\(\left(2-\sqrt{2}\right)\)x\(\left(-5\sqrt{2}\right)\)-\(\left(3\sqrt{2}-5\right)^2\)
Những câu hỏi liên quan
tínhsqrt{left(2-sqrt{5}right)^2}+sqrt{left(2sqrt{2}-sqrt{5}right)^2}sqrt{left(sqrt{7}-2sqrt{2}right)^2}+sqrt{left(3-2sqrt{2}right)^2}sqrt{left(x-3right)^2}left(x3right)sqrt{left(1-xright)^2}left(x1right)sqrt{9a^4}sqrt{100a^2}left(a 0right)
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tính
\(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
\(\sqrt{\left(\sqrt{7}-2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)^2}\)
\(\sqrt{\left(x-3\right)^2}\left(x>3\right)\)
\(\sqrt{\left(1-x\right)^2}\left(x>1\right)\)
\(\sqrt{9a^4}\)
\(\sqrt{100a^2}\left(a< 0\right)\)
\(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\\ =\left|2-\sqrt{5}\right|+\left|2\sqrt{2}-\sqrt{5}\right|\\ =\sqrt{5}-2+2\sqrt{2}-\sqrt{5}\\ =-2+\sqrt{2}\)
\(\sqrt{\left(\sqrt{7}-2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)}\\ =\left|\sqrt{7}-2\sqrt{2}\right|+\left|3-2\sqrt{2}\right|\\ =2\sqrt{2}-\sqrt{7}+3-2\sqrt{2}\\ =3-\sqrt{7}\)
\(\sqrt{\left(x-3\right)^2}\\ =\left|x-3\right|\\ =x-3\left(vì.x>3\right)\)
\(\sqrt{\left(1-x\right)^2}\\ =\left|1-x\right|\\ =x-1\left(vì.x>1\right)\)
\(\sqrt{9a^4}=\sqrt{\left(3a^2\right)^2}\\ =\left|3a^2\right|\\ =3a^2\)
\(\sqrt{100a^2}\\ =\sqrt{\left(10a\right)^2}\\ =\left|10a\right|\\ =-10a\left(vì.a< 0\right)\)
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Lời giải:
a. $=|2-\sqrt{5}|+|2\sqrt{2}-\sqrt{5}|$
$=(\sqrt{5}-2)+(2\sqrt{2}-\sqrt{5})=-2+2\sqrt{2}$
b. $=|\sqrt{7}-2\sqrt{2}|+|3-2\sqrt{2}|=2\sqrt{2}-\sqrt{7}+(3-2\sqrt{2})$
$=3-\sqrt{7}$
c.
$=|x-3|=x-3$
d.
$=|1-x|=x-1$
$=\sqrt{(3a^2)^2}=|3a^2|=3a^2$
e.
$=\sqrt{(10a)^2}=|10a|=-10a$
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a. \(\sqrt{x}\left(\sqrt{x}-3\right)-5\left(\sqrt{x}+3\right)\)
b. \(3\left(2+\sqrt{x}\right)+\left(\sqrt{x}+3\right)\left(2-\sqrt{x}\right)\)
c. \(\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-5\left(\sqrt{x}-1\right)\)
d. \(3\left(\sqrt{x}-2\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
a: Ta có: \(\sqrt{x}\left(\sqrt{x}-3\right)-5\left(\sqrt{x}+3\right)\)
\(=x-3\sqrt{x}-5\sqrt{x}-15\)
\(=x-8\sqrt{x}-15\)
b: Ta có: \(3\left(\sqrt{x}+2\right)+\left(\sqrt{x}+3\right)\left(2-\sqrt{x}\right)\)
\(=3\sqrt{x}+6+2\sqrt{x}-x+6-3\sqrt{x}\)
\(=-x+2\sqrt{x}+12\)
c: Ta có: \(\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-5\left(\sqrt{x}-1\right)\)
\(=x-9-5\sqrt{x}+5\)
\(=x-5\sqrt{x}-4\)
d: Ta có: \(3\left(\sqrt{x}-2\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(=3\sqrt{x}-6-x+1\)
\(=-x+3\sqrt{x}-5\)
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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...
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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)\(\left(3\sqrt{2}+1\right)\left(2\sqrt{3}+1\right)\left(6\sqrt{6}+1\right)\left(215-34\sqrt{3}-33\sqrt{2}\right)\)
Mình rút gọn như thế này đúng không nhỉ?Pleft(2-frac{sqrt{x}-1}{2sqrt{x}-3}right):left(frac{6sqrt{x}+1}{2x-sqrt{x}-3}+frac{sqrt{x}}{sqrt{x}+1}right)Pleft[frac{2left(2sqrt{x}-3right)}{2sqrt{x}-3}-frac{sqrt{x}-1}{2sqrt{x}-3}right]:left[frac{6sqrt{x}+1}{left(sqrt{x}+1right)left(2sqrt{x}-3right)}+frac{sqrt{x}left(2sqrt{x}-3right)}{left(sqrt{x}+1right)left(2sqrt{x}-3right)}right]Pleft(frac{4sqrt{x}-6}{2sqrt{x}-3}-frac{sqrt{x}-1}{2sqrt{x}-3}right):left(frac{6sqrt{x}+1}{left(sqrt{x}+1right)left(2sqrt{x...
