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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
\(C=\sqrt{6+2\sqrt{2\sqrt{3-\sqrt{4+2\sqrt{3}}}}}\)
\(C=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(C=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\) \(=\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\) \(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
\(D=\sqrt{\frac{8+2\sqrt{15}}{2}}-\sqrt{\frac{14-6\sqrt{5}}{2}}\) \(=\sqrt{\frac{\left(\sqrt{5}+\sqrt{3}\right)^2}{2}}-\sqrt{\frac{\left(3-\sqrt{5}\right)^2}{2}}\)
\(=\frac{\sqrt{5}+\sqrt{3}-3+\sqrt{5}}{\sqrt{2}}=\frac{2\sqrt{10}+\sqrt{6}-3\sqrt{2}}{2}\)
\(E=\sqrt{\frac{4+2\sqrt{3}}{2}}+\sqrt{\frac{4-2\sqrt{3}}{2}}\) \(=\sqrt{\frac{\left(\sqrt{3}+1\right)^2}{2}}+\sqrt{\frac{\left(\sqrt{3}-1\right)^2}{2}}\)
\(=\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
\(F=\sqrt{\frac{24-6\sqrt{7}}{2}}-\sqrt{\frac{24+6\sqrt{7}}{2}}\) \(=\sqrt{\frac{21-2\sqrt{21\cdot3}+3}{2}}-\sqrt{\frac{21+2\sqrt{21\cdot3}+3}{2}}\)
\(=\sqrt{\frac{\left(\sqrt{21}-\sqrt{3}\right)^2}{2}}-\sqrt{\frac{\left(\sqrt{21}+\sqrt{3}\right)^2}{2}}\)
\(=\frac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}=\frac{-2\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)
\(G=\left(3+\sqrt{3}\right)\cdot\sqrt{12-6\sqrt{3}}\) \(=\left(3+\sqrt{3}\right)\cdot\sqrt{\left(3-\sqrt{3}\right)^2}\)
\(=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)=9-3=6\)
\(H=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(3-\sqrt{5}\right)^2}\) \(=\sqrt{5}-2-3-\sqrt{5}=-5\)
\(I=\sqrt{\left(2\sqrt{2}-1\right)^2}-\sqrt{\left(2\sqrt{3}-1\right)^2}\)
\(=2\sqrt{2}-1-2\sqrt{3}+1=2\sqrt{2}-2\sqrt{3}\)
26 tháng 6 2019
\(\sqrt{13-4\sqrt{3}}+\sqrt{13+4\sqrt{3}}=2\sqrt{3}-1+2\sqrt{3}+1=4\sqrt{3}\)
\(\sqrt{19-6\sqrt{2}}-\sqrt{19+6\sqrt{2}}=3\sqrt{2}-1-3\sqrt{2}-1=-2\)
BT
2 tháng 8 2017
\(A=4-\sqrt{21-8\sqrt{5}}=4-\sqrt{4^2-8\sqrt{5}+\left(\sqrt{5}\right)^2}.\)
\(A=4-\sqrt{\left(4-\sqrt{5}\right)^2}=4-\left(4-\sqrt{5}\right)\)
=> \(A=\sqrt{5}\)
CW
1 tháng 8 2017
tớ ko chép lại đề, kí hiệu nhé
(1) \(=\left(\sqrt{6}-\sqrt{5}\right)^2-\sqrt{\left|\sqrt{6}+\sqrt{5}\right|^2}=\left(\sqrt{6}-\sqrt{5}\right)^2-\left(\sqrt{6}+\sqrt{5}\right)=1-2\sqrt{30}-\sqrt{6}-\sqrt{5}\)
ai ra đề mà để đáp án dài thế này mất thẩm mĩ quá!!!
