a) \(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\)
b) \(\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}\)
Rút gọn các biểu thức sau :
a) \(\sqrt{18\left(\sqrt{2}-\sqrt{3}\right)^2}\) - \(\sqrt{54}\)
b) \(\dfrac{a+\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\) - a\(\sqrt{\dfrac{1}{a}}\)
c) ( \(\sqrt{28}\) - 2\(\sqrt{3}\) +7) \(\sqrt{7}\) +\(\sqrt{84}\)
a: \(=3\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)-3\sqrt{6}\)
=3căn 6-6-3căn 6=-6
b: \(=\dfrac{a+\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\sqrt{a}\)
\(=\dfrac{a+\sqrt{ab}-a+\sqrt{ab}}{\sqrt{a}-\sqrt{b}}=\dfrac{2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\)
Bài 1:Tìn ĐKXĐ
a.\(\sqrt{\dfrac{2}{^{^{^{ }}}x^2}}\)
b.\(\sqrt{\dfrac{-3}{3x+5}}\)
Bài 2:
a.\(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}\)
b.\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)
c,\(\left(\sqrt{14}-3\sqrt{2}\right)^2+6\sqrt{28}\)
Trả lời giúp mình với ạ!Mình cảm ơn nhiều!
Bài 1:
a. Ta có \(\sqrt{\dfrac{2}{x^2}}=\dfrac{\sqrt{2}}{\left|x\right|}=\dfrac{\sqrt{2}}{x}\) ,để biểu thức có nghĩa thì \(x>0\)
b. Để biểu thức \(\sqrt{\dfrac{-3}{3x+5}}\) có nghĩa thì \(\dfrac{-3}{3x+5}\ge0\)
mà \(-3< 0\Rightarrow3x+5< 0\) \(\Rightarrow x< \dfrac{-5}{3}\)
Bài 2:
a. \(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}=\dfrac{\left(2+\sqrt{2}\right)\left(1-\sqrt{2}\right)}{1-2}=\dfrac{-\sqrt{2}}{-1}=\sqrt{2}\)
b. \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)
\(=14-14\sqrt{2}+7+14\sqrt{2}\)
\(=21\)
c. \(\left(\sqrt{14}-3\sqrt{2}\right)^2+6\sqrt{28}\)
\(=14-6\sqrt{28}+18+6\sqrt{28}\)
\(=32\)
Rút gọn các biểu thức sau:
a) \(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
b) \(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\)
c) \(\left(\sqrt{6}+\sqrt{5}\right)^2-\sqrt{120}\)
d) \(\left(\dfrac{1}{2}-\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\sqrt{200}\right):\dfrac{1}{8}\)
giúp nha
a) \(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
= \(2\sqrt{5}-3\sqrt{5}+9\sqrt{2}+6\sqrt{2}\)
= \(-\sqrt{5}+15\sqrt{2}\)
b) \(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\)
= \(\left(2\sqrt{7}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)
= \(2.7-2\sqrt{21}+7+2\sqrt{21}=14+7=21\)
c) \(\left(\sqrt{6}+\sqrt{5}\right)^2-\sqrt{120}\)
= \(6+2\sqrt{6}.\sqrt{5}+5-2\sqrt{30}\)
= \(11+2\sqrt{30}-2\sqrt{30}=11\)
d) \(\left(\dfrac{1}{2}-\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\sqrt{200}\right):\dfrac{1}{8}\)
= \(\left(\dfrac{1}{2}-\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\sqrt{2}+8\sqrt{2}\right).8\)
= \(4-4\sqrt{2}-12\sqrt{2}+64\sqrt{2}=4+48\sqrt{2}\)
Bài này dễ ẹc ( đâu có khó đâu :)) )
a) \(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
