\(\sqrt{\left(3-\sqrt{7}\right)^{ }2}\)
Những câu hỏi liên quan
Tính:1) ( 2sqrt{5}-sqrt{7} ) left(2sqrt{5}+sqrt{7}right)2) left(5sqrt{2}+2sqrt{3}right)left(2sqrt{3}-5sqrt{2}right)3) sqrt{left(sqrt{7}-2right)^2}+sqrt{left(sqrt{7}+2right)^2}4) sqrt{left(sqrt{3}+sqrt{2}right)^2}+sqrt{left(sqrt{3}-sqrt{2}right)^2}5) left(sqrt{5}-sqrt{6}right)^26) left(sqrt{3}-sqrt{5}right)^27) left(2sqrt{2}+sqrt{3}right)^2
Đọc tiếp
Tính:
1) ( \(2\sqrt{5}-\sqrt{7}\) ) \(\left(2\sqrt{5}+\sqrt{7}\right)\)
2) \(\left(5\sqrt{2}+2\sqrt{3}\right)\left(2\sqrt{3}-5\sqrt{2}\right)\)
3) \(\sqrt{\left(\sqrt{7}-2\right)^2}+\sqrt{\left(\sqrt{7}+2\right)^2}\)
4) \(\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
5) \(\left(\sqrt{5}-\sqrt{6}\right)^2\)
6) \(\left(\sqrt{3}-\sqrt{5}\right)^2\)
7) \(\left(2\sqrt{2}+\sqrt{3}\right)^2\)
\(1,=20-7=13\\ b,=12-50=-38\\ c,=\sqrt{7}-2+\sqrt{7}+2=2\sqrt{7}\\ d,=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\\ e,=11+2\sqrt{30}\\ f,=8-2\sqrt{15}\\ g,=11+2\sqrt{6}\)
Đúng 2
Bình luận (0)
1) \(=\left(2\sqrt{5}\right)^2-\left(\sqrt{7}\right)^2=20-7=13\)
2) \(=\left(2\sqrt{3}\right)^2-\left(5\sqrt{2}\right)^2=12-50=-38\)
3) \(=\sqrt{7}-2+\sqrt{7}+2=2\sqrt[]{7}\)
4) \(=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\)
5) \(=5+6-2\sqrt{5.6}=11-2\sqrt{30}\)
6) \(=3+5-2\sqrt{3.5}=8-4\sqrt{2}\)
7) \(=\left(2\sqrt{2}\right)^2+\left(\sqrt{3}\right)^2+2\sqrt{2\sqrt{2}.3}=11+2\sqrt{6\sqrt{2}}\)
Đúng 0
Bình luận (0)
Tính giá trị các biểu thức
A = \(\sqrt{\left(5-\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
B = \(\sqrt{\left(3-\sqrt{2}\right)^2}-\sqrt{\left(1-\sqrt{2}\right)^2}\)
C = \(\sqrt{\left(3+\sqrt{7}\right)^2}-\sqrt{\left(2-\sqrt{7}\right)^2}\)
D = \(\sqrt{4-2\sqrt{3}}+\sqrt{7+4\sqrt{3}}\)
`A=sqrt{(5-sqrt3)^2}+sqrt{(2-sqrt3)^2}`
`=5-sqrt3+2-sqrt3`
`=7-2sqrt3`
`B=sqrt{(3-sqrt2)^2}-sqrt{(1-sqrt2)^2}`
`=3-sqrt2-(sqrt2-1)`
`=4-2sqrt2`
`C=sqrt{(3+sqrt7)^2}-sqrt{(2-sqrt7)^2}`
`=3+sqrt7-(sqrt7-2)`
`=5`
`D=sqrt{4-2sqrt3}+sqrt{7+4sqrt3}`
`=sqrt{3-2sqrt3+1}+sqrt{4+2.2.sqrt3+3}`
`=sqrt{(sqrt3-1)^2}+sqrt{(2+sqrt3)^2}`
`=sqrt3-1+2+sqrt3=1+2sqrt3`
Đúng 2
Bình luận (0)
\(A=\left|5-\sqrt{3}\right|+\left|2-\sqrt{3}\right|=5-\sqrt{3}+2-\sqrt{3}=7-2\sqrt{3}\)
\(B=\left|3-\sqrt{2}\right|-\left|1-\sqrt{2}\right|=3-\sqrt{2}-\sqrt{2}+1=4-2\sqrt{2}\)
\(C=\left|3+\sqrt{7}\right|-\left|2-\sqrt{7}\right|=3+\sqrt{7}-\sqrt{7}+2=5\)
\(D=\sqrt{3-2\sqrt{3}+1}+\sqrt{4+2.2\sqrt{3}+3}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}=\left|\sqrt{3}-1\right|+\left|2+\sqrt{3}\right|\)
\(=\sqrt{3}-1+2+\sqrt{3}=1+2\sqrt{3}\)
Đúng 2
Bình luận (0)
F = \(\sqrt{\left(3-\sqrt{2}\right)^2}+\sqrt{\left(1-\sqrt{2}\right)^2}\)
G = \(\sqrt{\left(5+\sqrt{7}\right)^2}-\sqrt{\left(2-\sqrt{7}\right)^2}\)
