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12 tháng 6 2017
\(\sqrt{\frac{4}{9-4\sqrt{5}}}+\sqrt{\frac{9}{9+4\sqrt{5}}}=\sqrt{\frac{4}{4-4\sqrt{5}+5}}+\sqrt{\frac{9}{4+4\sqrt{5}+5}}\)
\(=\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}+\sqrt{\frac{9}{\left(2+\sqrt{5}\right)^2}}=\frac{2}{\sqrt{5}-2}+\frac{3}{2+\sqrt{5}}\)
\(=\frac{2\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}+\frac{3\left(2-\sqrt{5}\right)}{\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)}=\frac{2\sqrt{5}+4}{5-4}+\frac{6-3\sqrt{5}}{4-5}\)
\(=2\sqrt{5}+4+3\sqrt{5}-6=5\sqrt{5}-2\)
b) \(\left(5-4\sqrt{3}\right):\frac{2+\sqrt{3}}{2-\sqrt{3}}=\left(5-4\sqrt{3}\right).\frac{2-\sqrt{3}}{2+\sqrt{3}}\)
\(=\left(5-4\sqrt{3}\right).\frac{\left(2-\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\left(5-4\sqrt{3}\right).\frac{4-4\sqrt{3}+3}{4-3}\)
\(=\left(5-4\sqrt{3}\right)\left(7-4\sqrt{3}\right)=35-28\sqrt{3}-20\sqrt{3}+48\)
\(=73-48\sqrt{3}\)
Mình chịu câu c nha
AT
12 tháng 7 2021
1) \(\left(\sqrt{19}-3\right)\left(\sqrt{19}+3\right)=\left(\sqrt{19}\right)^2-3^2=19-9=10\)
2) \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\sqrt{\dfrac{8+2\sqrt{7}}{2}}-\sqrt{\dfrac{8-2\sqrt{7}}{2}}\)
\(=\sqrt{\dfrac{\left(\sqrt{7}\right)^2+2.\sqrt{7}.1+1^2}{2}}-\sqrt{\dfrac{\left(\sqrt{7}\right)^2-2.\sqrt{7}.1+1^2}{2}}\)
\(=\sqrt{\dfrac{\left(\sqrt{7}+1\right)^2}{2}}-\sqrt{\dfrac{\left(\sqrt{7}-1\right)^2}{2}}=\dfrac{\left|\sqrt{7}+1\right|}{\sqrt{2}}-\dfrac{\left|\sqrt{7}-1\right|}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}+1}{\sqrt{2}}-\dfrac{\sqrt{7}-1}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
3) \(\sqrt{8+\sqrt{60}}+\sqrt{45}-\sqrt{12}=\sqrt{8+\sqrt{4.15}}+\sqrt{9.5}-\sqrt{4.3}\)
\(=\sqrt{8+2\sqrt{15}}+3\sqrt{5}-2\sqrt{3}\)
\(=\sqrt{\left(\sqrt{5}\right)^2+2.\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2}+3\sqrt{5}-2\sqrt{3}\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}+3\sqrt{5}-2\sqrt{3}=\left|\sqrt{5}+\sqrt{3}\right|+3\sqrt{5}-2\sqrt{3}\)
\(\sqrt{5}+\sqrt{3}+3\sqrt{5}-2\sqrt{3}=4\sqrt{5}-\sqrt{3}\)
4) \(\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}\right)^2-2.2.\sqrt{5}+2^2}-\sqrt{\left(\sqrt{5}\right)^2+2.2.\sqrt{5}+2^2}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}=\left|\sqrt{5}-2\right|-\left|\sqrt{5}+2\right|\)
\(=\sqrt{5}-2-\sqrt{5}-2=-4\)
H
9 tháng 9 2023
\(\dfrac{4}{\sqrt{5}-3}-\dfrac{4}{\sqrt{5}+3}\\ =\dfrac{4\left(\sqrt{5}+3\right)}{5-9}-\dfrac{4\left(\sqrt{5}-3\right)}{5-9}\\ =\dfrac{4\left(\sqrt{5}+3\right)}{-4}-\dfrac{4\left(\sqrt{5}-3\right)}{-4}\\ =-\left(\sqrt{5}+3\right)+\sqrt{5}-3\\ =-\sqrt{5}-3+\sqrt{5}-3\\ =-6\)
ĐK: \(x\ge5;x\le1\)
PT trở thành:
\(\sqrt{4}.\sqrt{x-5}-\dfrac{3\sqrt{x-5}}{3}=\sqrt{1-x}\\ \Leftrightarrow2\sqrt{x-5}-\sqrt{x-5}=\sqrt{1-x}\\ \Leftrightarrow\sqrt{x-5}=\sqrt{1-x}\\ \Leftrightarrow x-5=1-x\\ \Leftrightarrow x-5-1+x=0\\ \Leftrightarrow2x-6=0\\ \Leftrightarrow x=3\left(loại\right)\)
Vậy PT vô nghiệm.
