\(\sqrt{11+\sqrt{7}}-\sqrt{\left(-1+\sqrt{7}\right)^2}\)
Những câu hỏi liên quan
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) \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=-2\)
Tính \(\left(\sqrt{7}+\sqrt{11}+\sqrt{13}\right)\left(\sqrt{11}+\sqrt{13}-\sqrt{7}\right)\left(\sqrt{7}+\sqrt{13}-\sqrt{11}\right)\left(\sqrt{7}+\sqrt{11}-\sqrt{13}\right)\)
Rút gọn:
A dfrac{4+sqrt{7}}{3sqrt{2}+sqrt{4+sqrt{7}}}+dfrac{4-sqrt{7}}{3sqrt{2}-sqrt{4-sqrt{7}}}
B dfrac{3sqrt{2}+sqrt{11}}{sqrt{2}+sqrt{6+sqrt{11}}}+dfrac{3sqrt{2}-sqrt{11}}{sqrt{2}-sqrt{6-sqrt{11}}}+18
C dfrac{1}{sqrt{3}+sqrt{5}}+dfrac{1}{sqrt{5}+sqrt{7}}+...+dfrac{1}{sqrt{2n+1}+sqrt{2n+3}}với n thuộc N*
D left(sqrt{3}+1right)left(sqrt{5}-1right)left(sqrt{15}-1right)left(7-2sqrt{3}+sqrt{5}right)
Edfrac{left(4+sqrt{3}right)}{sqrt[]{1}+sqrt{3}}+dfrac{left(8+sqrt{15}right)}{sqrt{3}+sqrt...
Đọc tiếp
Rút gọn:
A = \(\dfrac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\dfrac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}\)
B = \(\dfrac{3\sqrt{2}+\sqrt{11}}{\sqrt{2}+\sqrt{6+\sqrt{11}}}+\dfrac{3\sqrt{2}-\sqrt{11}}{\sqrt{2}-\sqrt{6-\sqrt{11}}}+18\)
C = \(\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+...+\dfrac{1}{\sqrt{2n+1}+\sqrt{2n+3}}\)với n thuộc N*
D = \(\left(\sqrt{3}+1\right)\left(\sqrt{5}-1\right)\left(\sqrt{15}-1\right)\left(7-2\sqrt{3}+\sqrt{5}\right)\)
E=\(\dfrac{\left(4+\sqrt{3}\right)}{\sqrt[]{1}+\sqrt{3}}+\dfrac{\left(8+\sqrt{15}\right)}{\sqrt{3}+\sqrt{5}}+...+\dfrac{2k+\sqrt{k^2-1}}{\sqrt{k-1}+\sqrt{k+1}}+...+\dfrac{240+\sqrt{14399}}{\sqrt{119}+\sqrt{121}}\)
F = \(\left(\dfrac{2a+1}{a\sqrt{a}-1}-\dfrac{\sqrt{a}}{a+\sqrt{a}+1}\right)\left(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\) với a >= 0 và a khác 1
14 tháng 11 2021 lúc 21:23
\(\sqrt{11+\sqrt{7}}\)\(-\sqrt{\left(-1+\sqrt{7}\right)^2}\)
Lời giải:
\(\sqrt{11+\sqrt{7}}-\sqrt{(-1+\sqrt{7})^2}=\sqrt{11+\sqrt{7}}-|-1+\sqrt{7}|\)
\(=\sqrt{11+\sqrt{7}}-(-1+\sqrt{7})=\sqrt{11+\sqrt{7}}+1-\sqrt{7}\)
Đúng 1
Bình luận (0)
21 tháng 7 2023 lúc 9:12
Rút gọn: (Giải chi tiết từng bước)
7) \(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)
8) \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)
`a, (sqrt 28 - sqrt 12 - sqrt 7) sqrt 7 + 2 sqrt 21`.
`= sqrt(28.7) - sqrt(12.7) - sqrt(7.7) + 2 sqrt 21`.
`= sqrt(4. 7.7) - sqrt (12.7) - 7 + 2 sqrt 21`.
`= 14 - sqrt(4.3.7) - 7 + 2 sqrt 21`.
`= 7`.
`b, (sqrt99-sqrt18-sqrt11)sqrt11+3sqrt22`
`= sqrt(99.11)- sqrt(18.11)-sqrt(11.11) +3sqrt22`
`= sqrt(9.11.11)-sqrt(2.9.11)-11+3sqrt22`
`= 33 - 11 = 22`.
