rút gọn:
A=(71997-71995) : (71997 - 7)
K
Khách
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KV
28 tháng 7 2023
\(\left(2^{17}+7^2\right).\left(9^{15}-3^{15}\right).\left(2^4-4^2\right)\)
\(=\left(2^{17}+7^2\right).\left(9^{15}-3^{15}\right).\left(16-16\right)\)
\(=\left(2^{17}+7^2\right).\left(9^{15}-3^{15}\right).0\)
\(=0\)
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\(\left(7^{1997}-7^{1995}\right):\left(7^{1994}.7\right)\)
\(=\left[7^{1995}\left(7^2-1\right)\right]:7^{1995}\)
\(=7^{1995}.48:7^{1995}\)
\(=48\)
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\(\left(1^2+2^3+3^4+4^5\right).\left(1^3+2^3+3^3+4^3\right).\left(3^8-81^2\right)\)
\(=\left(1^2+2^3+3^4+4^5\right).\left(1^3+2^3+3^3+4^3\right).\left(6561-6561\right)\)
\(=\left(1^2+2^3+3^4+4^5\right).\left(1^3+2^3+3^3+4^3\right).0\)
\(=0\)
-------------------------------------------------
\(\left(2^8+8^3\right):\left(2^5.2^3\right)\)
\(=\left[2^8+\left(2^3\right)^3\right]:2^8\)
\(=\left(2^8+2^9\right):2^8\)
\(=2^8.\left(1+2\right):2^8\)
\(=2^8.3:2^8\)
\(=3\)
TT
2
21 tháng 10 2021
Bài 2:
Ta có: \(x\left(x-4\right)-x^2+8=0\)
\(\Leftrightarrow x^2-4x-x^2+8=0\)
\(\Leftrightarrow-4x=-8\)
hay x=2
30 tháng 6 2021
`a)sqrt{(sqrt7-4)^2}+sqrt7`
`=|sqrt7-4|+sqrt7`
`=4-sqrt7+sqrt7=4`
`b)\sqrt{81a}-sqrt{144a}+sqrt{36a}(a>=0)`
`=9sqrta-12sqrta+6sqrta=3sqrta`
30 tháng 6 2021
a) Ta có: \(\sqrt{\left(\sqrt{7}-4\right)^2}+\sqrt{7}\)
\(=4-\sqrt{7}+\sqrt{7}\)
=4
b) Ta có: \(\sqrt{81a}-\sqrt{144a}+\sqrt{36a}\)
\(=9\sqrt{a}-12\sqrt{a}+6\sqrt{a}\)
\(=3\sqrt{a}\)
JR
3
H9
HT.Phong (9A5)
CTVHS
12 tháng 8 2023
a) \(A=\dfrac{2^6\cdot9^2}{6^4\cdot8}\)
\(=\dfrac{2^6\cdot\left(3^2\right)^2}{3^4\cdot2^4\cdot2^3}\)
\(=\dfrac{2^6\cdot3^4}{3^4\cdot2^7}\)
\(=\dfrac{1}{2}\)
b) \(B=\dfrac{2^{13}\cdot3^7}{2^{15}\cdot3^2\cdot9^2}\)
\(=\dfrac{2^{13}\cdot3^7}{2^{15}\cdot3^2\cdot\left(3^2\right)^2}\)
\(=\dfrac{2^{13}\cdot3^7}{2^{15}\cdot3^6}\)
\(=\dfrac{3}{2^2}\)
\(=\dfrac{3}{4}\)
10 tháng 9 2021
\(\left(x^2-2\right)\left(-x+3\right)\)
\(=-x^3+3x^2+2x-6\)
NL
1
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 5 2021
\(A=\dfrac{1}{2}sin4x-\dfrac{1}{2}sin2x+\dfrac{1}{2}sin12x+\dfrac{1}{2}sin2x\)
\(=\dfrac{1}{2}sin4x+\dfrac{1}{2}sin12x=sin8x.cos4x\)
\(B=\dfrac{1}{2}cos5x+\dfrac{1}{2}cos3x+\dfrac{1}{2}cosx-\dfrac{1}{2}cos5x\)
\(=\dfrac{1}{2}cos3x+\dfrac{1}{2}cosx=cos2x.cosx\)
12 tháng 1 2016
3A=1.2.3+2.3.(4-1)+3.4.(5-2)+............+2015.2016.(2017-2014)
3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+.............+2015.2016.2017-2014.2015.2016
3A=2015.2016.2017
A=2015.2016.2017:3
A=2015.672.2017
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 9 2021
\(A=\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}=\sqrt{3}\)
H9
HT.Phong (9A5)
CTVHS
4 tháng 7 2023
\(A=3+3^2+3^3+...+3^{2004}\)
\(\Rightarrow3A=3\left(3+3^2+3^3+...+3^{2004}\right)\)
\(\Rightarrow3A=3^2+3^3+3^4+...+3^{2005}\)
\(\Rightarrow3A-A=\left(3^2+3^3+3^4+...+3^{2005}\right)-\left(3+3^2+3^3+3^4+...+3^{2004}\right)\)
\(\Rightarrow2A=\left(3^2-3^2\right)+\left(3^3-3^3\right)+\left(3^4-3^4\right)+...+\left(3^{2004}-3^{2004}\right)+\left(3^{2005}-3\right)\)
\(\Rightarrow2A=3^{2005}-3\)
\(\Rightarrow A=\dfrac{3^{2005}+3}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 7 2021
\(A=\sqrt{3+\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}=\sqrt{3+\sqrt{5-\left(2\sqrt{3}+1\right)}}\)
\(=\sqrt{3+\sqrt{4-2\sqrt{3}}}=\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{3+\sqrt{3}-1}=\sqrt{2+\sqrt{3}}=\dfrac{1}{\sqrt{2}}\sqrt{4+2\sqrt{3}}\)
\(=\dfrac{1}{\sqrt{2}}\sqrt{\left(\sqrt{3}+1\right)^2}=\dfrac{\sqrt{3}+1}{\sqrt{2}}=\dfrac{\sqrt{2}+\sqrt{6}}{2}\)
LN
23 tháng 7 2021
\(A=\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}\\ =\sqrt{3+\sqrt{5-\sqrt{\left(1+2\sqrt{3}\right)^2}}}\\ =\sqrt{3+\sqrt{5-1+2\sqrt{3}}}\\ =\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}\\ =\sqrt{3+\sqrt{3}-1}\\ =\sqrt{2+\sqrt{3}}\)