Rút gọn : \(\sqrt{19-6\sqrt{2}}\) + \(\sqrt{9-4\sqrt{2}}\)
a) \(\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\)
b) \(\sqrt{17-2\sqrt{72}}+\sqrt{19+2\sqrt{18}}\)
c) \(\sqrt{12-2\sqrt{32}}+\sqrt{9+4\sqrt{2}}\)
đề bài là rút gọn biểu thức
giải chi tiết hộ mình ạ !!!
a: Ta có: \(\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\)
\(=\sqrt{5}+\sqrt{3}-\sqrt{5}-1\)
\(=\sqrt{3}-1\)
b: Ta có: \(\sqrt{17-2\sqrt{72}}+\sqrt{19+2\sqrt{18}}\)
\(=3-2\sqrt{2}+3\sqrt{2}+1\)
\(=4+\sqrt{2}\)
c: Ta có: \(\sqrt{12-2\sqrt{32}}+\sqrt{9+4\sqrt{2}}\)
\(=2\sqrt{2}-2+2\sqrt{2}+1\)
\(=4\sqrt{2}-1\)
a)
\(\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\\ =\sqrt{5+2\sqrt{5}\cdot\sqrt{3}+3}-\sqrt{5+2\sqrt{5}\cdot\sqrt{1}+1}\\ =\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{1}\right)^2}\\ =\sqrt{5}+\sqrt{3}-\sqrt{5}-\sqrt{1}\\ =\sqrt{3}-\sqrt{1}\)
b)
\(\sqrt{17-2\sqrt{72}}+\sqrt{19+2\sqrt{18}}\\ =\sqrt{9-2\sqrt{9}\cdot\sqrt{8}+8}+\sqrt{18+2\sqrt{18}\cdot\sqrt{1}+1}\\ =\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}+1\right)^2}\\ =3-2\sqrt{2}+3\sqrt{2}+1\\ =4+\sqrt{2}\)
c)
\(\sqrt{12-2\sqrt{32}}+\sqrt{9+4\sqrt{2}}\\ =\sqrt{8-2\sqrt{8}\cdot\sqrt{4}+4}+\sqrt{8+2\sqrt{8}\cdot\sqrt{1}+1}\\ =\sqrt{\left(2\sqrt{2}-2\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}\\ =2\sqrt{2}-2+2\sqrt{2}+1\\ =4\sqrt{2}-1\)
Rút gọn biểu thức
\(\sqrt{19-6\sqrt{2}}-\sqrt{9-4\sqrt{2}}\)
\(=\sqrt{\left(3\sqrt{2}-1\right)^2}-\sqrt{\left(2\sqrt{2}-1\right)^2}\)
\(=3\sqrt{2}-1-2\sqrt{2}+1=\sqrt{2}\)
Rút gọn biểu thức
1) \(\sqrt{9+4\sqrt{5}}\) - \(\sqrt{9-4\sqrt{5}}\)
2) \(\sqrt{12-6\sqrt{3}}\) + \(\sqrt{12+6\sqrt{3}}\)
1) \(\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{2^2+2\cdot2\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}-\sqrt{2^2-2\cdot2\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}\)
\(=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(=\left|2+\sqrt{5}\right|-\left|2-\sqrt{5}\right|\)
\(=2+\sqrt{5}+2-\sqrt{5}\)
\(=4\)
2) \(\sqrt{12-6\sqrt{3}}+\sqrt{12+6\sqrt{3}}\)
\(=\sqrt{3^2-2\cdot3\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}+\sqrt{3^2+2\cdot3\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(3-\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{3}\right)^2}\)
\(=\left|3-\sqrt{3}\right|+\left|3+\sqrt{3}\right|\)
\(=3-\sqrt{3}+3+\sqrt{3}\)
\(=6\)
Thực hiện phép tính (rút gọn biểu thức)
a) \(\sqrt{9+4\sqrt{5}}\) - \(\sqrt{9-4\sqrt{5}}\)
b) \(\sqrt{12-6\sqrt{3}}\) + \(\sqrt{12+6\sqrt{3}}\)
c) \(\sqrt{6\sqrt{2}+11}\) - \(\sqrt{11-6\sqrt{2}}\)
Lời giải:
a.