\(=\sqrt{2^2.5}-\sqrt{3^2.5}+3\sqrt{3^2.2}+\sqrt{6^2.2}\)
\(=2\sqrt{5}-3\sqrt{5}+9\sqrt{2}+6\sqrt{2}\)
\(=\left(2-3\right)\sqrt{5}+\left(9+6\right)\sqrt{2}\)
\(=15\sqrt{2}-\sqrt{5}\)
b) \(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\)
\(=\sqrt{2^2.7}.\sqrt{7}-2\sqrt{3}.\sqrt{7}+\sqrt{7}.\sqrt{7}+\sqrt{2^2.21}\)
\(=2.7-2\sqrt{21}+7+2\sqrt{21}\)
\(=14+7+\left(2-2\right)\sqrt{21}=21\)
c) \(\left(\sqrt{6}+\sqrt{5}\right)^2-\sqrt{120}\)
\(=6+2\sqrt{30}+5-\sqrt{2^2.30}\)
\(=6+5+2\sqrt{30}-2\sqrt{30}=11\)
d) \(\left(\dfrac{1}{2}\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\sqrt{200}\right):\dfrac{1}{8}\)
\(=\left(\dfrac{1}{2}\sqrt{\dfrac{2}{2^2}}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\sqrt{10^2.2}\right):\dfrac{1}{8}\)
\(=\left(\dfrac{1}{4}\sqrt{2}-\dfrac{3}{2}\sqrt{2}+8\sqrt{2}\right).8\)
\(=2\sqrt{2}-12\sqrt{2}+64\sqrt{2}=54\sqrt{2}\)
Hok tốt
\(a.\sqrt{20}-\sqrt{45}+3\sqrt{18}-\sqrt{72}\)
= \(2\sqrt{5}-3\sqrt{5}+9\sqrt{2}-6\sqrt{2}=3\sqrt{2}-\sqrt{5}\)
b. \(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\)
= \(\sqrt{196}-2\sqrt{21}+7+\sqrt{84}\)
= \(14-2\sqrt{21}+7+2\sqrt{21}=21\)
c. \(\left(\sqrt{6}+\sqrt{5}\right)^2-\sqrt{120}\)
= \(6+2\sqrt{30}+5-\sqrt{120}=6+2\sqrt{30}+5-2\sqrt{30}=11\)
d. \(\left(\dfrac{1}{2}-\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\sqrt{200}\right):\dfrac{1}{8}\)
= \(4-\dfrac{8}{\sqrt{2}}-\dfrac{3\sqrt{2}}{4}+\dfrac{32\sqrt{200}}{5}\)
= \(4-\dfrac{8}{\sqrt{2}}-\dfrac{3}{2\sqrt{2}}+\dfrac{32.5.\sqrt{8}}{5}=4-\dfrac{16}{2\sqrt{2}}+32\sqrt{8}\)
= \(\dfrac{8\sqrt{2}-16+256}{2\sqrt{2}}=\dfrac{240+8\sqrt{2}}{2\sqrt{2}}=60\sqrt{2}+4\)
a) \(\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}\)
b) \(\sqrt{150}+\sqrt{1,6}.\sqrt{60}+4,5\sqrt{2\dfrac{2}{3}}-\sqrt{6}\)
c) \(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\)
d) \(\left(\sqrt{6}+\sqrt{5}\right)^2-\sqrt{120}\)
rút gọn
a.\(\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}\)
b.\(\sqrt{150}+\sqrt{1,6}.\sqrt{60}+4,5.\sqrt{2\dfrac{2}{3}}-\sqrt{6}\)
c.\(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\)
d.\(\left(\sqrt{6}+\sqrt{5}\right)^2-\sqrt{120}\)
a, \(\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}\)
= \(2\sqrt{3}-10\sqrt{3}-\dfrac{\sqrt{3}\cdot\sqrt{11}}{\sqrt{11}}+5\sqrt{\dfrac{4}{3}}\)
= \(2\sqrt{3}-10\sqrt{3}-\sqrt{3}+5\sqrt{\dfrac{12}{3^2}}\)
= \(2\sqrt{3}-10\sqrt{3}-\sqrt{3}+5\dfrac{2\sqrt{3}}{3}\)
= \(2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\dfrac{10\sqrt{3}}{3}\)
= \(-9\sqrt{3}+\dfrac{10\sqrt{3}}{3}=\dfrac{-27\sqrt{3}}{3}+\dfrac{10\sqrt{3}}{3}=\dfrac{-17\sqrt{3}}{3}\)
b, \(\sqrt{150}+\sqrt{1,6}\cdot\sqrt{60}+4.5\sqrt{2\dfrac{2}{3}}-\sqrt{6}\)
= \(5\sqrt{6}+\dfrac{2\sqrt{10}}{5}\cdot2\sqrt{15}+4,5\sqrt{\dfrac{8}{3}}-\sqrt{6}\)
= \(5\sqrt{6}+4\sqrt{6}+4,5\sqrt{\dfrac{24}{3^2}}-\sqrt{6}\)
