\(\sqrt{3}.6+\sqrt{3}+\sqrt{3}-1\)
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H9
HT.Phong (9A5)
CTVHS
10 tháng 7 2023
\(\dfrac{6-\sqrt{6}}{\sqrt{6}-1}+\dfrac{6-\sqrt{6}}{\sqrt{6}}\)
\(=\dfrac{\sqrt{6}\cdot\sqrt{6}-\sqrt{6}}{\sqrt{6}-1}+\dfrac{\sqrt{6}\cdot\sqrt{6}-\sqrt{6}}{\sqrt{6}}\)
\(=\dfrac{\sqrt{6}\left(\sqrt{6}-1\right)}{\sqrt{6}-1}+\dfrac{\sqrt{6}\left(\sqrt{6}-1\right)}{\sqrt{6}}\)
\(=\dfrac{\sqrt{6}}{1}+\dfrac{\sqrt{6}-1}{1}\)
\(=\sqrt{6}+\sqrt{6}-1\)
\(=2\sqrt{6}-1\)
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\(\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{18}+2\sqrt{3}}\)
\(=\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{6}\cdot\sqrt{3}+\sqrt{6}\cdot\sqrt{2}}\)
\(=\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}\)
\(=\dfrac{\sqrt{6}\left(\sqrt{2}+\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}-\dfrac{3\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=\dfrac{\sqrt{6}\left(\sqrt{2}+\sqrt{3}\right)-3\left(\sqrt{2}-\sqrt{3}\right)}{-\sqrt{6}}\)
\(=\dfrac{2\sqrt{3}+3\sqrt{2}-3\sqrt{2}+3\sqrt{3}}{-\sqrt{6}}\)
\(=\dfrac{5\sqrt{3}}{-\sqrt{6}}=-\dfrac{5}{\sqrt{2}}\)
9 tháng 8 2022
1: \(=\sqrt{6}+\sqrt{6}+1=2\sqrt{6}+1\)
2: \(=\dfrac{6\left(1-\sqrt{3}\right)}{1-\sqrt{3}}+\dfrac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=6+3=9\)
3: \(=\sqrt{3}+1-\sqrt{3}=1\)
9 tháng 10 2021
5: Ta có: \(\dfrac{2-\sqrt{2}}{1-\sqrt{2}}+\dfrac{\sqrt{2}-\sqrt{6}}{\sqrt{3}-1}\)
\(=-\sqrt{2}-\sqrt{2}\)
\(=-2\sqrt{2}\)
6 tháng 8 2021
a)=\(\dfrac{3\sqrt{6}}{2}+\dfrac{2\sqrt{6}}{3}-\dfrac{4\sqrt{6}}{2}\)
\(=\dfrac{2\sqrt{6}}{3}-\dfrac{\sqrt{6}}{2} \)
=\(\dfrac{4\sqrt{6}}{6}-\dfrac{3\sqrt{6}}{6}=\dfrac{\sqrt[]{6}}{6}\)
6 tháng 8 2021
b)\(\dfrac{D}{\sqrt{3}}=\dfrac{\sqrt{\sqrt{3}+1}+1-\sqrt{\sqrt{3}+1}+1}{\sqrt{3}+1-1}\)
\(\dfrac{D}{\sqrt{3}}=\dfrac{2}{\sqrt{3}}\)
D=2
24 tháng 11 2021
\(a,=6\sqrt{2}-3-6\sqrt{2}=-3\\ b,=12\sqrt{3}-2\sqrt{5}-6\sqrt{3}+5\sqrt{5}=6\sqrt{3}+3\sqrt{5}\\ c,=\sqrt{3}-1-\sqrt{3}=-1\\ d,=\sqrt{6}-\dfrac{5\left(\sqrt{6}+1\right)}{5}=\sqrt{6}-\sqrt{6}-1=-1\)
TN
15 tháng 6 2017
Bài rút gọn
\(\sqrt{\left(x-1\right)^2}-x=\left|x-1\right|-x\)
\(=\left(x-1\right)-x=x-1-x=-1\left(x>1\right)\)
Bài gpt:
\(\sqrt{x^2-3x+2}+\sqrt{x^2-4x+3}=0\)
Đk:\(-1\le x\le3\)
\(pt\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-2}+\sqrt{x-3}\right)=0\)
Dễ thấy:\(\sqrt{x-2}+\sqrt{x-3}=0\) vô nghiệm
Nên \(\sqrt{x-1}=0\Rightarrow x-1=0\Rightarrow x=1\)
12 tháng 7 2021
2) Ta có: \(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}+\dfrac{12}{\sqrt{6}-3}-\sqrt{6}\)
\(=3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)-\sqrt{6}\)
\(=3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}-\sqrt{6}\)
\(=-11\)
3) Ta có: \(\left(\dfrac{3}{\sqrt{5}-\sqrt{2}}+\dfrac{4}{\sqrt{6}+\sqrt{2}}\right)\left(\sqrt{3}-1\right)^2\)
\(=\left(\sqrt{5}+\sqrt{2}+\sqrt{6}-\sqrt{2}\right)\left(4-2\sqrt{3}\right)\)
\(=\left(\sqrt{6}+\sqrt{5}\right)\left(4-2\sqrt{3}\right)\)
\(=4\sqrt{6}-6\sqrt{2}+4\sqrt{5}-2\sqrt{15}\)
27 tháng 10 2020
B=\(\sqrt{9+2.3\sqrt{6}+6}+\sqrt{9+2.3.2\sqrt{6}+24}=\sqrt{\left(3+\sqrt{6}\right)^2}+\sqrt{\left(3+2\sqrt{6}\right)^2}\)=\(=3+\sqrt{6}+2+2\sqrt{6}=5+3\sqrt{6}\)
\(\sqrt{3}.\sqrt{6}+\sqrt{3}+\sqrt{3^2}\)
\(=3\sqrt{2}+3+\sqrt{3}\)
\(=8,974691495\)