S
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
19 tháng 8 2023
1:
\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)
2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)
\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)
\(=\dfrac{20-6}{2}=7\)
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)
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\\ =\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}^2-\sqrt{5}^2\right)\\ =-\left(2-5\right)\\ =-\left(-3\right)\\ =3\)
KV
10 tháng 3 2019
\(A=\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)\) \(\left(\sqrt{6}+11\right)\)
\(A=\left(\frac{3.\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}{\sqrt{6}+1}\right)+\left(\frac{2.\left(\sqrt{6}+2\right)\left(\sqrt{6}-2\right)}{\sqrt{6}-2}\right)-\frac{4.\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}{3-\sqrt{6}}\)\(\left(\sqrt{6}+11\right)\)
\(A=\left(3.\left(\sqrt{6}-1\right)+2.\left(\sqrt{6}+2\right)-4.\left(3+\sqrt{6}\right)\right)\left(\sqrt{6}+11\right)\)
\(A=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)=-115\)
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
\(\sqrt{11-2\sqrt{30}}:\left(1-\frac{\sqrt{5}}{\sqrt{6}}\right)\)
\(\sqrt{11-2\sqrt{3}\sqrt{5}\sqrt{2}}:\frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}}\)
\(\sqrt{\left(\sqrt{3}\sqrt{2}\right)^2-2\sqrt{3}\sqrt{5}\sqrt{2}+\sqrt{5}^2}.\frac{\sqrt{6}}{\sqrt{6}-\sqrt{5}}\)
\(\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}.\frac{\sqrt{6}}{\sqrt{6}-\sqrt{5}}\)
\(\left(\sqrt{6}-\sqrt{5}\right).\frac{\sqrt{6}}{\sqrt{6}-\sqrt{5}}\)
\(=\sqrt{6}\)
\(\sqrt{11-2\sqrt{30}}:\left(1-\frac{\sqrt{5}}{\sqrt{6}}\right)\)
\(=\sqrt{6-2\sqrt{30}+5}:\left(1-\frac{\sqrt{5}}{\sqrt{6}}\right)\)
\(=\sqrt{\left(\sqrt{6}\right)^2-2\sqrt{30}+\left(\sqrt{5}\right)^2}:\left(1-\frac{\sqrt{5}}{\sqrt{6}}\right)\)
\(=\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}:\left(1-\frac{\sqrt{5}}{\sqrt{6}}\right)\)
\(=\left|\sqrt{6}-\sqrt{5}\right|:\left(\frac{\sqrt{6}}{\sqrt{6}}-\frac{\sqrt{5}}{\sqrt{6}}\right)\)
\(=\left(\sqrt{6}-\sqrt{5}\right):\left(\frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}}\right)\)
\(=\left(\sqrt{6}-\sqrt{5}\right).\frac{\sqrt{6}}{\sqrt{6}-\sqrt{5}}\)
\(=\sqrt{6}\)