\(=\sqrt{5+2.2\sqrt{5}+2^2}-\sqrt{5-2.2\sqrt{5}+2^2}\)
$=\sqrt{(\sqrt{5}+2)^2}-\sqrt{(\sqrt{5}-2)^2}$
$=|\sqrt{5}+2|-|\sqrt{5}-2|=(\sqrt{5}+2)-(\sqrt{5}-2)=4$
b.
$=\sqrt{3-2.3\sqrt{3}+3^2}+\sqrt{3+2.3.\sqrt{3}+3^2}$
$=\sqrt{(\sqrt{3}-3)^2}+\sqrt{(\sqrt{3}+3)^2}$
$=|\sqrt{3}-3|+|\sqrt{3}+3|$
$=(3-\sqrt{3})+(\sqrt{3}+3)=6$
c.
$=\sqrt{2+2.3\sqrt{2}+3^2}-\sqrt{2-2.3\sqrt{2}+3^2}$
$=\sqrt{(\sqrt{2}+3)^2}-\sqrt{(\sqrt{2}-3)^2}$
$=|\sqrt{2}+3|-|\sqrt{2}-3|$
$=(\sqrt{2}+3)-(3-\sqrt{2})=2\sqrt{2}$
\(\sqrt{3-2\sqrt{2}}-\sqrt{11+6\sqrt{2}}\)
\(\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}+\sqrt{19+8\sqrt{3}}\)
\(\sqrt{6-2\sqrt{5}}+\sqrt{9+4\sqrt{5}}-\sqrt{14-6\sqrt{5}}\)
\(\sqrt{11-4\sqrt{7}}+\sqrt{23-8\sqrt{7}}+\sqrt{\left(-2^6\right)}\)
rút gọn:giải chi tiết hộ mình nha
a) Ta có: \(\sqrt{3-2\sqrt{2}}-\sqrt{11+6\sqrt{2}}\)
\(=\sqrt{2}-1-3-\sqrt{2}\)
=-4
b) Ta có: \(\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}+\sqrt{19+8\sqrt{3}}\)
\(=\sqrt{3}-1-2+\sqrt{3}+4+\sqrt{3}\)
\(=3\sqrt{3}+1\)
c) Ta có: \(\sqrt{6-2\sqrt{5}}+\sqrt{9+4\sqrt{5}}-\sqrt{14-6\sqrt{5}}\)
\(=\sqrt{5}-1+\sqrt{5}-2-3+\sqrt{5}\)
\(=3\sqrt{5}-6\)
d) Ta có: \(\sqrt{11-4\sqrt{7}}+\sqrt{23-8\sqrt{7}}+\sqrt{\left(-2\right)^6}\)
\(=\sqrt{7}-2+4-\sqrt{7}+8\)
=10
Rút gọn:
A=\(\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
B=\(\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(A=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
\(A=\sqrt{5-2\sqrt{5}+1}-\sqrt{5+2\sqrt{5}+1}\)
\(A=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(A=\sqrt{5}-1-\sqrt{5}-1\)
\(A=-2\)
\(B=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(B=\sqrt{5+4\sqrt{5}+4}-\sqrt{5-4\sqrt{5}+4}\)
\(B=\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(B=\sqrt{5}+2-\sqrt{5}+2\)
\(B=4\)
Học tốt
Bài làm:
a) \(A=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
\(A=\sqrt{5-2\sqrt{5}+1}-\sqrt{5+2\sqrt{5}+1}\)
\(A=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(A=\sqrt{5}-1-\sqrt{5}-1=-2\)
b) \(B=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(B=\sqrt{4+4\sqrt{5}+5}-\sqrt{4-4\sqrt{5}+5}\)
\(B=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(B=2+\sqrt{5}-\sqrt{5}+2\)
\(B=4\)
Dạ là do bạn rút căn ra ấy ạ:
\(\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)
và \(\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}+1\)
\(\Rightarrow\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}=\left(\sqrt{5}-1\right)-\left(\sqrt{5}+1\right)\)
\(=\sqrt{5}-1-\sqrt{5}-1=-2\)
Chúc bn hc tốt!!!