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Mình rút gọn như thế này đúng không nhỉ?
\(P=\left(2-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1}{2x-\sqrt{x}-3}+\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(P=\left[\frac{2\left(2\sqrt{x}-3\right)}{2\sqrt{x}-3}-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right]:\left[\frac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(2\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right]\)
\(P=\left(\frac{4\sqrt{x}-6}{2\sqrt{x}-3}-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\frac{2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)
\(P=\left(\frac{4\sqrt{x}-6-\sqrt{x}+1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1+2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)
\(P=\frac{3\sqrt{x}-5}{2\sqrt{x}-3}:\frac{2x+3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\)
\(P=\frac{3\sqrt{x}-5}{2\sqrt{x}-3}.\frac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}{2x+3\sqrt{x}+1}\)
\(P=\left(3\sqrt{x}-5\right).\frac{\left(\sqrt{x}+1\right)}{2x+3\sqrt{x}+1}\)
\(P=\frac{3x+3\sqrt{x}-5\sqrt{x}-5}{2x+3\sqrt{x}+1}\)
\(P=\frac{3x-5\sqrt{x}-5}{2x+1}\)
từ dòng cuối là sai rồi bạn à
Bạn bỏ dòng cuối đi còn lại đúng rồi
Ở tử đặt nhân tử chung căn x chung rồi lại đặt căn x +1 chung
Ở mẫu tách 3 căn x ra 2 căn x +căn x rồi đặt nhân tử 2 căn x ra
rút gọn được \(\frac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
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Tính:
1.sqrt{left(sqrt{3}+sqrt{2}right)^2} 4.sqrt{left(sqrt{3}right)^2+2.left(sqrt{3}right).left(1right)+left(1right)^2}
2.sqrt{left(sqrt{5}-sqrt{6}right)^2} 5.sqrt{left(sqrt{5}right)^2+2.left(sqrt{5}right).left(sqrt{3}right)+left(sqrt{3}right)^2}
3.sqrt{left(2sqrt{2}+sqrt{3}right)^2}...
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Tính:
1.\(\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\) 4.\(\sqrt{\left(\sqrt{3}\right)^2+2.\left(\sqrt{3}\right).\left(1\right)+\left(1\right)^2}\)
2.\(\sqrt{\left(\sqrt{5}-\sqrt{6}\right)^2}\) 5.\(\sqrt{\left(\sqrt{5}\right)^2+2.\left(\sqrt{5}\right).\left(\sqrt{3}\right)+\left(\sqrt{3}\right)^2}\)
3.\(\sqrt{\left(2\sqrt{2}+\sqrt{3}\right)^2}\) 6.\(\sqrt{\left(\sqrt{6}\right)^2-2.\left(\sqrt{6}\right).\left(\sqrt{5}\right)+\left(\sqrt{5}\right)^2}\)
Tính
A = \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
B = \(\sqrt{\left(4-\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
C = \(\sqrt{\left(1-\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(A=\left|2-\sqrt{3}\right|+\left|1+\sqrt{3}\right|=2-\sqrt{3}+1+\sqrt{3}=3\)
\(B=\left|4-\sqrt{5}\right|-\left|2-\sqrt{5}\right|=4-\sqrt{5}-\sqrt{5}+2=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\)
\(C=\left|1-\sqrt{5}\right|-\left|2-\sqrt{5}\right|=\sqrt{5}-1-\sqrt{5}+2=1\)
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\(A=\left|2-\sqrt{3}\right|+\left|1+\sqrt{3}\right|=2-\sqrt{3}+1+\sqrt{3}=3\)
\(B=\left|4-\sqrt{5}\right|-\left|2-\sqrt{5}\right|=4-\sqrt{5}-\sqrt{5}+2=6-2\sqrt{5}\)
C=\(\left|1-\sqrt{5}\right|-\left|2-\sqrt{5}\right|=\sqrt{5}-1-\sqrt{5}+2=1\)
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Rút gọn biểu thức:
1) sqrt{left(1-sqrt{2}right)^2}+sqrt{left(sqrt{2}+3right)^2}
2) sqrt{left(sqrt{3}-2right)^2}+sqrt{left(sqrt{3}-1right)^2}
3) left(sqrt{19}-3right)left(sqrt{19}+3right)
4) 4x+sqrt{left(x-12right)^2}left(xge2right)
5) frac{sqrt{7}+sqrt{5}}{sqrt{7}-sqrt{5}}+frac{sqrt{7}-sqrt{5}}{sqrt{7}+sqrt{5}}
6) x+2y-sqrt{left(x^2-4xy+4y^2right)^2}left(xge2yright)
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Rút gọn biểu thức:
1) \(\sqrt{\left(1-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}+3\right)^2}\)
2) \(\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\)
