Rút Gọn
A= \(\frac{4^{10}+8^4}{4^5+8^6}\)
K
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NP
3
27 tháng 6 2021
`a)sqrt{8-2sqrt7}+sqrt{16-6sqrt7}`
`=sqrt{(sqrt7-1)^2}+sqrt{(3-sqrt7)^2}`
`=sqrt7-1+3-sqrt7=2`
`b)sqrt{(sqrt7-1)^2}-sqrt{11+4sqrt7}`
`=sqrt7-1-sqrt{(2+sqrt7)^2}`
`=sqrt7-1-2-sqrt7=-3`
27 tháng 6 2021
a, \(=\sqrt{7-2\sqrt{7}+1}+\sqrt{7-2.3\sqrt{7}+9}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}+\sqrt{\left(3-\sqrt{7}\right)^2}=\left|\sqrt{7}-1\right|+\left|3-\sqrt{7}\right|\)
\(=\sqrt{7}-1+3-\sqrt{7}=2\)
\(b,=\left|\sqrt{7}-1\right|-\sqrt{7+2.2\sqrt{7}+4}\)
\(=\left|\sqrt{7}-1\right|-\sqrt{\left(\sqrt{7}+2\right)^2}=\left|\sqrt{7}-1\right|-\left|\sqrt{7}+2\right|\)
\(=\sqrt{7}-1-\sqrt{7}-2=-3\)
NA
9 tháng 2 2022
1:
a: Vì \(\dfrac{-4}{3}=\dfrac{-4\cdot3}{3\cdot3}=\dfrac{-12}{9}=\dfrac{12}{9}\\ \Rightarrow\dfrac{-4}{3}=\dfrac{12}{9}\)
b: Vì : \(-2\cdot3=-6\\ -6\cdot8=-48\)
nên 2 p/s ko bằng nhau
HC
2
Y
20 tháng 6 2023
\(1,\)
\(a,\sqrt{6-2\sqrt{5}}=\sqrt{\sqrt{5^2}-2.\sqrt{5}.1+1}=\sqrt{\left(\sqrt{5}-1\right)^2}=\left|\sqrt{5}-1\right|=\sqrt{5}-1\)
\(b,\sqrt{8+2\sqrt{7}}=\sqrt{\sqrt{7^2}+2.\sqrt{7}.1+1}=\sqrt{\left(\sqrt{7}+1\right)^2}=\left|\sqrt{7}+1\right|=\sqrt{7}+1\)
\(2,\)
\(a,\sqrt{\left(\sqrt{10}-3\right)^2}-\sqrt{10}\)
\(=\left|\sqrt{10}-3\right|-\sqrt{10}\)
\(=\sqrt{10}-\sqrt{10}-3\)
\(=-3\)
\(b,\sqrt{\left(5+\sqrt{7}\right)^2}-\sqrt{8-2\sqrt{7}}\)
\(=\left|5+\sqrt{7}\right|-\sqrt{\left(\sqrt{7}-1\right)^2}\)
\(=5+\sqrt{7}-\left|\sqrt{7}-1\right|\)
\(=5+\sqrt{7}-\sqrt{7}+1\)
\(=6\)
DT
0
10 tháng 8 2016
\(A=\frac{4^6.9^5+69.120}{8^4.3^{12}+6^{11}}=\frac{2^{12}.3^{10}+2^3.3^2.115}{2^{12}.3^{12}+\left(2.3\right)^{11}}=\frac{3^2.2^3\left(115+2^9.3^8\right)}{6^{11}\left(6+1\right)}=\frac{115+2^9.3^8}{6^8.3.7}\)
\(B=\frac{10^4+5.10^3+5^4}{25}=\frac{\left(10^2\right)^2+2.5^2.10^2+\left(5^2\right)^2}{25}=\frac{\left(10^2+5^2\right)^2}{25}=\frac{125^2}{25}=\frac{25.625}{25}=625\)
\(C=\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{4^{10}.2^{10}+4^{10}}{4^4.4^7+4^4.2^4}=\frac{4^{10}\left(2^{10}+1\right)}{4^4.2^4\left(2^{10}+1\right)}=\frac{4^6}{2^4}=256\)
VD
4
S
1 tháng 7 2018
\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=\frac{-2}{6}=-\frac{1}{3}\)
1 tháng 7 2018
Ta có:\
\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
\(A=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}\)
\(A=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(A=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}\)
\(A=-\frac{2}{6}=-\frac{1}{3}\)
MT
1
TH
27 tháng 9 2018
\(M=\frac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(M=\frac{\left(2^2\right)^5\cdot\left(3^2\right)^4-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\)
\(M=\frac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)
\(M=\frac{2^{10}\cdot3^8\cdot\left(1-3\right)}{2^{10}\cdot3^8\cdot\left(1+5\right)}\)
\(M=\frac{-2}{6}=\frac{-1}{3}\)
Y
25 tháng 6 2023
\(a,\dfrac{\sqrt{10}-\sqrt{15}}{\sqrt{8}-\sqrt{12}}\\ =\dfrac{\sqrt{2}.\sqrt{5}-\sqrt{3}.\sqrt{5}}{\sqrt{2}.\sqrt{4}-\sqrt{3}.\sqrt{4}}\\ =\dfrac{\sqrt{5}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{4}\left(\sqrt{2}-\sqrt{3}\right)}\\ =\dfrac{\sqrt{5}}{\sqrt{2^2}}\\ =\dfrac{\sqrt{5}}{2}\)
\(b,\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}\\ =\dfrac{\sqrt{5}.\sqrt{3}-\sqrt{3}.\sqrt{2}}{\sqrt{5}.\sqrt{7}-\sqrt{7}.\sqrt{2}}\\ =\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}\\ =\dfrac{\sqrt{3}}{\sqrt{7}}\)
\(c,\dfrac{5+\sqrt{5}}{\sqrt{10}+\sqrt{2}}\\ =\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}=\dfrac{\sqrt{5}}{\sqrt{2}}\)
H
25 tháng 6 2023
\(a,=\dfrac{\sqrt{2}.\sqrt{5}-\sqrt{3}.\sqrt{5}}{2\sqrt{2}-2\sqrt{3}}\\ =\dfrac{\sqrt{5}.\left(\sqrt{2}-\sqrt{3}\right)}{2\left(\sqrt{2}-\sqrt{3}\right)}\\ =\dfrac{\sqrt{5}}{2}\)
\(b,=\dfrac{\sqrt{3}.\sqrt{5}-\sqrt{2}.\sqrt{3}}{\sqrt{5}.\sqrt{7}-\sqrt{2}.\sqrt{7}}\\ =\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}\\ =\dfrac{\sqrt{3}}{\sqrt{7}}=\dfrac{\sqrt{21}}{7}\)
\(c,=\dfrac{\sqrt{5}.\sqrt{5}+\sqrt{5}}{\sqrt{2}.\sqrt{5}+\sqrt{2}}\\ =\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}\\ =\dfrac{\sqrt{5}}{\sqrt{2}}=\dfrac{\sqrt{10}}{2}\)
UA
1
cho mình sửa lại kết quả là A = 4
8^10 + 4^10 / 8^4 + 4^11
= (2^3)^10 + (2^2)^10 / (2^3)^4 + (2^2)^11
=2^30 + 2^20 / 2^12 +2^22
= 2^20(2^10 +1) / 2^12( 2^10 +1)
=2^20/210
= 2^8 = 256