có ai biết giải bài này không giúp mình với mình đang cần gấp, xin cảm ơn
Bài 20: rút gọn
1, \(\sqrt{9-4\sqrt{5}}.\sqrt{9+4\sqrt{5}}\)
2, \(\left(2\sqrt{2}-6\right).\sqrt{11+6\sqrt{2}}\)
3, \(\sqrt{2}.\sqrt{2-\sqrt{3}}\left(\sqrt{3}+1\right)\)
4, \(\sqrt{2-\sqrt{3}}\left(\sqrt{6}-\sqrt{2}\right).\left(2+\sqrt{3}\right)\)
5, \(\sqrt{27+10\sqrt{2}}:\dfrac{1}{\sqrt{\left(\sqrt{2}-5\right)^2}}\)
Bài 21: rút gọn
1, \(5\sqrt{\dfrac{1}{5}}\) 2, \(\dfrac{12}{5}\sqrt{\dfrac{5}{4}}\)
3, \(\dfrac{30}{5\sqrt{6}}\) 4, \(\dfrac{20}{2\sqrt{5}}\)
5, \(\dfrac{2-\sqrt{2}}{\sqrt{2}}\)
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
Rút gọn : ( giúp với )
a) \(\dfrac{\sqrt{6}+\sqrt{10}}{\sqrt{21}+\sqrt{35}}\)
b) \(\dfrac{\sqrt{405}+3\sqrt{27}}{3\sqrt{3}+\sqrt{45}}\)
c) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{6}-\sqrt{9}-\sqrt{12}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
d) \(\dfrac{\sqrt{6-2\sqrt{5}}}{\sqrt{5}-1}\)
a. \(\dfrac{\sqrt{2}.\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{7}.\left(\sqrt{3}+\sqrt{5}\right)}=\dfrac{\sqrt{2}}{\sqrt{7}}=\sqrt{\dfrac{2}{7}}\)
d. \(\dfrac{\sqrt{6-2\sqrt{5}}}{\sqrt{5}-1}=\dfrac{\sqrt{5-2\sqrt{5}+1}}{\sqrt{5}-1}=\dfrac{\left(\sqrt{5}-1\right)^2}{\sqrt{5}-1}=\sqrt{5}-1\)
Rút gọn:
A=\(\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
B=\(\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
C=\(\sqrt{14-6\sqrt{5}-\sqrt{14+6\sqrt{5}}}\)
\(A=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
\(A=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(A=\sqrt{5}-1-\sqrt{5}-1\)
\(A=-2\)
\(B=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(B=\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(B=\sqrt{5}+2-\sqrt{5}+2\)
\(B=4\)
Sửa đề :
\(C=\sqrt{14-6\sqrt{5}}-\sqrt{14+6\sqrt{5}}\)
\(C=\sqrt{\left(3-\sqrt{5}\right)^2}-\sqrt{\left(3+\sqrt{5}\right)^2}\)
\(C=3-\sqrt{5}-3-\sqrt{5}\)
\(C=-2\sqrt{5}\)
Rút gọn các biểu thức sau :
a,\(\dfrac{\sqrt{6}+\sqrt{10}}{\sqrt{21}+\sqrt{35}}\)
b,\(\dfrac{\sqrt{405}+3\sqrt{27}}{3\sqrt{3}+\sqrt{45}}\)
c,\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{6}-\sqrt{9}-\sqrt{12}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
d, D=\(\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{9\left(x^2+2xy+y^2\right)}{4}}\) \(\left(vớix\ne y,x\ne-y\right)\)
d: \(D=\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{9\left(x^2+2xy+y^2\right)}{4}}\)
\(=\dfrac{2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{3\left(x+y\right)}{2}\)
\(=\dfrac{3}{x-y}\)