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AH
Akai Haruma
Giáo viên
5 tháng 8 2020
2.1
\(A=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}=\sqrt{5+2\sqrt{5.1}+1}-\sqrt{5-2\sqrt{5.1}+1}\)
\(=\sqrt{(\sqrt{5}+1)^2}-\sqrt{(\sqrt{5}-1)^2}=|\sqrt{5}+1|-|\sqrt{5}-1|=2\)
2.2
\(B\sqrt{2}=\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}-2\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{3+2\sqrt{3.5}+5}+\sqrt{3-2\sqrt{3.5}+5}-2\sqrt{5-2\sqrt{5.1}+1}\)
\(=\sqrt{(\sqrt{3}+\sqrt{5})^2}+\sqrt{(\sqrt{3}-\sqrt{5})^2}-2\sqrt{(\sqrt{5}-1)^2}\)
\(=|\sqrt{3}+\sqrt{5}|+|\sqrt{3}-\sqrt{5}|-2|\sqrt{5}-1|=2\)
$\Rightarrow B=\sqrt{2}$
AH
Akai Haruma
Giáo viên
5 tháng 8 2020
Bài 1:
1. ĐKXĐ: \(\left\{\begin{matrix} 2x-1\geq 0\\ x-3\geq 0\\ 5-x>0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ x\geq 3\\ x< 5\end{matrix}\right.\Leftrightarrow 3\leq x< 5\)
2.
ĐKXĐ: \(\left\{\begin{matrix} x-1\geq 0\\ 2-x\geq 0\\ x+1>0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ x\leq 2\\ x>-1\end{matrix}\right.\Leftrightarrow 1\leq x\leq 2\)
D
26 tháng 7 2018
\(A=\sqrt{19-3\sqrt{40}}-\sqrt{19+3\sqrt{40}}=\sqrt{19-2\sqrt{90}}-\sqrt{19+2\sqrt{90}}=\sqrt{10-2.\sqrt{10}.3+9}-\sqrt{10+2.\sqrt{10}.3+9}=\sqrt{\left(\sqrt{10}-3\right)^2}-\sqrt{\left(\sqrt{10}+3\right)^2}=\sqrt{10}-3-\sqrt{10}-3=-6\)\(B=\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{18}}-2\sqrt{6+3\sqrt{3}}=\sqrt{18-2.\sqrt{18}.\sqrt{3}+3}+\sqrt{6+2.\sqrt{3}.\sqrt{6}+3}-\sqrt{24+12\sqrt{3}}=\sqrt{\left(\sqrt{18}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{6}+\sqrt{\sqrt{3}}\right)^2}-\sqrt{\left(\sqrt{18}+\sqrt{6}\right)^2}=\sqrt{18}-\sqrt{3}+\sqrt{6}+\sqrt{3}-\sqrt{18}-\sqrt{6}=0\)
Y
4 tháng 7 2019
\(C=\sqrt{6+2\sqrt{2\sqrt{3-\sqrt{4+2\sqrt{3}}}}}\)
\(C=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(C=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\) \(=\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\) \(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
\(D=\sqrt{\frac{8+2\sqrt{15}}{2}}-\sqrt{\frac{14-6\sqrt{5}}{2}}\) \(=\sqrt{\frac{\left(\sqrt{5}+\sqrt{3}\right)^2}{2}}-\sqrt{\frac{\left(3-\sqrt{5}\right)^2}{2}}\)
\(=\frac{\sqrt{5}+\sqrt{3}-3+\sqrt{5}}{\sqrt{2}}=\frac{2\sqrt{10}+\sqrt{6}-3\sqrt{2}}{2}\)
\(E=\sqrt{\frac{4+2\sqrt{3}}{2}}+\sqrt{\frac{4-2\sqrt{3}}{2}}\) \(=\sqrt{\frac{\left(\sqrt{3}+1\right)^2}{2}}+\sqrt{\frac{\left(\sqrt{3}-1\right)^2}{2}}\)
\(=\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
\(F=\sqrt{\frac{24-6\sqrt{7}}{2}}-\sqrt{\frac{24+6\sqrt{7}}{2}}\) \(=\sqrt{\frac{21-2\sqrt{21\cdot3}+3}{2}}-\sqrt{\frac{21+2\sqrt{21\cdot3}+3}{2}}\)