H = \(\sqrt{\left(3-\sqrt{10}\right)^2}+\sqrt{\left(2-\sqrt{10}\right)^2}\)
`F=sqrt{(3-sqrt2)^2}+sqrt{(1-sqrt2)^2}``
`=3-sqrt2+sqrt2-1=2`
`G=sqrt{(5+sqrt7)^2}-sqrt{(2-sqrt7)^2}`
`=5+sqrt7-(sqrt7-2)`
`=5+sqrt7-sqrt7+2=2`
`H=sqrt{(3-sqrt{10})^2}+sqrt{(2-sqrt{10})^2}`
`=sqrt{10}-3+sqrt{10}-2`
`=2\sqrt{10}-5`
Đúng 2
Bình luận (0)
\(F=\left|3-\sqrt{2}\right|+\left|1-\sqrt{2}\right|=3-\sqrt{3}+\sqrt{2}-1=2\)
\(G=\left|5+\sqrt{7}\right|-\left|2-\sqrt{7}\right|=5+\sqrt{7}-\sqrt{7}+2=7\)
\(H=\left|3-\sqrt{10}\right|+\left|2-\sqrt{10}\right|=\sqrt{10}-3+\sqrt{10}-2=2\sqrt{10}-5\)
Đúng 1
Bình luận (0)
rút gọna) left(-7sqrt{7}right)left(-2sqrt{8}right)b) -sqrt{33}.3sqrt{3}c) left(3sqrt{5}right).left(-10sqrt{3}right)d) dfrac{1}{2}sqrt{5}.left(-6sqrt{2}right)e) dfrac{2}{3}sqrt{7}.left(-dfrac{9}{16}sqrt{3}right)f) 15sqrt{6}:5sqrt{3}g) -25sqrt{12}:left(-5sqrt{6}right)h) 36sqrt{8}:12sqrt{2}i) 4sqrt{27}:left(-2sqrt{3}right)
Đọc tiếp
rút gọn
a) \(\left(-7\sqrt{7}\right)\left(-2\sqrt{8}\right)\)
b) \(-\sqrt{33}.3\sqrt{3}\)
c) \(\left(3\sqrt{5}\right).\left(-10\sqrt{3}\right)\)
d) \(\dfrac{1}{2}\sqrt{5}.\left(-6\sqrt{2}\right)\)
e) \(\dfrac{2}{3}\sqrt{7}.\left(-\dfrac{9}{16}\sqrt{3}\right)\)
f) \(15\sqrt{6}:5\sqrt{3}\)
g) \(-25\sqrt{12}:\left(-5\sqrt{6}\right)\)
h) \(36\sqrt{8}:12\sqrt{2}\)
i) \(4\sqrt{27}:\left(-2\sqrt{3}\right)\)
i: =-12*căn 3/2căn 3=-6
h: =72căn 2/12căn 2=6
g: =25căn 12/5căn 6=5căn 2
f: =(15:5)*căn 6:3=3căn 2
d: =-1/2*6*căn 10=-3căn 10
Đúng 0
Bình luận (0)
a : \(\sqrt{5+2\sqrt{6}}-\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}\)
b : \(\dfrac{\sqrt{7}-2\sqrt{7}}{2-\sqrt{7}}+\dfrac{6}{\sqrt{7}+1}+\left(3\sqrt{2}-2\sqrt{3}\right)\left(3\sqrt{2}+2\sqrt{3}\right)\)
a: \(\sqrt{5+2\sqrt{6}}-\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}\)
\(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\left|\sqrt{2}-\sqrt{3}\right|\)
\(=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)
b: Sửa đề: \(\dfrac{7-2\sqrt{7}}{2-\sqrt{7}}+\dfrac{6}{\sqrt{7}+1}+\left(3\sqrt{2}-2\sqrt{3}\right)\left(3\sqrt{2}+2\sqrt{3}\right)\)
\(=\dfrac{\sqrt{7}\left(\sqrt{7}-2\right)}{-\left(\sqrt{7}-2\right)}+\dfrac{6\left(\sqrt{7}-1\right)}{6}+18-12\)
\(=-\sqrt{7}+\sqrt{7}-1+6=5\)
Đúng 2
Bình luận (0)
Rút gọn :
dfrac{sqrt{x+sqrt{4left(x-1right)}}-sqrt{x-sqrt{4left(x-1right)}}}{sqrt{x^2-4left(x-1right)}}.left(sqrt{x-1}-dfrac{1}{sqrt{x-1}}right)
b)left(sqrt{2}+1right)left(sqrt{3}+1right)left(sqrt{6}+1right)left(5-2sqrt{2}-sqrt{3}right)
c)left(sqrt{5}+1right)left(sqrt{7}+1right)left(sqrt{35}+1right)left(34-4sqrt{7}-6sqrt{5}right)
d) left(sqrt{7}+1right)left(2sqrt{2}-1right)left(2sqrt{14}-1right)left(55+12sqrt{2}-7sqrt{7}right)
e)left(3sqrt{2}+1right)left(2sqrt{3}+1right)left(6sqrt{6}+1...
