\(6\sqrt{2}\) + \(\sqrt{6-\sqrt{11}}\)\(-\sqrt{6+\sqrt{11}}\)
thu gon
mik can gap gip mik vs
Những câu hỏi liên quan
6) Thực hiện các phép tính
a. \(\sqrt{2}\left(\sqrt{8}+\sqrt{32}-\sqrt{98}\right)\)
b. \(\dfrac{2}{\sqrt{5}+2}+\dfrac{2}{2-\sqrt{5}}\)
c. \(\left(2+\sqrt{3}\right)\sqrt{11-6\sqrt{2}}\)
Mng giúp mik vs ạ >.<!!
b, \(\dfrac{2}{\sqrt{5}+2}+\dfrac{2}{2-\sqrt{5}}\)
\(=\dfrac{2\left(\sqrt{5}-2\right)}{5-4}-\dfrac{2\left(\sqrt{5}+2\right)}{5-4}\)
\(=2\sqrt{5}-4-2\sqrt{5}-4=-8\)
Đúng 2
Bình luận (0)
a, \(\sqrt{2}\left(\sqrt{8}+\sqrt{32}-\sqrt{98}\right)\)
\(=\sqrt{2}\left(2\sqrt{2}+4\sqrt{2}-7\sqrt{2}\right)\)
\(=\sqrt{2}.\left(-\sqrt{2}\right)=-2\)
Đúng 1
Bình luận (0)
c, \(\left(2+\sqrt{3}\right)\sqrt{11-6\sqrt{2}}\)
\(=\left(2+\sqrt{3}\right)\sqrt{\left(\sqrt{2}-3\right)^2}\)
\(=\left(2+\sqrt{3}\right)\left(3-\sqrt{2}\right)\)
\(=6-2\sqrt{2}+3\sqrt{3}-\sqrt{6}\)
Đúng 1
Bình luận (0)
Xem thêm câu trả lời
chứng minh :a) 11+6sqrt{2} (3+sqrt{2})^2 b) sqrt{11+6sqrt{2}}+sqrt{11-6sqrt{2}}6 c) sqrt{8-2sqrt{7}}-sqrt{8+2sqrt{7}} -2 d) sqrt{49-12sqrt{5}}-sqrt{49+12sqrt{5}}-4
Đọc tiếp
chứng minh :a) 11+6\(\sqrt{2}\)= (3+\(\sqrt{2}\))\(^2\)
b) \(\sqrt{11+6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)=6
c) \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)= -2
d) \(\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)=-4
a: \(\left(3+\sqrt{2}\right)^2=3^2+2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2\)
\(=9+6\sqrt{2}+2=11+6\sqrt{2}\)
b: \(\sqrt{11+6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=3+\sqrt{2}+3-\sqrt{2}=6\)
c: \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\sqrt{7}-1-\sqrt{7}-1=-2\)
d: \(\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)
\(=\sqrt{45-2\cdot3\sqrt{5}\cdot2+4}-\sqrt{45+2\cdot3\sqrt{5}\cdot2+4}\)
\(=\sqrt{\left(3\sqrt{5}-2\right)^2}-\sqrt{\left(3\sqrt{5}+2\right)^2}\)
\(=3\sqrt{5}-2-3\sqrt{5}-2=-4\)
Đúng 3
Bình luận (0)
a) \(\left(3+\sqrt{2}\right)^2=9+6\sqrt{2}+2=11+6\sqrt{2}\)
b) \(\sqrt{11+6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=3+\sqrt{2}+3-\sqrt{2}=6\)
c) \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\sqrt{7}-1-\sqrt{7}-1=-2\)
d) \(\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)
\(=\sqrt{\left(3\sqrt{5}-2\right)^2}-\sqrt{\left(3\sqrt{5}+2\right)^2}\)
\(=3\sqrt{5}-2-3\sqrt{5}-2=-4\)
Đúng 1
Bình luận (0)
20 tháng 6 2019 lúc 20:49
Thu gọn biểu thức sau:
C=\(\sqrt{11-6\sqrt{2}}-\sqrt{11+6\sqrt{2}}\)
\(C=\sqrt{9-2\cdot3\sqrt{2}+2}-\sqrt{9+2\cdot3\cdot\sqrt{2}+2}\)
\(C=\sqrt{\left(3-\sqrt{2}\right)^2}-\sqrt{\left(3+\sqrt{2}\right)^2}\)
\(C=3-\sqrt{2}-\left(3+\sqrt{2}\right)=-2\sqrt{2}\)
Đúng 0
Bình luận (0)
