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a: \(=\sqrt{5}-3\sqrt{5}-4\sqrt{3}+15\sqrt{3}=-2\sqrt{5}+11\sqrt{3}\)
b: \(=3\sqrt{10}-\sqrt{5}+6-\sqrt{2}\)
c; \(=15\sqrt{2}-10\sqrt{3}-12\sqrt{2}-\sqrt{3}=-11\sqrt{3}+3\sqrt{2}\)
d: \(=3-\sqrt{3}+\sqrt{3}-1=2\)
f: \(=\sqrt{10}-\sqrt{10}-2-2\sqrt{10}=-2-2\sqrt{10}\)
Bài 2:
a: \(=\sqrt{5}-2\)
b: \(=2\sqrt{3}+4\sqrt{3}-5\sqrt{3}-9\sqrt{3}=-8\sqrt{3}\)
c: \(=\sqrt{4+2\sqrt{2}}\cdot\sqrt{4-2\sqrt{2}}=\sqrt{16-8}=2\sqrt{2}\)
d: \(=\sqrt{2}+1-2+\sqrt{2}=2\sqrt{2}-1\)
e: \(=\dfrac{8-2\sqrt{15}+8+2\sqrt{15}}{2}-\dfrac{6+2\sqrt{5}}{4}\)
\(=\dfrac{16-3-\sqrt{5}}{2}=\dfrac{13-\sqrt{5}}{2}\)
f: \(=\sqrt{5\sqrt{3+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)
\(=\sqrt{5\sqrt{3+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3+5\left(5-\sqrt{3}\right)}}\)
\(=\sqrt{5\sqrt{3+25-5\sqrt{3}}}\)
\(=\sqrt{5\sqrt{28-5\sqrt{3}}}\)
a. \(2\sqrt{16}+\sqrt{2}.\sqrt{0,02}-\dfrac{\sqrt{12,1}}{\sqrt{0,1}}=2.4+\sqrt{0,04}-\sqrt{\dfrac{12,1}{0,1}}=8+0,2-11=-2,8\)b. \(5\sqrt{20}-4\sqrt{45}+\dfrac{15}{\sqrt{5}}=10\sqrt{5}-12\sqrt{5}+3\sqrt{5}=\sqrt{5}\)
c. \(\left(\dfrac{\sqrt{6}-\sqrt{3}}{5\sqrt{2}-5}+\dfrac{\sqrt{5}}{5}\right):\dfrac{2}{\sqrt{5}-\sqrt{3}}=\left(\dfrac{\sqrt{3}\left(\sqrt{2}-1\right)}{5\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{5}}{5}\right).\dfrac{\sqrt{5}-\sqrt{3}}{2}=\dfrac{\sqrt{3}+\sqrt{5}}{5}.\dfrac{\sqrt{5}-\sqrt{3}}{2}=\dfrac{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{5.2}=\dfrac{5-3}{10}=\dfrac{2}{10}=\dfrac{1}{5}\)d. \(\dfrac{\sqrt{6}-3}{\sqrt{3}-\sqrt{2}}-\dfrac{4}{\sqrt{3}+1}+3\sqrt{3}=\dfrac{-\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-\dfrac{4}{\sqrt{3}+1}+3\sqrt{3}=3\sqrt{3}-\sqrt{3}-\dfrac{4}{\sqrt{3}+1}=\dfrac{\left(\sqrt{3}+1\right).2\sqrt{3}-4}{\sqrt{3}+1}=\dfrac{6+2\sqrt{3}-4}{\sqrt{3}+1}=\dfrac{2+2\sqrt{3}}{\sqrt{3}+1}=\dfrac{ 2\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=2\)
a) \(\dfrac{\sqrt{6}+\sqrt{10}}{\sqrt{21}+\sqrt{35}}=\dfrac{\sqrt{2}\sqrt{3}+\sqrt{2}\sqrt{5}}{\sqrt{7}\sqrt{3}+\sqrt{7}\sqrt{5}}\)
= \(\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{7}\left(\sqrt{3}+\sqrt{5}\right)}=\dfrac{\sqrt{2}}{\sqrt{7}}=\sqrt{\dfrac{2}{7}}\)
b) \(\dfrac{\sqrt{405}+3\sqrt{27}}{3\sqrt{3}+\sqrt{45}}=\dfrac{9\sqrt{5}+9\sqrt{3}}{3\sqrt{3}+3\sqrt{5}}=3\dfrac{3\sqrt{3}+3\sqrt{5}}{3\sqrt{3}+3\sqrt{5}}=3.1=3\)
c) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{6}-\sqrt{9}-\sqrt{12}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)-\sqrt{3}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(1-\sqrt{3}\)
P/s: bạn làm thêm bước nữa nha, mình lười, hehe
d) \(\dfrac{\sqrt{6-2\sqrt{5}}}{\sqrt{5}-1}=\dfrac{\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}.1+1^2}}{\sqrt{5}-1}\)
\(=\dfrac{\sqrt{\left(\sqrt{5-1}\right)^2}}{\sqrt{5}-1}=\dfrac{\left|\sqrt{5}-1\right|}{\sqrt{5}-1}=\dfrac{\sqrt{5}-1}{\sqrt{5}-1}=1\)
Câu a : \(\left(\sqrt{80}+\sqrt{20}\right):\sqrt{45}=\sqrt{80}:\sqrt{45}+\sqrt{20}:\sqrt{45}=\sqrt{\dfrac{16}{9}}+\sqrt{\dfrac{4}{9}}=\dfrac{4}{3}+\dfrac{2}{3}=\dfrac{6}{3}=2\)