(2) \(=\sqrt{\left|\sqrt{5}+\sqrt{3}\right|^2}-\sqrt{\left|\sqrt{5}-\sqrt{3}\right|^2}=\left(\sqrt{5}+\sqrt{3}\right)-\left(\sqrt{5}-\sqrt{3}\right)=2\sqrt{3}\)
(3) \(=\sqrt{\left|\sqrt{7}+2\right|^2}-\sqrt{\left|3-\sqrt{5}\right|^2}=\sqrt{7}+2-3+\sqrt{5}=\sqrt{7}+\sqrt{5}-1\)
lại thêm 1 phép tính không đẹp....
(4) \(=\sqrt{\left|3\sqrt{2}-2\right|^2}-\sqrt{\left|3\sqrt{2}+1\right|^2}=3\sqrt{2}-2-3\sqrt{2}-1=-3\)
(5) \(=\sqrt{\left|2\sqrt{3}-1\right|^2}+\sqrt{\left|2\sqrt{3}-3\right|^2}=2\sqrt{3}-1+2\sqrt{3}-3=4\sqrt{3}-4\)
kiểm tra lại kết quả nhé ^^! Cảm ơn!
1 tháng 8 2018
\(a.\sqrt{19-6\sqrt{2}}=\sqrt{18-2.3\sqrt{2}+1}=3\sqrt{2}-1\)
\(b.\sqrt{21+12\sqrt{3}}=\sqrt{12+2.2\sqrt{3}.3+9}=2\sqrt{3}+3\)
\(c.\sqrt{57-40\sqrt{2}}=\sqrt{32-2.4\sqrt{2}.5+25}=4\sqrt{2}-5\)
\(d.\sqrt{\left(5-2\sqrt{6}\right)\left(4-2\sqrt{3}\right)}=\sqrt{3-2\sqrt{3}.\sqrt{2}+2}.\sqrt{3-2\sqrt{3}+1}=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}-1\right)\) \(e.\sqrt{21+6\sqrt{6}}+\sqrt{21-6\sqrt{6}}=\sqrt{18+2.3\sqrt{2}.\sqrt{3}+3}+\sqrt{18-2.3\sqrt{2}.\sqrt{3}+3}=3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}=6\sqrt{2}\) \(g.\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}=\sqrt{4-2.2\sqrt{3}+3}-\sqrt{4+2.2\sqrt{3}+3}=2-\sqrt{3}-2-\sqrt{3}=-2\sqrt{3}\)
HT
1 tháng 8 2018
a)
=\(\sqrt{18-2.3\sqrt{2}.1+1}\)
\(=\sqrt{\left(3\sqrt{2}-1\right)^2}\)
\(=3\sqrt{2}-1\)
b)
=\(\sqrt{12+2.2\sqrt{3}.3+9}\)
=\(\sqrt{\left(2\sqrt{3}+3\right)^2}\)
=\(2\sqrt{3}+3\)
c)
=\(\sqrt{25-2.5.4\sqrt{2}+32}\)
=\(\sqrt{\left(5-4\sqrt{2}\right)^2}\)
=\(4\sqrt{2}-5\)
d)
\(=\sqrt{\left(3-2.\sqrt{3}.\sqrt{2}+2\right)\left(3-2\sqrt{3}+1\right)}\\ =\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2\left(\sqrt{3}-1\right)^2}\\ =\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}-1\right)\\ =3-\sqrt{3}-\sqrt{6}+\sqrt{2}\)
e)
\(=\sqrt{18+2.3\sqrt{2}.\sqrt{3}+3}+\sqrt{18-2.3\sqrt{2}.\sqrt{3}+3}\\ =\sqrt{\left(3\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\\ =3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}\\ =6\sqrt{2}\)
g)
\(=\sqrt{4-2.2.\sqrt{3}+3}-\sqrt{4+2.2.\sqrt{3}+3}\\ =\sqrt{\left(2-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\\ =2-\sqrt{3}-2-\sqrt{3}\\ =-2\sqrt{3}\)
