Rút gọn các biểu thức sau: 27 . 48 . 1 - a 2 v ớ i a > 1
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20 tháng 12 2021
a: \(=3\sqrt{3}-2\sqrt{3}+4\sqrt{3}-5\sqrt{3}=2\sqrt{3}\)
CP
2
H9
HT.Phong (9A5)
CTVHS
20 tháng 7 2023
\(2\sqrt{27}-\sqrt{\dfrac{16}{3}}-\sqrt{48}-\sqrt{8\dfrac{1}{3}}\)
\(=6\sqrt{3}-4\sqrt{\dfrac{1}{3}}-4\sqrt{3}-5\sqrt{\dfrac{1}{3}}\)
\(=2\sqrt{3}-9\sqrt{\dfrac{1}{3}}\)
\(=2\sqrt{3}-3\sqrt{9\cdot\dfrac{1}{3}}\)
\(=2\sqrt{3}-3\sqrt{3}\)
\(=-\sqrt{3}\)
________________________
\(\left(\sqrt{125}-\sqrt{12}-2\sqrt{5}\right)\left(3\sqrt{5}-\sqrt{3}+\sqrt{27}\right)\)
\(=\left(5\sqrt{5}-2\sqrt{3}-2\sqrt{5}\right)\left(3\sqrt{5}-\sqrt{3}+3\sqrt{3}\right)\)
\(=\left(3\sqrt{5}-2\sqrt{3}\right)\left(3\sqrt{5}+2\sqrt{3}\right)\)
\(=\left(3\sqrt{5}\right)^2-\left(2\sqrt{3}\right)^2\)
\(=15-12\)
\(=3\)
XO
16 tháng 12 2023
a) \(A=\sqrt{18}.\sqrt{2}-\sqrt{48}:\sqrt{3}=\sqrt{18.2}-\sqrt{48:3}\)
\(=\sqrt{36}-\sqrt{16}=6-4=2\)
b) \(B=\dfrac{8}{\sqrt{5}-1}+\dfrac{8}{\sqrt{5}+1}=\dfrac{8\sqrt{5}+8+8\sqrt{5}-8}{\left(\sqrt{5}-1\right).\left(\sqrt{5}+1\right)}=\dfrac{16\sqrt{5}}{4}=4\sqrt{5}\)
19 tháng 1 2022
a) \(E=2\sqrt{40\sqrt{12}}+3\sqrt{5\sqrt{48}}-2\sqrt{\sqrt{75}}-4\sqrt{15\sqrt{27}}.\)
\(=8\sqrt{5\sqrt{3}}+6\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}-12\sqrt{5\sqrt{3}}}\)
\(=0\)
19 tháng 1 2022
b) \(F=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}.\)
Vì \(=\frac{5}{12}-\frac{1}{\sqrt{6}}=\frac{5-2\sqrt{6}}{12}=\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}\)
\(\frac{1}{\sqrt{3}}+\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}=\frac{2\sqrt{3}+\sqrt{2}}{6}\)
Nên \(F=\frac{2\sqrt{3}+\sqrt{2}}{6}+\frac{1}{\sqrt{3}}\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}}=\frac{2\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{6}=\frac{3\sqrt{3}}{6}=\frac{\sqrt{3}}{2}\)
H9
HT.Phong (9A5)
CTVHS
14 tháng 7 2023
a) \(5\sqrt{48}-4\sqrt{27}-2\sqrt{57}+\sqrt{108}\)
\(=20\sqrt{3}-12\sqrt{3}-2\sqrt{57}+6\sqrt{3}\)
\(=\left(20-12+6\right)\sqrt{3}-2\sqrt{57}\)
\(=14\sqrt{3}-2\sqrt{57}\)
b) \(2\sqrt{24}-2\sqrt{54}+3\sqrt{6}-\sqrt{150}\)
\(=4\sqrt{6}-6\sqrt{6}+3\sqrt{6}-5\sqrt{6}\)
\(=\left(4-6+3-5\right)\sqrt{6}\)
\(=-4\sqrt{6}\)
F
20 tháng 8 2021
1.
A= \(2\sqrt{6}\) + \(6\sqrt{6}\) - \(8\sqrt{6}\)
A= 0
2.
A= \(12\sqrt{3}\) + \(5\sqrt{3}\) - \(12\sqrt{3}\)
A= 0
3.
A= \(3\sqrt{2}\) - \(10\sqrt{2}\) + \(6\sqrt{2}\)
A= -\(\sqrt{2}\)
4.
A= \(3\sqrt{2}\) + \(4\sqrt{2}\) - \(\sqrt{2}\)
A= \(6\sqrt{2}\)
5.
M= \(2\sqrt{5}\) - \(3\sqrt{5}\) + \(\sqrt{5}\)
M= 0
6.
A= 5 - \(3\sqrt{5}\) + \(3\sqrt{5}\)
A= 5
This literally took me a while, pls sub :D
https://www.youtube.com/channel/UC4U1nfBvbS9y_Uu0UjsAyqA/featured
28 tháng 10 2016
a) \(87^2+26\cdot87+13^2=87^2+2\cdot87\cdot13+13^2=\left(87+13\right)^2=100^2=10000\)
19 tháng 10 2021
\(A=-\dfrac{3+\sqrt{5}+3-\sqrt{5}}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\cdot\dfrac{\sqrt{5}}{5}\\ A=\dfrac{-6}{4}\cdot\dfrac{\sqrt{5}}{5}=\dfrac{-3\sqrt{5}}{10}\)
H9
HT.Phong (9A5)
CTVHS
11 tháng 10 2023
1) \(A=3\sqrt{\dfrac{1}{3}}-\dfrac{5}{2}\sqrt{12}-\sqrt{48}\)
\(=3\cdot\dfrac{\sqrt{1}}{\sqrt{3}}-\dfrac{5\sqrt{12}}{2}-\sqrt{4^2\cdot3}\)
\(=\dfrac{3\cdot1}{\sqrt{3}}-\dfrac{5\cdot2\sqrt{3}}{2}-4\sqrt{3}\)
\(=\sqrt{3}-5\sqrt{3}-4\sqrt{3}\)
\(=-8\sqrt{3}\)
2) \(A=\sqrt{12-4x}\) có nghĩa khi:
\(12-4x\ge0\)
\(\Leftrightarrow4x\le12\)
\(\Leftrightarrow x\le\dfrac{12}{4}\)
\(\Leftrightarrow x\le3\)
3) \(\dfrac{2x-2\sqrt{x}}{x-1}\)
\(=\dfrac{2\sqrt{x}\cdot\sqrt{x}-2\sqrt{x}}{\left(\sqrt{x}\right)^2-1^2}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2\sqrt{\text{x}}}{\sqrt{x}+1}\)
VT
2
CX
7 tháng 12 2017
Vì A = 2^50+2^49+2^48+...+2^2+2+1
Suy ra: 2A=2^51+2^50+2^49+...+2^3+2^2+2
A=(2^51+2^50+2^49+...+2^3+2^2+2)-(2^50+2^49+2^48+...+2^2+2+1)
A=2^51-1
7 tháng 12 2017
2A = 2^51+2^50+....+2^2+2
A = 2A - A = (2^51+2^50+....+2)-(2^50+2^51+....+2+1) = 2^51-1
k mk nha
27 . 48 1 - a 2 = 9 . 3 . 3 . 16 . 1 - a 2