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5 tháng 7 2021
Bài 2 :
Ta có : \(\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{5-2\sqrt{5}\sqrt{3}+3}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\)
\(=\left(4+\sqrt{15}\right)\left(5+3-2\sqrt{15}\right)\)
\(=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)\)
\(=2\left(16-15\right)=2.1=2\)
5 tháng 7 2021
Bài 1 :
a, ĐKXĐ : \(x\ge0\)
Ta có : \(PT\Leftrightarrow3\sqrt{5x}-4\sqrt{5x}+8\sqrt{5x}=21\)
\(\Leftrightarrow7\sqrt{5x}=21\)
\(\Leftrightarrow\sqrt{5x}=3\)
\(\Leftrightarrow x=\dfrac{9}{5}\left(TM\right)\)
Vậy ...
b, Ta có : \(PT\Leftrightarrow\sqrt{\left(x-5\right)^2}=4\)
\(\Leftrightarrow\left|x-5\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=4\\x-5=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)
Vậy ....
H9
HT.Phong (9A5)
CTVHS
24 tháng 9 2023
a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\))
\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{2}{2}\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\left(tm\right)\)
b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))
\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
\(\Leftrightarrow x+1=16\)
\(\Leftrightarrow x=15\left(tm\right)\)
13 tháng 8 2021
\(6\sqrt{\frac{3}{4}}+10\sqrt{\frac{12}{25}}-15\sqrt{\frac{16}{3}}+9\sqrt{\frac{4}{3}}\)
\(=6\cdot\frac{\sqrt{3}}{2}+10\cdot\frac{2\sqrt{3}}{5}-15\cdot\frac{4}{\sqrt{3}}+9\cdot\frac{2}{\sqrt{3}}\)
\(=3\sqrt{3}+4\sqrt{3}-20\sqrt{3}+6\sqrt{3}=-7\sqrt{3}\)
QA
14 tháng 8 2021
Trả lời:
\(6\sqrt{\frac{3}{4}}+10\sqrt{\frac{12}{25}}-15\sqrt{\frac{16}{3}}+9\sqrt{\frac{4}{3}}\)
\(=6.\frac{\sqrt{3}}{\sqrt{4}}+10.\frac{\sqrt{12}}{\sqrt{25}}-15.\frac{\sqrt{16}}{\sqrt{3}}+9.\frac{\sqrt{4}}{\sqrt{3}}\)
\(=6.\frac{\sqrt{3}}{2}+10.\frac{\sqrt{2^2.3}}{5}-15.\frac{4}{\sqrt{3}}+9.\frac{2}{\sqrt{3}}\)
\(=3\sqrt{3}+10.\frac{2\sqrt{3}}{5}-15.\frac{4\sqrt{3}}{3}+9.\frac{2\sqrt{3}}{3}\)
\(=3\sqrt{3}+4\sqrt{3}-20\sqrt{3}+6\sqrt{3}\)
\(=\left(3+4-20+6\right).\sqrt{3}=-7\sqrt{3}\)
22 tháng 7 2017
a) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(1+\sqrt{2}\)
b) \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
= \(\dfrac{8+2\sqrt{5}}{2}.\left(\sqrt{5}-\sqrt{3}\right).\sqrt{2}.\sqrt{4-\sqrt{15}}\)
=\(\dfrac{\left(\sqrt{5}+\sqrt{3}\right)^2}{2}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
= \(\dfrac{\left(\sqrt{5}+\sqrt{3}\right)^2}{2}.\left(\sqrt{5}-\sqrt{3}\right)^2\)
= \(\dfrac{\left(5-3\right)^2}{2}=2\)
NB
0
NB
0
CB
0
HN
0
B
23 tháng 7 2019
\(a,\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right).\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right).\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right).\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right).\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=\left(4+\sqrt{15}\right).2\left(4-\sqrt{15}\right)\)
\(=2\)
\(b,\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{4}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)