3) \(\left(\sqrt{19}-3\right)\left(\sqrt{19}+3\right)\)
4) \(4x+\sqrt{\left(x-12\right)^2}\left(x\ge2\right)\)
5) \(\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\)
6) \(x+2y-\sqrt{\left(x^2-4xy+4y^2\right)^2}\left(x\ge2y\right)\)
Gấp lắm . Giúp mình cảm ơn ạBài 1 2sqrt{left(1+sqrt{3}right)^{ }3}-sqrt{left(2sqrt{3}-3right)^2}left(1+sqrt{3}-sqrt{5}right).left(1+sqrt{3}+sqrt{5}right)left(sqrt[]{dfrac{8}{3}}-sqrt{5}right)xsqrt{6}left(5+4sqrt{2}right).left(3+2sqrt{1}+sqrt{2}right).left(3-2sqrt{1}+2right)sqrt{3+2sqrt{2}}-sqrt{3-2sqrt{2}}
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Gấp lắm . Giúp mình cảm ơn ạ
Bài 1
\(2\sqrt{\left(1+\sqrt{3}\right)^{ }3}-\sqrt{\left(2\sqrt{3}-3\right)^2}\)
\(\left(1+\sqrt{3}-\sqrt{5}\right).\left(1+\sqrt{3}+\sqrt{5}\right)\)
\(\left(\sqrt[]{\dfrac{8}{3}}-\sqrt{5}\right)x\sqrt{6}\)
\(\left(5+4\sqrt{2}\right).\left(3+2\sqrt{1}+\sqrt{2}\right).\left(3-2\sqrt{1}+2\right)\)
\(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)
e) Ta có: \(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{2}+1-\sqrt{2}+1\)
=2
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27 tháng 7 2021 lúc 16:36
Bài 2: a) sqrt{left(3-2sqrt{2}right)^2}+sqrt{left(3+2sqrt{2}right)^2} b) sqrt{left(5-2sqrt{6}right)^2}-sqrt{left(5+2sqrt{6}right)^2}c) sqrt{left(sqrt{5}-sqrt{2}right)^2}+sqrt{left(sqrt{5}+sqrt{2}right)^2} d) sqrt{left(sqrt{2}+1right)^2}-sqrt{left(sqrt{2}-5right)^2}e) sqrt{left(2-sqrt{3}right)^2}+sqrt{left(1-sqrt{3}right)^2} f) sqrt{left(3+sqrt{2}right)^2}-sqrt{left(1-sqrt{2}right)^2}Hộ mk vs ạ
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Bài 2:
a) \(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}\) b) \(\sqrt{\left(5-2\sqrt{6}\right)^2}-\sqrt{\left(5+2\sqrt{6}\right)^2}\)
c) \(\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\) d) \(\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}-5\right)^2}\)
e) \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1-\sqrt{3}\right)^2}\) f) \(\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(1-\sqrt{2}\right)^2}\)
Hộ mk vs ạ
a)\(\sqrt{\left(3-2\sqrt{2}\right)^2}\) + \(\sqrt{\left(3+2\sqrt{2}\right)^2}\) = \(\left(3-2\sqrt{2}\right)+\left(3+2\sqrt{2}\right)\) =\(3-2\sqrt{2}+3+2\sqrt{2}\) =\(6\)
b)\(\sqrt{\left(5-2\sqrt{6}\right)^2}-\sqrt{\left(5+2\sqrt{6}\right)^2}=\left(5-2\sqrt{6}\right)-\left(5+2\sqrt{6}\right)=5-2\sqrt{6}-5-2\sqrt{6}\)\(=-4\sqrt{6}\)
c)\(\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}=\sqrt{5}-\sqrt{2}+\sqrt{5}+\sqrt{2}=2\sqrt{5}\)
d)\(\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}-5\right)^2}=\sqrt{2}+1-5+\sqrt{2}=2\sqrt{2}-4\)
e)\(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1-\sqrt{3}\right)^2}=2-\sqrt{3}+\sqrt{3}-1=1\)
f)\(\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(1-\sqrt{2}\right)^2}=3+\sqrt{2}-\sqrt{2}+1=4\)
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a) \(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}\)
\(=3-2\sqrt{2}+3+2\sqrt{2}\)
=6
c) Ta có: \(\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\)
\(=\sqrt{5}-\sqrt{2}+\sqrt{5}+\sqrt{2}\)
\(=2\sqrt{5}\)
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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
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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) \(\left(2\sqrt{2}+\sqrt{3}\right)^2\)
\(1,=20-7=13\\ b,=12-50=-38\\ c,=\sqrt{7}-2+\sqrt{7}+2=2\sqrt{7}\\ d,=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\\ e,=11+2\sqrt{30}\\ f,=8-2\sqrt{15}\\ g,=11+2\sqrt{6}\)
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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) \(=\left(2\sqrt{2}\right)^2+\left(\sqrt{3}\right)^2+2\sqrt{2\sqrt{2}.3}=11+2\sqrt{6\sqrt{2}}\)
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