Đọc tiếp
Rút gọn :
\(\dfrac{\sqrt{x+\sqrt{4\left(x-1\right)}}-\sqrt{x-\sqrt{4\left(x-1\right)}}}{\sqrt{x^2-4\left(x-1\right)}}.\left(\sqrt{x-1}-\dfrac{1}{\sqrt{x-1}}\right)\)
b)\(\left(\sqrt{2}+1\right)\left(\sqrt{3}+1\right)\left(\sqrt{6}+1\right)\left(5-2\sqrt{2}-\sqrt{3}\right)\)
c)\(\left(\sqrt{5}+1\right)\left(\sqrt{7}+1\right)\left(\sqrt{35}+1\right)\left(34-4\sqrt{7}-6\sqrt{5}\right)\)
d) \(\left(\sqrt{7}+1\right)\left(2\sqrt{2}-1\right)\left(2\sqrt{14}-1\right)\left(55+12\sqrt{2}-7\sqrt{7}\right)\)
e)\(\left(3\sqrt{2}+1\right)\left(2\sqrt{3}+1\right)\left(6\sqrt{6}+1\right)\left(215-34\sqrt{3}-33\sqrt{2}\right)\)
Thực hiện phép tính (rút gọn biểu thức) a)sqrt{left(3+sqrt{2}right)^2}-sqrt{left(3-2sqrt{2}right)^2}b) sqrt{left(sqrt{7}-2sqrt{2}right)^2}-sqrt{left(sqrt{7}+2sqrt{2}right)^2}c)sqrt{left(3+sqrt{5}right)^2}+sqrt{left(3-sqrt{5}right)^2}d) sqrt{left(2-sqrt{3}right)^2}-sqrt{left(2+sqrt{3}right)^2}
Đọc tiếp
Thực hiện phép tính (rút gọn biểu thức)
a)\(\sqrt{\left(3+\sqrt{2}\right)^2}\)-\(\sqrt{\left(3-2\sqrt{2}\right)^2}\)
b) \(\sqrt{\left(\sqrt{7}-2\sqrt{2}\right)^2}\)-\(\sqrt{\left(\sqrt{7}+2\sqrt{2}\right)^2}\)
c)\(\sqrt{\left(3+\sqrt{5}\right)^2}\)+\(\sqrt{\left(3-\sqrt{5}\right)^2}\)
d) \(\sqrt{\left(2-\sqrt{3}\right)^2}\)-\(\sqrt{\left(2+\sqrt{3}\right)^2}\)
Lời giải:
a. $=|3+\sqrt{2}|-|3-2\sqrt{2}|=(3+\sqrt{2})-(3-2\sqrt{2})$
$=3\sqrt{2}$
b. $=|\sqrt{7}-2\sqrt{2}|-|\sqrt{7}+2\sqrt{2}|$
$=(2\sqrt{2}-\sqrt{7})-(\sqrt{7}+2\sqrt{2})$
$=-2\sqrt{7}$
c.
$=|3+\sqrt{5}|+|3-\sqrt{5}|=(3+\sqrt{5})+(3-\sqrt{5})=6$
d.
$=|2-\sqrt{3}|-|2+\sqrt{3}|=(2-\sqrt{3})-(2+\sqrt{3})=-2\sqrt{3}$
Đúng 1
Bình luận (0)
tính
\(\sqrt{5}.\left(\sqrt{3}+\sqrt{2}\right)\)
\(\left(\sqrt{3}-\sqrt{5}\right)^2\)
\(\left(\sqrt{7}-\sqrt{2}\right).\left(\sqrt{7}+\sqrt{2}\right)\)
\(\left(\sqrt{3}+\sqrt{5}\right)^2\)
\(\left(\sqrt{8+3\sqrt{7}}\right)+\left(\sqrt{8-3\sqrt{7}}\right)\)
a : \(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
b : \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
c : \(\sqrt{\left(2\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)
d : \(\sqrt{52-16\sqrt{3}}+\sqrt{\left(4\sqrt{3}-7\right)^2}\)
a.
