rút gọn biểu thức
\(2\sqrt{\frac{16}{5}}-5\sqrt{\frac{9}{125}}-6\sqrt{\frac{121}{45}}\)
K
Khách
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Những câu hỏi liên quan
10 tháng 10 2020
Bài 1:
a) \(5\sqrt{\frac{1}{5}}+\frac{1}{3}\sqrt{45}+\frac{5-\sqrt{5}}{\sqrt{5}}\)
\(=\sqrt{5}+\sqrt{5}+\sqrt{5}-1\)
\(=3\sqrt{5}-1\)
b) \(\sqrt{48}-6\sqrt{\frac{1}{3}}+\frac{\sqrt{3}-3}{\sqrt{3}}\)
\(=4\sqrt{3}-2\sqrt{3}+1-\sqrt{3}\)
\(=\sqrt{3}+1\)
c) \(\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{5}{\sqrt{5}}\right)\div\left(\frac{1}{\sqrt{5}-\sqrt{2}}\right)\)
\(=\left(-\sqrt{2}-\sqrt{5}\right)\div\frac{\sqrt{5}+\sqrt{2}}{5-2}\)
\(=-\left(\sqrt{2}+\sqrt{5}\right)\cdot\frac{3}{\sqrt{5}+\sqrt{2}}\)
\(=-3\)
10 tháng 10 2020
Bài 2:
đk: \(x\ge1\)
Ta có: \(\sqrt{4x+4}-\sqrt{9x-9}-8\sqrt{\frac{x+1}{16}}=5\)
\(\Leftrightarrow2\sqrt{x+1}-3\sqrt{x-1}-2\sqrt{x+1}=5\)
\(\Leftrightarrow-3\sqrt{x-1}=5\)
\(\Rightarrow\sqrt{x-1}=-\frac{5}{3}\) (vô lý)
=> PT vô nghiệm
23 tháng 5 2021
Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé
KT
13 tháng 8 2018
\(B=\frac{9\sqrt{5}+3\sqrt{27}}{\sqrt{5}+\sqrt{3}}=\frac{9\sqrt{5}+9\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\frac{9\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{5}+\sqrt{3}}=9\)
\(C=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{4}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}.\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(\sqrt{2}+1\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\sqrt{2}+1\)
mik chỉnh lại đề
\(D=\frac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}=\frac{6\sqrt{2}-4\sqrt{3}+2\sqrt{5}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}\)
\(=\frac{2\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}{3\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}=\frac{2}{3}\)
H9
HT.Phong (9A5)
CTVHS
23 tháng 8 2023
a) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\)
\(=\left[-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\cdot\left(\sqrt{2}-\sqrt{5}\right)\)
\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)
\(=-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)
\(=-\left(2-5\right)\)
\(=-\left(-3\right)\)
\(=3\)
b) Ta có:
\(x^2-x\sqrt{3}+1\)
\(=x^2-2\cdot\dfrac{\sqrt{3}}{2}\cdot x+\left(\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
\(=\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
Mà: \(\left(x-\dfrac{\sqrt{3}}{2}\right)^2\ge0\forall x\) nên
\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x\)
Dấu "=" xảy ra:
\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{\sqrt{3}}{2}\)
Vậy: GTNN của biểu thức là \(\dfrac{1}{4}\) tại \(x=\dfrac{\sqrt{3}}{2}\)
H
23 tháng 8 2023
a)
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\\ =\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}^2-\sqrt{5}^2\right)\\ =-\left(2-5\right)\\ =-\left(-3\right)\\ =3\)
PK
rút gọn biểu thức
\(D=\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}+\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}}\)
0
TH
rút gọn biểu thức
\(D=\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}+\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)
1
2 tháng 10 2016
\(D=\sqrt{\frac{\left(5+2\sqrt{6}\right)^2}{25-24}}+\sqrt{\frac{\left(5-2\sqrt{6}\right)^2}{25-24}}=5+2\sqrt{6}+5-2\sqrt{6}=10\)
NT
0
14 tháng 11 2019
\(\frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}+\sqrt{5}}+\frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}-\sqrt{5}}\)
\(=\frac{\left(\sqrt{6}-\sqrt{5}\right)^2}{\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)}+\frac{\left(\sqrt{6}+\sqrt{5}\right)^2}{\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)}\)
\(=\frac{\left(\sqrt{6}-\sqrt{5}\right)^2+\left(\sqrt{6}+\sqrt{5}\right)^2}{\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)}\)
\(=\frac{11-2\sqrt{30}+11+2\sqrt{30}}{\left(\sqrt{6}\right)^2-\left(\sqrt{5}\right)^2}\)
\(=\frac{22}{1}=22\)
14 tháng 11 2019
\(\frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}+\sqrt{5}}+\frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}-\sqrt{5}}\)
\(=\frac{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)+\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}{\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)}\)
\(=\frac{\left(\sqrt{6}-\sqrt{5}\right)^2+\left(\sqrt{6}+\sqrt{5}\right)^2}{\sqrt{6}^2+\sqrt{5}^2}\)
\(=\sqrt{6}^2-2\sqrt{6}.\sqrt{5}+\sqrt{5}^2+\sqrt{6}^2+2\sqrt{6}.\sqrt{5}+\sqrt{5}^2\)
\(=6+5+6+5=22\)
\(=2\frac{4}{\sqrt{5}}-5\frac{3}{5\sqrt[]{5}}-6\frac{11}{3\sqrt{5}}\)
\(=\frac{2.4.15-5.3.3-6.11.5}{15\sqrt{5}}\)
\(=\frac{-255}{15\sqrt{5}}=\frac{-17\sqrt{5}}{5}\)