Tính
a/ A=\(\sqrt{6-2\sqrt{5}}\) - \(\sqrt{5}\)
b/ B=\(\sqrt{17\sqrt{2}-24}\)+ 4
c/ C=\(\sqrt{12+6\sqrt{3}-4}\)- \(\sqrt{3}\)
Bài 3: Thực hiện các phép tính sau:
a) \(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)
b) \(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\)
c) \(\sqrt{6-4\sqrt{2}}+\)\(\sqrt{22-12\sqrt{2}}\)
hộ mk với
a) \(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)
\(=2\sqrt{5}+2+\sqrt{5}-2\)
\(=3\sqrt{5}\)
b) \(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\)
\(=3-2\sqrt{2}+2\sqrt{2}-1\)
=2
c) \(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\)
\(=2-\sqrt{2}+3\sqrt{2}-2\)
\(=2\sqrt{2}\)
Tính
a/ A=\(\sqrt{6-2\sqrt{5}}\) - \(\sqrt{5}\)
b/ B=\(\sqrt{17\sqrt{2}-24}\)+ 4
c/ C=\(\sqrt{12+6\sqrt{3}-4}\)- \(\sqrt{3}\)
a/ \(A=\sqrt{6-2\sqrt{5}}-\sqrt{5}\)\(=\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}+1^2}-\sqrt{5}\)\(=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{5}\)\(=\sqrt{5}-1-\sqrt{5}\)\(=-1.\)
Bạn kiểm tra lại câu b với c đi, hình như sai đề rồi.
Thực hiện phép tính:
a)\(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)
b)\(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)
c)\(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\)
d)\(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\)
a,\(\sqrt{\left(\sqrt{3}-1\right)^2}\) \(+\sqrt{\left(\sqrt{3}+1\right)^2}=2\sqrt{3}\)
b. \(\sqrt{\left(2\sqrt{5}+2\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}=3\sqrt{5}\)
c,\(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}=4\)
d.\(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-2\right)^2}=2\sqrt{2}\)
thực hiện phép tính
a)\(\sqrt{6+2\sqrt{5}}-\sqrt{62\sqrt{5}}\)
b)\(\sqrt{24-8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)
c) \(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\)
d)\(\sqrt{41+12\sqrt{5}}-\sqrt{46-6\sqrt{ }5}\)
e)\(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\)
f)\(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\)
g) \(\sqrt{43+24\sqrt{3}}-\sqrt{49-8\sqrt{3}}\)
cho cách làm dạng bài này luôn. Chỗ nào chưa hiểu thì nói tớ sẽ giải thích thêm (cần góp ý để hoàn thiện thêm phần hướng dẫn đó mà. Cảm ơn cậu).
Phương Nam Phim (à quên, Từ Hạ) hân hạnh giới thiệu bộ phim...
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.
$=\sqrt{2+2.3\sqrt{2}+3^2}-\sqrt{2-2.3\sqrt{2}+3^2}$
$=\sqrt{(\sqrt{2}+3)^2}-\sqrt{(\sqrt{2}-3)^2}$
$=|\sqrt{2}+3|-|\sqrt{2}-3|$
$=(\sqrt{2}+3)-(3-\sqrt{2})=2\sqrt{2}$
Bài 1: Chứng minh:
a) \(\sqrt{9+\sqrt{17}}-\sqrt{9-\sqrt{17}}-\sqrt{2}=0\)
b) \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}=\sqrt{5}+1\)
c) \(\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}< 24\)
d) \(\sqrt{6+\sqrt{6+\sqrt{6+,,,+\sqrt{6}}}}=3\)
Bài 1: Tính
A=\(\sqrt{46-6\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
B=\(\sqrt{13-\sqrt{160}-\sqrt{53+4\sqrt{90}}}\)
C=\(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)
D=\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
E= \(\sqrt{4-\sqrt{7}}+\sqrt{4+\sqrt{7}}\)
F= \(\sqrt{3+\sqrt{11+6\sqrt{2}}}-\sqrt{5+2\sqrt{6}}\)
G=\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
Bài 2: so sánh
a) \(\sqrt{24}+\sqrt{45}\) và 12
b) \(\sqrt{37}-\sqrt{15}\) và 2
c) \(\sqrt{16}\) và \(\sqrt{15}\times\sqrt{17}\)
d) 8 và \(\sqrt{15}+\sqrt{17}\)
Bài 2 :
a,\(\sqrt{24}+\sqrt{45}< \sqrt{25}+\sqrt{49}=5+7=12=>\sqrt{24}+\sqrt{45}< 12\)
b. \(\sqrt{37}-\sqrt{15}>\sqrt{36}-\sqrt{16}=6-4=2=>\sqrt{37}-\sqrt{15}>2\)
c, \(\sqrt{15}.\sqrt{17}>\sqrt{15}.\sqrt{16}>\sqrt{16}=>\sqrt{15}.\sqrt{17}>\sqrt{16}\)
a : \(\sqrt{\left(2\sqrt{2}-1\right)^2}-\sqrt{17+12\sqrt{2}}\)
b : \(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\)
c : \(\sqrt{\left(4-3\sqrt{2}\right)^2}-\sqrt{19+6\sqrt{2}}\)
\(\sqrt{\left(2\sqrt{2-1}\right)^2}-\sqrt{17+12\sqrt{2}}\\ =\left|2\sqrt{2}-1\right|-\sqrt{9+2\cdot3\cdot2\sqrt{2}+\left(2\sqrt{2}\right)^2}\\ =2\sqrt{2}-1-\sqrt{\left(3+2\sqrt{2}\right)^2}\\=2\sqrt{2}-1-\left(3+2\sqrt{2}\right)\\ =2\sqrt{2}-1-3-2\sqrt{2}\\ =-4\)
__
\(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\\ =\left|2-\sqrt{5}\right|+\sqrt{9-2\cdot3\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}\\ =2-\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}\\ =2-\sqrt{5}+3-\sqrt{5}\\ =5-2\sqrt{5}\)
__
\(\sqrt{\left(4-3\sqrt{2}\right)^2}-\sqrt{19+6\sqrt{2}}\\ =\left|4-3\sqrt{2}\right|-\sqrt{18+2\cdot3\cdot\sqrt{2}+1}\\ =4-3\sqrt{2}-\sqrt{\left(3\sqrt{2}+1\right)^2}\\ =4-3\sqrt{2}-3\sqrt{2}-1\\ =3-6\sqrt{2}\)
Tính thu gọn :
a , \(\sqrt{17-12\sqrt{2}}-\sqrt{17+12\sqrt{2}}\)
b , \(\sqrt{27+12\sqrt{5}}-\sqrt{27-12\sqrt{5}}\)
c , \(\sqrt{15-6\sqrt{6}}+\sqrt{15+\sqrt{6\sqrt{6}}}\)
d , \(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
e , \(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
f , \(\sqrt{5+\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
f, \(\sqrt{\sqrt{5}+\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}+\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}=\sqrt{\sqrt{5}+\sqrt{3-2\sqrt{5}+3}}=\sqrt{\sqrt{5}+\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}+\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{\sqrt{5}+\sqrt{5}-1}=\sqrt{2\sqrt{5}-1}\)
mik sửa lại câu f , tí nhé :
f , \(\sqrt{\sqrt{5}+\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
a,\(=\sqrt{9-2.3.2\sqrt{2}+8}-\sqrt{9+2.3.2\sqrt{2}+8}\)
\(=\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(3+2\sqrt{2}\right)^2}\) \(=3-2\sqrt{2}-3-2\sqrt{2}=-4\sqrt{2}\)