= \(5\sqrt{6}+4\sqrt{6}+4,5\cdot\dfrac{2\sqrt{6}}{3}-\sqrt{6}\)
= \(5\sqrt{6}+4\sqrt{6}+3\sqrt{6}-\sqrt{6}=11\sqrt{6}\)
c, \(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\cdot\sqrt{7}+\sqrt{84}\)
= \(\left(2\sqrt{7}-2\sqrt{3}+\sqrt{7}\right)\cdot\sqrt{7}+2\sqrt{21}\)
= \(\left(3\sqrt{7}-2\sqrt{3}\right)\cdot\sqrt{7}+2\sqrt{21}\)
= \(21-2\sqrt{21}+2\sqrt{21}=21\)
d, \(\left(\sqrt{6}+\sqrt{5}\right)^2-\sqrt{120}\)
= \(6+2\sqrt{30}+5-2\sqrt{30}=11\)
* 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.\(\left(\dfrac{6-2\sqrt{2}}{3-\sqrt{2}}-\dfrac{5}{\sqrt{5}}\right).\dfrac{1}{2-\sqrt{5}}\)
d.\(\sqrt{\left(2-\sqrt{5}\right)^2-\sqrt{5}}\)
a) \(\text{2}\sqrt{\text{18}}-9\sqrt{50}+3\sqrt{8}\)
= \(\text{6}\sqrt{\text{2}}-45\sqrt{2}+6\sqrt{2}\)
= \(-33\sqrt{2}\)
b) = \(7-2.\sqrt{7}.\sqrt{3}+3+7.2\sqrt{21}\)
= \(10-2\sqrt{21}+14\sqrt{21}\)
= \(10+12\sqrt{21}\)
* 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
a : \(\sqrt{5+2\sqrt{6}}-\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}\)
b : \(\dfrac{\sqrt{7}-2\sqrt{7}}{2-\sqrt{7}}+\dfrac{6}{\sqrt{7}+1}+\left(3\sqrt{2}-2\sqrt{3}\right)\left(3\sqrt{2}+2\sqrt{3}\right)\)
a: \(\sqrt{5+2\sqrt{6}}-\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}\)
\(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\left|\sqrt{2}-\sqrt{3}\right|\)
\(=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)
b: Sửa đề: \(\dfrac{7-2\sqrt{7}}{2-\sqrt{7}}+\dfrac{6}{\sqrt{7}+1}+\left(3\sqrt{2}-2\sqrt{3}\right)\left(3\sqrt{2}+2\sqrt{3}\right)\)
\(=\dfrac{\sqrt{7}\left(\sqrt{7}-2\right)}{-\left(\sqrt{7}-2\right)}+\dfrac{6\left(\sqrt{7}-1\right)}{6}+18-12\)
\(=-\sqrt{7}+\sqrt{7}-1+6=5\)
tính
1.\(\sqrt{147}+\sqrt{54}-4\sqrt{27}\)
2.\(\sqrt{28}-4\sqrt{63}+7\sqrt{112}\)
3.\(\sqrt{49}-5\sqrt{28}+\dfrac{1}{2}\sqrt{63}\)
4.\(\left(2\sqrt{6}-4\sqrt{3}-\dfrac{1}{4}\sqrt{8}\right).3\sqrt{6}\)
5.(\(2\sqrt{1\dfrac{9}{16}}-5\sqrt{5\dfrac{1}{16}}\)):\(\sqrt{16}\)
6.\(\left(\sqrt{48}-3\sqrt{27}-\sqrt{147}\right):\sqrt{3}\)
7.\(\left(\sqrt{50}-3\sqrt{49}\right):\sqrt{2}-\sqrt{162}:\sqrt{2}\)
8.\(\left(2\sqrt{1\dfrac{9}{10}}-\sqrt{5\dfrac{1}{10}}\right):\sqrt{10}\)
9.\(2\sqrt{\dfrac{16}{3}}-3\sqrt{\dfrac{1}{27}}-6\sqrt{\dfrac{4}{75}}\)
10.\(2\sqrt{27}-6\sqrt{\dfrac{4}{3}}+\dfrac{3}{5}\sqrt{75}\)
11.\(\dfrac{\sqrt{18}}{\sqrt{2}}-\dfrac{\sqrt{12}}{\sqrt{3}}\)
12.\(\dfrac{\sqrt{27}}{\sqrt{3}}+\dfrac{\sqrt{98}}{\sqrt{2}}-\sqrt{175}:\sqrt{7}\)
13.\(\left(\dfrac{\sqrt{8}}{\sqrt{2}}-\dfrac{\sqrt{180}}{\sqrt{5}}\right).\sqrt{5}-\sqrt{\dfrac{81}{11}}.\sqrt{11}\)
14.\(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)
15.\(\left(3\sqrt{2}-2\sqrt{3}\right)\left(3\sqrt{2}+2\sqrt{3}\right)\)
16.\(\left(1+\sqrt{5}-\sqrt{3}\right)\left(1+\sqrt{5}+\sqrt{3}\right)\)