\(\sqrt{\left(1+\sqrt{2}\right)^2}-1\)
\(\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt[3]{\left(\sqrt{2}-5\right)^3}\)
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23 tháng 7 2021
e) Ta có: \(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{2}+1-\sqrt{2}+1\)
=2
AO
4 tháng 12 2017
\(\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)\)
\(=\left(1+\sqrt{3}\right)^2-2\)
\(=3+2\sqrt{3}+1-2\)
\(=2\sqrt{3}+2\)
\(=2\left(\sqrt{3}+1\right)\)
\(\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2\)
\(=\left(\sqrt{3-\sqrt{5}}\right)^2+2.\left(\sqrt{3-\sqrt{5}}\right).\left(\sqrt{3+\sqrt{5}}\right)+\)\(\left(\sqrt{3+\sqrt{5}}\right)^2\)
\(=3-\sqrt{5}+2.\left(3-\sqrt{5}\right)+3+\sqrt{5}\)
\(=6+6-2\sqrt{5}\)
\(=12-2\sqrt{5}\)
\(=2\left(6-\sqrt{5}\right)\)
29 tháng 6 2021
Ta có: \(\dfrac{8+x\left(1+\sqrt{x-2\sqrt{x}+1}\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{x-3\sqrt{x}}{2\left(x-\sqrt{x}-6\right)}\)
\(=\dfrac{8+x\left(1+\sqrt{x}-1\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x\sqrt{x}+8}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}+2}{x-4}+\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{1}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2\left(\sqrt{x}+2\right)+\sqrt{x}\left(\sqrt{x}-2\right)}{2\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2\sqrt{x}+4+x-2\sqrt{x}}{2\left(x-4\right)}\)
\(=\dfrac{x+4}{2x-8}\)
4 tháng 9 2023
\(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\\ =\left|2-\sqrt{5}\right|+\left|2\sqrt{2}-\sqrt{5}\right|\\ =\sqrt{5}-2+2\sqrt{2}-\sqrt{5}\\ =-2+\sqrt{2}\)
\(\sqrt{\left(\sqrt{7}-2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)}\\ =\left|\sqrt{7}-2\sqrt{2}\right|+\left|3-2\sqrt{2}\right|\\ =2\sqrt{2}-\sqrt{7}+3-2\sqrt{2}\\ =3-\sqrt{7}\)
\(\sqrt{\left(x-3\right)^2}\\ =\left|x-3\right|\\ =x-3\left(vì.x>3\right)\)
\(\sqrt{\left(1-x\right)^2}\\ =\left|1-x\right|\\ =x-1\left(vì.x>1\right)\)
\(\sqrt{9a^4}=\sqrt{\left(3a^2\right)^2}\\ =\left|3a^2\right|\\ =3a^2\)
\(\sqrt{100a^2}\\ =\sqrt{\left(10a\right)^2}\\ =\left|10a\right|\\ =-10a\left(vì.a< 0\right)\)
AH
Akai Haruma
Giáo viên
4 tháng 9 2023
Lời giải:
a. $=|2-\sqrt{5}|+|2\sqrt{2}-\sqrt{5}|$
$=(\sqrt{5}-2)+(2\sqrt{2}-\sqrt{5})=-2+2\sqrt{2}$
b. $=|\sqrt{7}-2\sqrt{2}|+|3-2\sqrt{2}|=2\sqrt{2}-\sqrt{7}+(3-2\sqrt{2})$
$=3-\sqrt{7}$
c.
$=|x-3|=x-3$
d.
$=|1-x|=x-1$
$=\sqrt{(3a^2)^2}=|3a^2|=3a^2$
e.
$=\sqrt{(10a)^2}=|10a|=-10a$
27 tháng 6 2021
\(A=\left|2-\sqrt{3}\right|+\left|1+\sqrt{3}\right|=2-\sqrt{3}+1+\sqrt{3}=3\)
\(B=\left|4-\sqrt{5}\right|-\left|2-\sqrt{5}\right|=4-\sqrt{5}-\sqrt{5}+2=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\)
\(C=\left|1-\sqrt{5}\right|-\left|2-\sqrt{5}\right|=\sqrt{5}-1-\sqrt{5}+2=1\)
27 tháng 6 2021
\(A=\left|2-\sqrt{3}\right|+\left|1+\sqrt{3}\right|=2-\sqrt{3}+1+\sqrt{3}=3\)
\(B=\left|4-\sqrt{5}\right|-\left|2-\sqrt{5}\right|=4-\sqrt{5}-\sqrt{5}+2=6-2\sqrt{5}\)
C=\(\left|1-\sqrt{5}\right|-\left|2-\sqrt{5}\right|=\sqrt{5}-1-\sqrt{5}+2=1\)
H9
HT.Phong (9A5)
CTVHS
1 tháng 9 2023
\(\sqrt{2-\sqrt{3}}\left(\sqrt{6}-\sqrt{2}\right)\left(2+\sqrt{3}\right)\)
\(=\sqrt{2-\sqrt{3}}\cdot\sqrt{2}\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)\)
\(=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)\)
\(=\sqrt{\left(\sqrt{3}\right)^2-2\sqrt{3}\cdot1+1}\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)\)
\(=\left(\sqrt{3}-1\right)^2\left(2+\sqrt{3}\right)\)
\(=\left(4-2\sqrt{3}\right)\left(2+\sqrt{3}\right)\)
\(=8+4\sqrt{3}-4\sqrt{3}-6\)
\(=2\)
\(\sqrt{\left(1+\sqrt{2}\right)^2}-1=|1+\sqrt{2}|-1=1+\sqrt{2}-1=\sqrt{2}\)
\(\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt[3]{\left(\sqrt{2}-5\right)^3}=|\sqrt{2}-3|+\sqrt{2}-5=3-\sqrt{2}+\sqrt{2}-5=-2\)