a) 11+6\(\sqrt{2}\) = \(\left(3+\sqrt{2}\right)^2\)
b) 8-2\(\sqrt{7}\)=\(\left(\sqrt{7}-1\right)^2\)
c)\(\sqrt{11+6\sqrt{2}}=\sqrt{11-6\sqrt{2}}=6\)
d) \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=-2\)
a)\(\sqrt{\left(2\sqrt{2}-3\right)^2+\sqrt{15}}\)
b)\(\sqrt{\left(\sqrt{10}-3\right)}^2+\sqrt{\left(\sqrt{10}-4\right)^2}\)
c)\(11+6\sqrt{2}=\left(3+\sqrt{2}\right)^2\)
d)\(\sqrt{11}+6\sqrt{2}+\sqrt{11-6\sqrt{2}=6}\)
b: =căn 10-3+4-căn 10=1
a: \(=\sqrt{11-4\sqrt{6}+\sqrt{15}}\)
Đúng 0
Bình luận (0)
Rút gọn: ( làm hộ vs mình đúng cho =)))
\(A=\frac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}\)
\(A=\frac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}=\frac{\sqrt{3}+\sqrt{9+2.3\sqrt{3}+2}-\sqrt{3+2\sqrt{3}\sqrt{2}+2}}{\sqrt{2}+\sqrt{5+2\sqrt{5}.1+1}-\sqrt{5+2\sqrt{5}\sqrt{2}+2}}\)
\(=\frac{\sqrt{3}+\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}}{\sqrt{2}+\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}=\frac{\sqrt{3}+3+\sqrt{2}-\sqrt{3}-\sqrt{2}}{\sqrt{2}+\sqrt{5}+1-\sqrt{5}-\sqrt{2}}\)
\(=\frac{3}{1}=3\)
Đúng 0
Bình luận (0)
A=\(\frac{\sqrt{3}+3+\sqrt{2}-\sqrt{2}-\sqrt{3}}{\sqrt{2}+1+\sqrt{5}-\sqrt{2}-\sqrt{5}}=\frac{3}{1}=3\)
Đúng 0
Bình luận (0)
\(\sqrt{\left(\sqrt{7}-5\right)^2}+\sqrt{\left(2-\sqrt{7}\right)^2}\)
\(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
Lời giải:
a. \(=|\sqrt{7}-5|+|2-\sqrt{7}|=5-\sqrt{7}+(\sqrt{7}-2)=3\)
b. \(=\sqrt{(3+\sqrt{2})^2}-\sqrt{(3-\sqrt{2})^2}=|3+\sqrt{2}|-|3-\sqrt{2}|\)
\(=(3+\sqrt{2})-(3-\sqrt{2})=2\sqrt{2}\)
c.
\(=\sqrt{(3+2\sqrt{2})^2}+\sqrt{(3-2\sqrt{2})^2}=|3+2\sqrt{2}|+|3-2\sqrt{2}|\)
$=(3+2\sqrt{2})+(3-2\sqrt{2})=6$
d.
$=\sqrt{(\sqrt{5}+1)^2}-\sqrt{(\sqrt{5}-1)^2}$
$=|\sqrt{5}+1|-|\sqrt{5}-1|=\sqrt{5}+1-(\sqrt{5}-1)=2$
Đúng 3
Bình luận (0)
a)\(\sqrt{2+\sqrt{2+\sqrt[]{2+\sqrt{2+...}}}}\)
b)\(\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+...}}}}\)
MINH CAN GAP A
a/ \(A=\sqrt{2+\sqrt{2+...}}\)
\(\Leftrightarrow A^2=2+A\)
\(\Leftrightarrow\left(A-2\right)\left(A+1\right)=0\)
\(\Leftrightarrow A=2\)
b/ \(B=\sqrt{6+\sqrt{6+...}}\)
\(\Leftrightarrow B^2=6+B\)
\(\Leftrightarrow\left(B-3\right)\left(B+2\right)=0\)
\(\Leftrightarrow B=3\)
Đúng 0
Bình luận (0)
C/m: \(\sqrt{11+6\sqrt{2}}\) + \(\sqrt{11-6\sqrt{2}}\) = 6
Ta có VT:
\(VT=\sqrt{11+6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{3^2+2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}+\sqrt{3^2-2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=\left|3+\sqrt{2}\right|+\left|3-\sqrt{2}\right|\)
\(=3+\sqrt{2}+3-\sqrt{2}\)
\(=6=VP\left(dpcm\right)\)
Đúng 2
Bình luận (0)
\(VT=\sqrt{9+2\cdot3\cdot\sqrt{2}+2}+\sqrt{9-2\cdot3\cdot\sqrt{2}+2}\)
\(=3+\sqrt{2}+3-\sqrt{2}\)
=6=VP
Đúng 0
Bình luận (0)