VN
1 tháng 5 2017
Làm tới dòng thứ 3 máy đơ, 2 lần rồi T,T
Mình chia làm 2 phần tính nhé
\(A=\frac{4\sqrt{2}}{\sqrt{10-2\sqrt{21}}}+\frac{3}{\sqrt{15+6\sqrt{6}}}-\frac{1}{\sqrt{19-6\sqrt{10}}}\)
\(A=\frac{4\sqrt{2}}{\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}}+\frac{3}{\sqrt{\left(\sqrt{9}+\sqrt{6}\right)^2}}-\frac{1}{\sqrt{\left(\sqrt{10}-\sqrt{9}\right)^2}}\)
\(A=\frac{4\sqrt{2}}{\sqrt{7}-\sqrt{3}}+\frac{3}{3+\sqrt{6}}-\frac{1}{\sqrt{10}-3}\)
\(A=\frac{4\sqrt{2}\left(\sqrt{7}+\sqrt{3}\right)}{7-3}+\frac{3\left(3-\sqrt{6}\right)}{9-6}-\frac{1\left(\sqrt{10}+3\right)}{10-9}\)
\(A=\frac{4\sqrt{14}+4\sqrt{6}}{4}+\frac{9-3\sqrt{6}}{3}-\sqrt{10}-3\)
\(A=\sqrt{14}+\sqrt{6}+3-\sqrt{6}-\sqrt{10}-3\)
\(A=\sqrt{14}-\sqrt{10}\)
\(B=\sqrt{6+\sqrt{35}}\)
\(B=\frac{\sqrt{2}\left(\sqrt{6+\sqrt{35}}\right)}{\sqrt{2}}\)
\(B=\frac{\sqrt{12+2\sqrt{35}}}{\sqrt{2}}\)
\(B=\frac{\sqrt{\left(\sqrt{7}+\sqrt{5}\right)^2}}{\sqrt{2}}\)
\(B=\frac{\sqrt{7}+\sqrt{5}}{\sqrt{2}}\)
\(\Rightarrow M=A.B=\left(\sqrt{14}-\sqrt{10}\right).\frac{\sqrt{7}+\sqrt{5}}{\sqrt{2}}\)
\(M=\sqrt{2}\left(\sqrt{7}-\sqrt{5}\right).\frac{\sqrt{7}+\sqrt{5}}{\sqrt{2}}\)
\(M=\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
\(M=\left(\sqrt{7}\right)^2-\left(\sqrt{5}\right)^2\)
\(M=7-5=2\)
29 tháng 7 2017
a. \(\sqrt{13+6\sqrt{4+\sqrt{9-4\sqrt{2}}}}=\sqrt{13+6\sqrt{4+\sqrt{\left(\sqrt{8}-1\right)^2}}}=\sqrt{13+6\sqrt{4+\sqrt{8}-1}}=\sqrt{13+6\sqrt{3+\sqrt{8}}}=\sqrt{13+6\sqrt{\left(\sqrt{2}+1\right)^2}}=\sqrt{13+6\left(\sqrt{2}+1\right)}=\sqrt{13+6\sqrt{2}+6}=\sqrt{19+6\sqrt{2}}=\sqrt{\left(3\sqrt{2}+1\right)^2}=1+3\sqrt{2}\)
b. \(\left(\sqrt{3}-1\right)\sqrt{2\sqrt{19+8\sqrt{3}}-4}=\left(\sqrt{3}-1\right)\sqrt{2\sqrt{\left(4+\sqrt{3}\right)^2}-4}=\left(\sqrt{3}-1\right)\sqrt{8+2\sqrt{3}-4}=\left(\sqrt{3}-1\right)\sqrt{\left(\sqrt{3}+1\right)^2}=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=3-1=2\)
c. \(\sqrt{5+2\sqrt{6}}+\sqrt{14-4\sqrt{6}}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}+2\sqrt{3}-\sqrt{2}=3\sqrt{3}\)