`HaNa♬D`
9 tháng 9 2023
a: \(=\dfrac{4\left(\sqrt{5}+3\right)-4\left(\sqrt{5}-3\right)}{5-9}=\dfrac{4\left(\sqrt{5}+3-\sqrt{5}+3\right)}{-4}=-6\)
b: ĐKXĐ: x-5>=0 và 1-x<=0
=>x>=5 và x<=1
=>Không có x thỏa mãn ĐKXĐ
=>PT vô nghiệm
25 tháng 11 2023
2: ĐKXĐ: x>=0
\(\sqrt{3x}-2\sqrt{12x}+\dfrac{1}{3}\cdot\sqrt{27x}=-4\)
=>\(\sqrt{3x}-2\cdot2\sqrt{3x}+\dfrac{1}{3}\cdot3\sqrt{3x}=-4\)
=>\(\sqrt{3x}-4\sqrt{3x}+\sqrt{3x}=-4\)
=>\(-2\sqrt{3x}=-4\)
=>\(\sqrt{3x}=2\)
=>3x=4
=>\(x=\dfrac{4}{3}\left(nhận\right)\)
3:
ĐKXĐ: x>=0
\(3\sqrt{2x}+5\sqrt{8x}-20-\sqrt{18}=0\)
=>\(3\sqrt{2x}+5\cdot2\sqrt{2x}-20-3\sqrt{2}=0\)
=>\(13\sqrt{2x}=20+3\sqrt{2}\)
=>\(\sqrt{2x}=\dfrac{20+3\sqrt{2}}{13}\)
=>\(2x=\dfrac{418+120\sqrt{2}}{169}\)
=>\(x=\dfrac{209+60\sqrt{2}}{169}\left(nhận\right)\)
4: ĐKXĐ: x>=-1
\(\sqrt{16x+16}-\sqrt{9x+9}=1\)
=>\(4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>\(\sqrt{x+1}=1\)
=>x+1=1
=>x=0(nhận)
5: ĐKXĐ: x<=1/3
\(\sqrt{4\left(1-3x\right)}+\sqrt{9\left(1-3x\right)}=10\)
=>\(2\sqrt{1-3x}+3\sqrt{1-3x}=10\)
=>\(5\sqrt{1-3x}=10\)
=>\(\sqrt{1-3x}=2\)
=>1-3x=4
=>3x=1-4=-3
=>x=-3/3=-1(nhận)
6: ĐKXĐ: x>=3
\(\dfrac{2}{3}\sqrt{x-3}+\dfrac{1}{6}\sqrt{x-3}-\sqrt{x-3}=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}\cdot\left(\dfrac{2}{3}+\dfrac{1}{6}-1\right)=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}\cdot\dfrac{-1}{6}=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}=\dfrac{2}{3}:\dfrac{1}{6}=\dfrac{2}{3}\cdot6=\dfrac{12}{3}=4\)
=>x-3=16
=>x=19(nhận)
15 tháng 7 2021
1) \(\dfrac{\sqrt{5}+2}{\sqrt{5}-2}=9+4\sqrt{5}\)
2) \(\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{2}-\sqrt{5}}=\dfrac{\sqrt{10}\left(\sqrt{5}-\sqrt{2}\right)}{-\left(\sqrt{5}-\sqrt{2}\right)}=-\sqrt{10}\)
3) \(\dfrac{\sqrt{20}-3\sqrt{10}}{3-\sqrt{5}}=\dfrac{\sqrt{10}\left(\sqrt{5}-3\right)}{-\left(\sqrt{5}-3\right)}=-\sqrt{10}\)
4) \(\dfrac{6-2\sqrt{5}}{3+\sqrt{5}}=\dfrac{\left(6-2\sqrt{5}\right)\left(3-\sqrt{5}\right)}{4}=\dfrac{18-6\sqrt{5}-6\sqrt{5}+10}{4}=\dfrac{28-12\sqrt{5}}{4}=7-3\sqrt{5}\)
5)\(\dfrac{9+4\sqrt{5}}{\sqrt{5}+2}=\sqrt{5}+2\)
6 tháng 8 2021
Bài 20:
a) \(\sqrt{9-4\sqrt{5}}\cdot\sqrt{9+4\sqrt{5}}=\sqrt{81-80}=1\)
b) \(\left(2\sqrt{2}-6\right)\cdot\sqrt{11+6\sqrt{2}}=2\left(\sqrt{2}-3\right)\left(3+\sqrt{2}\right)\)
\(=2\left(2-9\right)=2\cdot\left(-7\right)=-14\)
c: \(\sqrt{2}\cdot\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
=2