\(=\sqrt{\frac{\left(\sqrt{21}-\sqrt{3}\right)^2}{2}}-\sqrt{\frac{\left(\sqrt{21}+\sqrt{3}\right)^2}{2}}\)
\(=\frac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}=\frac{-2\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)
\(G=\left(3+\sqrt{3}\right)\cdot\sqrt{12-6\sqrt{3}}\) \(=\left(3+\sqrt{3}\right)\cdot\sqrt{\left(3-\sqrt{3}\right)^2}\)
\(=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)=9-3=6\)
\(H=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(3-\sqrt{5}\right)^2}\) \(=\sqrt{5}-2-3-\sqrt{5}=-5\)
\(I=\sqrt{\left(2\sqrt{2}-1\right)^2}-\sqrt{\left(2\sqrt{3}-1\right)^2}\)
\(=2\sqrt{2}-1-2\sqrt{3}+1=2\sqrt{2}-2\sqrt{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 4 2019
a/ \(A=\frac{30\left(\sqrt{6}-1\right)}{5}+\frac{2\left(\sqrt{6}+2\right)}{2}-\frac{6\left(3+\sqrt{6}\right)}{3}=6\sqrt{6}-6+\sqrt{6}+2-6-2\sqrt{6}\)
\(A=5\sqrt{6}-10\)
\(B=\sqrt{17-6\sqrt{2}+\sqrt{8+4\sqrt{2}+1}}\)
\(B=\sqrt{17-6\sqrt{2}+\sqrt{\left(2\sqrt{2}+1\right)^2}}=\sqrt{18-4\sqrt{2}}\)
Đến đây ko rút gọn được nữa, nhưng nếu đề là:
\(B=\sqrt{17+6\sqrt{2}+\sqrt{8+4\sqrt{2}+1}}=\sqrt{18+8\sqrt{2}}=4+\sqrt{2}\)
c/
\(C=\sqrt{8-2\sqrt{7}}+\sqrt{8+2\sqrt{7}}=\sqrt{\left(\sqrt{7}-1\right)^2}+\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(C=\sqrt{7}-1+\sqrt{7}+1=2\sqrt{7}\)
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 4 2019
\(D=\sqrt{a-2\sqrt{a}+1}-\sqrt{a-8\sqrt{a}+16}\)
\(D=\sqrt{\left(\sqrt{a}-1\right)^2}-\sqrt{\left(4-\sqrt{a}\right)^2}=\sqrt{a}-1-\left(4-\sqrt{a}\right)=2\sqrt{a}-5\)
\(E=\sqrt{a-2+2\sqrt{a-2}+1}+\sqrt{a-2-2\sqrt{a-2}+1}\) (\(a\ge2\))
\(E=\sqrt{\left(\sqrt{a-2}+1\right)^2}+\sqrt{\left(\sqrt{a-2}-1\right)^2}\)
\(E=\sqrt{a-2}+1+\left|\sqrt{a-2}-1\right|\)
\(\Rightarrow\left[{}\begin{matrix}E=2\sqrt{a-2}\left(a\ge3\right)\\E=2\left(2\le a\le3\right)\end{matrix}\right.\)
\(F=\sqrt[3]{10+6\sqrt{3}}-\sqrt{3}=\sqrt[3]{1+3.1.\sqrt{3}+3.1.\sqrt{3}^2+\sqrt{3}^3}-\sqrt{3}\)
\(F=\sqrt[3]{\left(1+\sqrt{3}\right)^3}-\sqrt{3}=1+\sqrt{3}-\sqrt{3}=1\)
\(G=\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\Rightarrow G^3=\left(\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\right)^3\)
\(\Rightarrow G^3=14+3\left(\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\right)\left(\sqrt[3]{49-50}\right)\)
\(\Rightarrow G^3=14-3G\Rightarrow G^3+3G-14=0\)
\(\Rightarrow G=2\)
HT
31 tháng 7 2018
a)
=\(\sqrt{15^2-2\cdot15\cdot\sqrt{2}+2}+\sqrt{11^2+2\cdot11\cdot\sqrt{2}+2}\)
=\(\sqrt{\left(15-\sqrt{2}\right)^2}+\sqrt{\left(11+\sqrt{2}\right)}^2\)
=\(15-\sqrt{2}+11+\sqrt{2}\)
=26
c)