d. \(\sqrt{5-2\sqrt{6}}+\sqrt{11-4\sqrt{6}}=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{3}\right)^2}=\sqrt{3}-\sqrt{2}+2\sqrt{2}-\sqrt{3}=\sqrt{2}\)
MA
3 tháng 1 2019
1) \(\sqrt{12}\)+\(5\sqrt{3}-\sqrt{48}\)
= \(2\sqrt{3}+5\sqrt{3}-4\sqrt{3}\)
= (2+5-4).\(\sqrt{3}\)
= \(3\sqrt{3}\)
2)\(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\)
= \(5\sqrt{5}+2\sqrt{5}-3.3\sqrt{5}\)
= \(5\sqrt{5}+2\sqrt{5}-9\sqrt{5}\)
= \(\left(5+2-9\right).\sqrt{5}\)
= -2\(\sqrt{2}\)
3)\(3\sqrt{32}+4\sqrt{8}-5\sqrt{18}\)
= \(3.4\sqrt{2}+4.2\sqrt{2}-5.3\sqrt{2}
\)
= 12\(\sqrt{2}\) \(+8\sqrt{2}\) \(-15\sqrt{2}\)
= \(\left(12+8-15\right).\sqrt{2}\)
= \(5\sqrt{2}\)
4)\(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)
= \(3.2\sqrt{3}-4.3\sqrt{3}+5.4\sqrt{3}\)
= \(6\sqrt{3}-12\sqrt{3}+20\sqrt{3}\)
= \(\left(6-12+20\right).\sqrt{3}\)
= \(14\sqrt{3}\)
5)\(\sqrt{12}+\sqrt{75}-\sqrt{27}\)
= \(2\sqrt{3}+5\sqrt{3}-3\sqrt{3}\)
= \(\left(2+5-3\right).\sqrt{3}\)
= \(4\sqrt{3}\)
6) \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\)
= \(2.3\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= 6\(\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= \(\left(6-7+9\right).\sqrt{2}\)
= 8\(\sqrt{2}\)
7)\(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\)
= \(3.2\sqrt{5}-2.3\sqrt{5}+4\sqrt{5}\)
= \(6\sqrt{5}-6\sqrt{5}+4\sqrt{5}\)
= \(4\sqrt{5}\)
8)\(\left(\sqrt{2}+2\right).\sqrt{2}-2\sqrt{2}\)
= \(\left(\sqrt{2}\right)^2+2\sqrt{2}-2\sqrt{2}\)
= 2
14 tháng 6 2018
a) \(\sqrt{6-4\sqrt{2}}+\sqrt{19-6\sqrt{2}}\)\(=\sqrt{4-4\sqrt{2}+2}+\sqrt{18-2.3\sqrt{2}.1+1}=\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-1\right)^2}\)\(=\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-1\right)^2}\)
= / 2 - \(\sqrt{2}\) / + / 3\(\sqrt{2}\) - 1/
= 2 - \(\sqrt{2}\) + 3\(\sqrt{2}\) - 1
= 2\(\sqrt{2}\) + 1
b) \(\sqrt{46-6\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
\(=\sqrt{45-2.3.\sqrt{5}+1}-\sqrt{20-2.3.2.\sqrt{5}+9}\)
\(=\sqrt{\left(3\sqrt{5}-1\right)^2}-\sqrt{\left(2\sqrt{5}-3\right)^2}\)
= / 3\(\sqrt{5}\) - 1/ - / 2\(\sqrt{5}\) - 3/
= 3\(\sqrt{5}\) - 1 - 2\(\sqrt{5}\) + 3
= \(\sqrt{5}\) + 2