$A=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}$
$A\sqrt{2}=\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}$
$A\sqrt{2}=\sqrt{(\sqrt{3}-1)^2}+\sqrt{(\sqrt{3}+1)^2}$
$=|\sqrt{3}-1|+|\sqrt{3}+1|=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}$
$\Rightarrow A=2\sqrt{3}: \sqrt{2}=\sqrt{6}$
---------------------
$B=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}$
$B\sqrt{2}=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}$
$B\sqrt{2}=\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}$
$=|\sqrt{7}-1|-|\sqrt{7}+1|=\sqrt{7}-1-(\sqrt{7}+1)=-2$
$\Rightarrow B=-2:\sqrt{2}=-\sqrt{2}$
Đúng 1
Bình luận (0)
\(a,\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
\(A-\sqrt{2}=\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)\cdot\sqrt{2}\\ =\sqrt{2-\sqrt{3}}\cdot\sqrt{2}-\sqrt{2+\sqrt{3}}\cdot\sqrt{2}\\ =\sqrt{\left(2-\sqrt{3}\right)\cdot2}-\sqrt{\left(2+\sqrt{3}\right)\cdot2}\\ =\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}\\ =\sqrt{3-2\sqrt{3}+1}-\sqrt{3+2\sqrt{3}+1}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}\\ =\left|\sqrt{3}-1\right|-\left|\sqrt{3}+1\right|\\ =\sqrt{3}-1-\sqrt{3}-1\\ =-2\)
Ta có :
\(A-\sqrt{2}=-2\\ \Leftrightarrow A=\dfrac{-2}{\sqrt{2}}=\dfrac{-\left(\sqrt{2}\right)^2}{\sqrt{2}}=-\sqrt{2}\)
__
C làm giống câu a, nhé.
__
\(\sqrt{\left(2\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\\ =\left|2\sqrt{5}+1\right|-\left|\sqrt{5}-2\right|\\ =2\sqrt{5}+1-\sqrt{5}+2\\ =3+\sqrt{5}\)
__
\(\sqrt{52-16\sqrt{3}}+\sqrt{\left(4\sqrt{3}-7\right)^2}\\ =\sqrt{48-2\cdot4\cdot\sqrt{3}\cdot2+4}+\left|4\sqrt{3}-7\right|\\ =\sqrt{\left(4\sqrt{3}\right)^2-2\cdot4\cdot\sqrt{3}\cdot2+2^2}+4\sqrt{3}-7\\ =\sqrt{\left(4\sqrt{3}-2\right)^2}+4\sqrt{3}-7\\ =4\sqrt{3}-2+4\sqrt{3}-7\\ =8\sqrt{3}-9\)
Đúng 0
Bình luận (1)
c.
$C=\sqrt{(2\sqrt{5}+1)^2}-\sqrt{(\sqrt{5}-2)^2}$
$=|2\sqrt{5}+1|-|\sqrt{5}-2|=2\sqrt{5}+1-(\sqrt{5}-2)=\sqrt{5}+3$
d.
$D=\sqrt{52-16\sqrt{3}}+\sqrt{4\sqrt{3}-7)^2}$
$=\sqrt{(4\sqrt{3})^2-2.4\sqrt{3}.2+2^2}+|4\sqrt{3}-7|$
$=\sqrt{(4\sqrt{3}-2)^2}+|4\sqrt{3}-7|$
$=|4\sqrt{3}-2|+|4\sqrt{3}-7|$
$=4\sqrt{3}-2+7-4\sqrt{3}=5$
Đúng 0
Bình luận (0)
Rút gọn các biểu thức sau:a) left(sqrt{8}-3sqrt{2}+sqrt{10}right)left(sqrt{2}-3sqrt{0,4}right) b) left(sqrt{28}-2sqrt{14}+sqrt{7}right).sqrt{7}+7sqrt{8}c) 2sqrt{left(sqrt{2}-3right)^2}+sqrt{2left(-3right)^2}-5sqrt{left(-1right)^4} d) left(frac{sqrt{14}-sqrt{7}}{1-sqrt{2}}+frac{sqrt{15}-sqrt{5}}{1-sqrt{3}}right):frac{1}{sqrt{7}-sqrt{5}}
Đọc tiếp
Rút gọn các biểu thức sau:
a) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-3\sqrt{0,4}\right)\) b) \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}\)
c) \(2\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{2\left(-3\right)^2}-5\sqrt{\left(-1\right)^4}\) d) \(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)