftrightarrow\sqrt{27}-\dfrac{4\sqrt{3}}{3}-\dfrac{1}{5}.\dfrac{5\sqrt{3}}{2}\)
\(\Leftrightarrow3\sqrt{3}-\dfrac{4\sqrt{3}}{3}-\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow\dfrac{7\sqrt{3}}{6}\)
\(b,\left(1+\dfrac{5-\sqrt{5}}{1-\sqrt{5}}\right).\left(\dfrac{5+\sqrt{5}}{1+\sqrt{5}}+1\right)\)
\(\Leftrightarrow\)\(\left[1+\dfrac{\left(5-\sqrt{5}\right)\left(1+\sqrt{5}\right)}{-4}\right].\left[\dfrac{\left(5+\sqrt{5}\right).\left(1-\sqrt{5}\right)}{-4}+1\right]\)
\(\Leftrightarrow\)\(\left(1+\dfrac{5+5\sqrt{5}-\sqrt{5}-5}{-4}\right).\left(\dfrac{5-5\sqrt{5}+\sqrt{5}-5}{-4}+1\right)\)
\(\Leftrightarrow\)\(\left(1+\dfrac{4\sqrt{5}}{-4}\right)\left(\dfrac{-4\sqrt{5}}{-4}+1\right)\)
\(\Leftrightarrow\left(1-\sqrt{5}\right)\left(\sqrt{5}+1\right)\)
\(\Leftrightarrow\left(1-\sqrt{5}\right).\left(1+\sqrt{5}\right)\)
<=> 1-5
=-4
tìm x để các biểu thức sau có nghĩa:
a)\(\sqrt{\left(x-2\right)}\)+\(\dfrac{1}{x-5}\) b)\(\sqrt{\left(2x-6\right)\left(7-x\right)}\) c)\(\sqrt{4x^2-25}\)
d)\(\dfrac{2}{x^2-9}\)-\(\sqrt{5-2x}\) e)\(\dfrac{x}{x^2-4}\)+\(\sqrt{x-2}\)
a) \(\sqrt{x-2}+\dfrac{1}{x-5}\) có nghĩa khi:
\(\left\{{}\begin{matrix}x-2\ge0\\x-5\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ne5\end{matrix}\right.\)
b) \(\sqrt{\left(2x-6\right)\left(7-x\right)}=\sqrt{2\left(x-3\right)\left(7-x\right)}\) có nghĩa khi:
\(\left(x-3\right)\left(7-x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3\ge0\\7-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3\le0\\7-x\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge3\\x\le7\end{matrix}\right.\\\left\{{}\begin{matrix}x\le3\\x\ge7\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow3\le x\le7\)
c) \(\sqrt{4x^2-25}=\sqrt{\left(2x-5\right)\left(2x+5\right)}\) có nghĩa khi:
\(\left(2x-5\right)\left(2x+5\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-5\ge0\\2x+5\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-5\le0\\2x+5\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\x\ge-\dfrac{5}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{5}{2}\\x\le-\dfrac{5}{2}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{5}{2}\\x\le-\dfrac{5}{2}\end{matrix}\right.\)
d) \(\dfrac{2}{x^2-9}-\sqrt{5-2x}=\dfrac{2}{\left(x+3\right)\left(x-3\right)}-\sqrt{5-2x}\) có nghĩa khi:
\(\left\{{}\begin{matrix}x+3\ne0\\x-3\ne0\\5-2x\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm3\\x\le\dfrac{5}{2}\end{matrix}\right.\)
e) \(\dfrac{x}{x^2-4}+\sqrt{x-2}=\dfrac{x}{\left(x+2\right)\left(x-2\right)}+\sqrt{x-2}\) có nghĩa khi:
\(\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\\x-2\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm2\\x\ge2\end{matrix}\right.\)
\(\Leftrightarrow x>2\)