=\(\sqrt{\dfrac{\left(\sqrt{3}-1\right)^2}{2}}\left(\sqrt{5}+2\right)\)
=\(\dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{2}}\)
NT
18 tháng 6 2019
a)
\(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\\ =\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}\\ =\sqrt{5}+1-\sqrt{5}+1\\ =2\)
b) Sửa đề:
\(\sqrt{7+2\sqrt{6}}+\sqrt{7-2\sqrt{6}}-2\sqrt{6}\\ =\sqrt{\left(\sqrt{6}+1\right)^2}+\sqrt{\left(\sqrt{6}-1\right)^2}-2\sqrt{6}\\ =\sqrt{6}+1+\sqrt{6}-1-2\sqrt{6}\\ =0\)
c)
\(\sqrt{9+4\sqrt{5}}+\sqrt{9-4\sqrt{5}}-2\sqrt{5}\\ =\sqrt{9+2\sqrt{20}}+\sqrt{9-2\sqrt{20}}-2\sqrt{5}\\ =\sqrt{\left(\sqrt{5}+2\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}-2\sqrt{5}\\ =\sqrt{5}+2+\sqrt{5}-2-2\sqrt{5}\\ =0\)
16 tháng 5 2019
câu b:
(\(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\))^2
\(=\left(5+2\sqrt{6}\right)-\left(5-2\sqrt{6}\right)\)\(-2\sqrt{5+2\sqrt{6}}\sqrt{5-2\sqrt{6}}\)
\(=4\sqrt{6}-2\sqrt{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)}\)
\(=4\sqrt{6}-2\sqrt{5^2-\left(2\sqrt{6}\right)^2}\)
\(=4\sqrt{6}-2\sqrt{25-24}=4\sqrt{6}-2\)
mấy câu khác tương tự
9 tháng 7 2022
Bài 1:
a: \(=\sqrt{32.4}=\dfrac{9}{5}\sqrt{10}\)
b: \(=\sqrt{5\cdot5\cdot7\cdot7\cdot11\cdot11}=5\cdot7\cdot11=385\)
c: \(=5-2\sqrt{6}\)
d: \(=18-1=17\)
e: \(=3\sqrt{2}-2\sqrt{3}+7\sqrt{3}-7\sqrt{2}=-4\sqrt{2}+5\sqrt{3}\)
AH
Akai Haruma
Giáo viên
30 tháng 6 2019
a)
\((\sqrt{3}-2\sqrt{12}+2\sqrt{4})(\sqrt{27}+\sqrt{144}-2\sqrt{16})\)
\(=(\sqrt{3}-4\sqrt{3}+4)(3\sqrt{3}+12-8)\)
\(=(-3\sqrt{3}+4)(3\sqrt{3}+4)=4^2-(3\sqrt{3})^2=16-27=-11\)
b)
\((2\sqrt{5}+2\sqrt{3})^2-4\sqrt{60}\)
\(=(2\sqrt{5})^2+2.2\sqrt{5}.2\sqrt{3}+(2\sqrt{3})^2-8\sqrt{15}\)
\(=32+8\sqrt{15}-8\sqrt{15}=32\)
c)
\(\sqrt{6}(3\sqrt{12}-4\sqrt{3}+\sqrt{48}-5\sqrt{6})\)
\(=3\sqrt{72}-4\sqrt{18}+\sqrt{6.48}-5.\sqrt{36}\)
\(=18\sqrt{2}-12\sqrt{2}+12\sqrt{2}-30=18\sqrt{2}-30\)
AH
Akai Haruma
Giáo viên
30 tháng 6 2019
d)
\((\sqrt{2}-\sqrt{3})(\sqrt{6}+\sqrt{2})(\sqrt{2}+\sqrt{3})\)
\(=(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})(\sqrt{6}+\sqrt{2})\)
\(=(2-3)(\sqrt{6}+\sqrt{2})=-(\sqrt{6}+\sqrt{2})\)
e) Biểu thức bên trong căn lớn âm nên biểu căn bậc 2 không có nghĩa
f)
\((\frac{2}{\sqrt{3}-1}+\frac{3}{\sqrt{3}-2}+\frac{15}{3-\sqrt{3}}).\frac{1}{\sqrt{3}+5}\)
\(=(\frac{2\sqrt{3}+15}{3-\sqrt{3}}+\frac{3}{\sqrt{3}-2}).\frac{1}{\sqrt{3}+5}\)
\(=\frac{2\sqrt{3}+15)(\sqrt{3}-2)+3(3-\sqrt{3})}{(3-\sqrt{3})(\sqrt{3}-2)}.\frac{1}{\sqrt{3}+5}\)
\(=\frac{-15+8\sqrt{3}}{(-9+5\sqrt{3})(\sqrt{3}+5)}=\frac{-15+8\sqrt{3}}{-30+16\sqrt{3}}=\frac{-15+8\sqrt{3}}{2(-15+8\sqrt{3})}=\frac{1}{2}\)