Câu b : \(\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{18}+\sqrt{27}\right)=\sqrt{54}+\sqrt{81}-\sqrt{36}-\sqrt{54}=\sqrt{81}-\sqrt{36}=9-6=3\)
Câu c : \(\dfrac{5\sqrt{3}-3\sqrt{5}}{\sqrt{5}-\sqrt{3}}-\dfrac{6}{\sqrt{15+3}}=\dfrac{\sqrt{15}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}-\dfrac{6}{\sqrt{18}}\)
\(=\sqrt{15}-\dfrac{6}{\sqrt{18}}=\dfrac{\sqrt{270}-6}{3\sqrt{2}}=\dfrac{3\sqrt{30}-6}{3\sqrt{2}}=\dfrac{3\left(\sqrt{30}-6\right)}{3\sqrt{2}}=\dfrac{\sqrt{30}-2}{\sqrt{2}}=\sqrt{15}-\sqrt{2}\)
a: \(=2\cdot4+0.2-11=8-11+0.2=-2.8\)
b: \(=10\sqrt{5}-12\sqrt{5}+3\sqrt{5}=\sqrt{5}\)
c: \(=\left(\dfrac{\sqrt{3}\left(\sqrt{2}-1\right)}{5\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{5}}{5}\right)\cdot\dfrac{\sqrt{5}-\sqrt{3}}{2}\)
\(=\dfrac{\sqrt{5}+\sqrt{3}}{5}\cdot\dfrac{\sqrt{5}-\sqrt{3}}{2}=\dfrac{1}{2}\)
d: \(=-\sqrt{3}+2-2\sqrt{3}+3\sqrt{3}=2\)
a: ĐKXĐ: x-5>=0
=>x>=5
\(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\cdot\sqrt{9x-45}=4\)
=>\(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)
=>\(2\sqrt{x-5}=4\)
=>x-5=4
=>x=9(nhận)
b: ĐKXĐ: x-1>=0
=>x>=1
\(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}=4\)
=>\(\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}=4\)
=>\(-2\sqrt{x-1}=4\)
=>\(\sqrt{x-1}=-2\)(vô lý)
Vậy: Phương trình vô nghiệm
c: ĐKXĐ: x-2>=0
=>x>=2
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot\sqrt{9x-18}+6\cdot\sqrt{\dfrac{x-2}{81}}=-4\)
=>\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot3\sqrt{x-2}+6\cdot\dfrac{\sqrt{x-2}}{9}=-4\)
=>\(\sqrt{x-2}\left(\dfrac{1}{3}-2+\dfrac{2}{3}\right)=-4\)
=>\(-\sqrt{x-2}=-4\)
=>x-2=16
=>x=18(nhận)
d: ĐKXĐ: x+3>=0
=>x>=-3
\(\sqrt{9x+27}+4\sqrt{x+3}-\dfrac{3}{4}\cdot\sqrt{16x+48}=0\)
=>\(3\sqrt{x+3}+4\sqrt{x+3}-\dfrac{3}{4}\cdot4\sqrt{x+3}=0\)
=>\(4\sqrt{x+3}=0\)
=>x+3=0
=>x=-3(nhận)
a) \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\)
= \(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\)
= \(2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
= \(2\sqrt{x-5}=4\)
= \(\sqrt{x-5}=2\)
= \(\left|x-5\right|=4\)
=> \(x-5=\pm4\)
\(x=\pm4+5\)
\(x=9;x=1\)
Vậy x=9; x=1
\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)
\(ĐK:x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)
\(\Leftrightarrow4x^2-9=4x+12\)
\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)
\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(ĐK:x\ge5\)
\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)
\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)
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 \(\dfrac{5}{3}+\sqrt{x+3}>0\))
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)
a: \(\left(\sqrt{\dfrac{4}{3}}+\sqrt{3}\right)\cdot\sqrt{6}\)
\(=\sqrt{\dfrac{4}{3}\cdot6}+\sqrt{3\cdot6}\)
\(=\sqrt{8}+\sqrt{18}=2\sqrt{2}+3\sqrt{2}=5\sqrt{2}\)
b: \(\left(1-2\sqrt{5}\right)^2=\left(2\sqrt{5}-1\right)^2\)
\(=\left(2\sqrt{5}\right)^2-2\cdot2\sqrt{5}\cdot1+1\)
\(=21-4\sqrt{5}\)
c: \(2\sqrt{3}-\sqrt{27}=2\sqrt{3}-3\sqrt{3}=-\sqrt{3}\)
d: \(\sqrt{45}-\sqrt{20}+\sqrt{5}\)
\(=3\sqrt{5}-2\sqrt{5}+\sqrt{5}\)
\(=4\sqrt{5}-2\sqrt{5}=2\sqrt{5}\)