c) \(\sqrt{7-2\sqrt{10}}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5-2\sqrt{5}.\sqrt{2}+2}-\sqrt{5-2\sqrt{5}+1}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}\)
= / \(\sqrt{5}\) - \(\sqrt{2}\) / - / \(\sqrt{5}\) - 1 /
= 1 - \(\sqrt{2}\)
14 tháng 6 2018
a) \(\sqrt{6-4\sqrt{2}}+\sqrt{19-6\sqrt{2}}\)
\(=\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-1\right)^2}\)
\(=2-\sqrt{2}+3\sqrt{2}-1\)
\(=2\sqrt{2}+1\)
AH
Akai Haruma
Giáo viên
6 tháng 10 2019
a)
\((4+\sqrt{15})(\sqrt{10}-\sqrt{6})\sqrt{4-\sqrt{15}}=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{3+5-2\sqrt{3.5}}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})^2=(4+\sqrt{15})(8-2\sqrt{15})=2(4+\sqrt{15})(4-\sqrt{15})\)
\(=2(4^2-15)=2\)
b)
\(\sqrt{10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}=\sqrt{(8+2\sqrt{15})+2+2(\sqrt{6}+\sqrt{10})}\)
\(=\sqrt{(\sqrt{5}+\sqrt{3})^2+2\sqrt{2}(\sqrt{3}+\sqrt{5})+2}\)
\(=\sqrt{(\sqrt{5}+\sqrt{3}+\sqrt{2})^2}=\sqrt{5}+\sqrt{3}+\sqrt{2}\)
AH
Akai Haruma
Giáo viên
6 tháng 10 2019
c)
\((\sqrt{5+2\sqrt{9\sqrt{5}-19}}-\sqrt{7-\sqrt{5}}):(2\sqrt{\sqrt{5}-2})\)
\(=(\sqrt{(5+2\sqrt{9\sqrt{5}-19})(\sqrt{5}+2)}-\sqrt{(7-\sqrt{5})(\sqrt{5}+2)}):(2\sqrt{(\sqrt{5}-2)(\sqrt{5}+2)})\)
\(=[\sqrt{10+5\sqrt{5}+2\sqrt{(9\sqrt{5}-19)(9+4\sqrt{5})}}-\sqrt{9+5\sqrt{5}}]:2\)
\(=[\sqrt{10+5\sqrt{5}+2\sqrt{9+5\sqrt{5}}}-\sqrt{9+5\sqrt{5}}]:2\)
\(=[\sqrt{(9+5\sqrt{5})+2\sqrt{9+5\sqrt{5}}+1}-\sqrt{9+5\sqrt{5}}]:2\)
\(=[\sqrt{(\sqrt{9+5\sqrt{5}}+1)^2}-\sqrt{9+5\sqrt{5}}]:2\)
\(=[\sqrt{9+5\sqrt{5}}+1-\sqrt{9+5\sqrt{5}}]:2=\frac{1}{2}\)
d)
\((\sqrt{9+\sqrt{5}}+\sqrt{9-\sqrt{5}})^2=18+2\sqrt{(9+\sqrt{5})(9-\sqrt{5})}=18+4\sqrt{19}\)
\(\Rightarrow \sqrt{9+\sqrt{5}}+\sqrt{9-\sqrt{5}}=\sqrt{18+4\sqrt{19}}\)
Do đó:
\(\frac{\sqrt{9+\sqrt{5}}+\sqrt{9-\sqrt{5}}}{\sqrt{9+2\sqrt{19}}}-\sqrt{3-2\sqrt{2}}=\frac{\sqrt{18+4\sqrt{19}}}{\sqrt{9+2\sqrt{19}}}-\sqrt{2+1-2\sqrt{2.1}}\)
\(=\frac{\sqrt{2}.\sqrt{9+2\sqrt{19}}}{\sqrt{9+2\sqrt{19}}}-\sqrt{(\sqrt{2}-1)^2}=\sqrt{2}-(\sqrt{2}-1)=1\)
\(\sqrt{19-6\sqrt{2}}\)
\(=\sqrt{18-2.3\sqrt{2}+1}\)
\(=\sqrt{\left(3\sqrt{2}\right)^2-2.3\sqrt{2}+1}\)
\(=\sqrt{\left(3\sqrt{2}-1\right)^2}\)
\(=3\sqrt{2}-1\)