d) \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}-\sqrt{2}\right)\left(2+\sqrt{3}\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)\)
\(=\left(4-2\sqrt{3}\right)\left(2+\sqrt{3}\right)\)
\(=8+4\sqrt{3}-4\sqrt{3}-6\)
=2
AH
Akai Haruma
Giáo viên
26 tháng 6 2021
\(A=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{3+1+2\sqrt{3.1}}-\sqrt{3+1-2\sqrt{3.1}}\)
\(=\sqrt{(\sqrt{3}+1)^2}-\sqrt{(\sqrt{3}-1)^2}=|\sqrt{3}+1|-|\sqrt{3}-1|=2\)
\(B=\sqrt{4+5-2\sqrt{4.5}}+\sqrt{4+5+2\sqrt{4.5}}=\sqrt{(\sqrt{4}-\sqrt{5})^2}+\sqrt{(\sqrt{4}+\sqrt{5})^2}\)
\(=|\sqrt{4}-\sqrt{5}|+|\sqrt{4}+\sqrt{5}|=2\sqrt{5}\)
AH
Akai Haruma
Giáo viên
26 tháng 6 2021
\(C\sqrt{2}=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7+1-2\sqrt{7.1}}-\sqrt{7+1+2\sqrt{7.1}}\)
\(=\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}\)
\(=|\sqrt{7}-1|-|\sqrt{7}+1|=-2\Rightarrow C=-\sqrt{2}\)
----------------------------
\(7+4\sqrt{3}=(2+\sqrt{3})^2\Rightarrow 10\sqrt{7+4\sqrt{3}}=10(2+\sqrt{3})\)
\(\Rightarrow \sqrt{48-10\sqrt{7+4\sqrt{3}}}=\sqrt{28-10\sqrt{3}}=\sqrt{(5-\sqrt{3})^2}=5-\sqrt{3}\)
\(\Rightarrow 3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}=3+5(5-\sqrt{3})=28-5\sqrt{3}\)
\(\Rightarrow D=\sqrt{5\sqrt{28-5\sqrt{3}}}\)
AT
12 tháng 7 2021
\(\sqrt{7-4\sqrt{3}}=\sqrt{2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(2-\sqrt{3}\right)^2}=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)
\(\sqrt{9+4\sqrt{5}}=\sqrt{2^2+2.2.\sqrt{5}+\left(\sqrt{5}\right)^2}+\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2+\sqrt{5}\right|=2+\sqrt{5}\)
\(\sqrt{11-4\sqrt{7}}=\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}.2+2^2}=\sqrt{\left(\sqrt{7}-2\right)^2}=\left|\sqrt{7}-2\right|=\sqrt{7}-2\)
12 tháng 7 2021
\(\sqrt{7-4\sqrt{3}}=2-\sqrt{3}\)
\(\sqrt{9+4\sqrt{5}}=\sqrt{5}+2\)
\(\sqrt{11-4\sqrt{7}}=\sqrt{7}-2\)
Y
26 tháng 6 2023
\(c,\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\\ =\sqrt{\sqrt{3^2}+2\sqrt{3}.1+1}+\sqrt{\sqrt{3^2}-2\sqrt{3}.1+1}\\ =\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =\left|\sqrt{3}+1\right|+\left|\sqrt{3}-1\right|\\ =\sqrt{3}+1+\sqrt{3}-1\\ =2\sqrt{3}\)
\(d,\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\\ =\sqrt{\sqrt{5^2}+2.2\sqrt{5}+2^2}-\sqrt{\sqrt{5^2}-2.2\sqrt{5} +2^2}\\ =\sqrt{\left(\sqrt{5}+2\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}\\ =\left|\sqrt{5}+2\right|-\left|\sqrt{5}-2\right|\\ =\sqrt{5}+2-\sqrt{5}+2\\ =4\)