Đúng 2
Bình luận (1)
\(\sqrt{\left(\sqrt{7}-5\right)^2}+\sqrt{\left(2-\sqrt{7}\right)^2}\)
\(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
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)
Câu 11) Tínha) sqrt{25}+sqrt{49} b) sqrt{121}-sqrt{81}2) Với x -2 thì sqrt{2x+1} có nghĩa không3) Rút gọn biểu thức sau :a) sqrt{left(sqrt{3}-1right)^2}-sqrt{3} b) left(sqrt{28}-2sqrt{14}+sqrt{7}right)sqrt{7}+7sqrt{8} c) dfrac{sqrt{27}-sqrt{108}+sqrt{12}}{sqrt{3}}
Đọc tiếp
Câu 1
1) Tính
a) \(\sqrt{25}+\sqrt{49}\) b) \(\sqrt{121}-\sqrt{81}\)
2) Với x > -2 thì \(\sqrt{2x+1}\) có nghĩa không
3) Rút gọn biểu thức sau :
a) \(\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\) b) \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\) c) \(\dfrac{\sqrt{27}-\sqrt{108}+\sqrt{12}}{\sqrt{3}}\)
1:
a: \(\sqrt{25}+\sqrt{49}=5+7=12\)
b: \(\sqrt{121}-\sqrt{81}=11-9=2\)
2: x>-2
=>2x>-4
=>2x+1>-3
=>Với x>-2 thì \(\sqrt{2x+1}\) chưa chắc có nghĩa
3:
a: \(\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\)
\(=\left|\sqrt{3}-1\right|-\sqrt{3}\)
\(=\sqrt{3}-1-\sqrt{3}=-1\)
b: \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)
\(=\left(3\sqrt{7}-2\sqrt{14}\right)\cdot\sqrt{7}+14\sqrt{2}\)
\(=21-14\sqrt{2}+14\sqrt{2}=21\)
c:
\(\dfrac{\sqrt{27}-\sqrt{108}+\sqrt{12}}{\sqrt{3}}\)
\(=\dfrac{3\sqrt{3}-6\sqrt{3}+2\sqrt{3}}{\sqrt{3}}=3+2-6=-1\)
Đúng 1
Bình luận (0)
rút gọnAleft(sqrt{28}-2sqrt{14}+sqrt{7}right)cdotsqrt{7}+7sqrt{8}Bsqrt{6+2sqrt{5}}-sqrt{6-2sqrt{5}}Cleft(sqrt{7}-sqrt{10}right)^2+sqrt{280}Ddfrac{sqrt{99}}{sqrt{11}}+sqrt{7}cdotsqrt{63}-sqrt{sqrt{81}}Esqrt{27}left(s-sqrt{5}right)^2cdotleft(3sqrt{48}right)giải chi tiết ra giúp mik nha,cảm ơn nhiều
Đọc tiếp
rút gọn
A=\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)
B=\(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
C=\(\left(\sqrt{7}-\sqrt{10}\right)^2+\sqrt{280}\)
D=\(\dfrac{\sqrt{99}}{\sqrt{11}}+\sqrt{7}\cdot\sqrt{63}-\sqrt{\sqrt{81}}\)
E=\(\sqrt{27}\left(s-\sqrt{5}\right)^2\cdot\left(3\sqrt{48}\right)\)
giải chi tiết ra giúp mik nha,cảm ơn nhiều
\(\dfrac{5}{4-\sqrt{11}}+\dfrac{1}{3+\sqrt{7}}-\dfrac{6}{\sqrt{7}-2}-\dfrac{\sqrt{7}-5}{2}=4+\sqrt{11}-3\sqrt{7}\)
\(\dfrac{\sqrt{x}+\sqrt{y}}{2\left(\sqrt{x}-\sqrt{y}\right)}-\dfrac{\sqrt{x}-\sqrt{y}}{2\left(\sqrt{x}+\sqrt{y}\right)}-\dfrac{y+x}{y-x}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(\dfrac{5\left(4+\sqrt{11}\right)}{\left(4+\sqrt{11}\right)\left(4-\sqrt{11}\right)}+\dfrac{3-\sqrt{7}}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}-\dfrac{6\left(\sqrt{7}+2\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}-\dfrac{\sqrt{7}-5}{2}\)\(=\dfrac{\left(4+\sqrt{11}\right)5}{16-11}+\dfrac{3-\sqrt{7}}{9-7}-\dfrac{6\left(\sqrt{7}+2\right)}{7-4}-\dfrac{\sqrt{7}-5}{2}\)
\(=4+\sqrt{11}-\dfrac{3-\sqrt{7}}{2}-2\left(\sqrt{7}+2\right)-\dfrac{\sqrt{7}-5}{2}=\dfrac{8+2\sqrt{11}-3+\sqrt{7}-4\sqrt{7}-8-\sqrt{7}+5}{2}=\dfrac{2\sqrt{11}-4\sqrt{7}+2}{2}=1+\sqrt{11}-2\sqrt{7}\)
Đúng 0
Bình luận (1)
Thực hiên phép tính
a, \(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)
b, \(2\sqrt{3}\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)\)
c, \(\left(1+\sqrt{3-\sqrt{2}}\right)\left(1+\sqrt{3}+\sqrt{2}\right)\)
d,\(\left(\sqrt{\sqrt{11}+\sqrt{7}}\right)-\left(\sqrt{\sqrt{11}-\sqrt{7}}\right)^2\)
giải giùm nha
a) \(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)
\(=\left(2+6+15-36\right)\sqrt{3}=-13\sqrt{3}\)
b) \(2\sqrt{3}\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)=6\left(3+8-5\right)=36\)
Đúng 0
Bình luận (0)
a)\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)
\(=\sqrt{4\cdot3}+2\sqrt{9\cdot3}+3\sqrt{25\cdot3}-9\sqrt{16\cdot3}\)
\(=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)
\(=-13\sqrt{3}\)
b)\(2\sqrt{3}\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)\)
\(=2\sqrt{3}\left(\sqrt{9\cdot3}+2\sqrt{16\cdot3}-\sqrt{25\cdot3}\right)\)
\(=2\sqrt{3}\left(3\sqrt{3}+8\sqrt{3}-5\sqrt{3}\right)\)
\(2\sqrt{3}\cdot6\sqrt{3}=12\cdot3=36\)
Đúng 0
Bình luận (1)
