Rút gọn biểu thức:
a) \(M=2\sqrt{27}-6\sqrt{\frac{4}{3}}+\frac{3}{5}\sqrt{75};\)
b) \(P=\frac{2}{x-1}\sqrt{\frac{x^2-2x+1}{4x^2}};\)với 0<x<1.
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Những câu hỏi liên quan
20 tháng 5 2019
a)\(\)https://www.cymath.com/answer?q=2sqrt(27)-6sqrt(4%2F3)%2B3%2F5sqrt(75)
20 tháng 5 2019
\(M=2\sqrt{27}-6\sqrt{\frac{4}{3}}+\frac{3}{5}\sqrt{75}=2\sqrt{3^2.3}-6\sqrt{\frac{2^2.3}{3^2}}+\frac{3}{5}\sqrt{5^2.3}=.\)
\(=6\sqrt{3}-4\sqrt{3}+3\sqrt{3}=5\sqrt{3}\)
\(P=\frac{2}{x-1}\sqrt{\frac{x^2-2x+1}{4x^2}}.Với...0< x< 1\Leftrightarrow\) \(P=\frac{2}{x-1}\sqrt{\frac{\left(x-1\right)^2}{\left(2x\right)^2}}=\frac{2}{(x-1)}.\frac{\left(1-x\right)}{2x}=\frac{-1}{x}.\)
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
\(D=\frac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}=\frac{6\sqrt{2}-4\sqrt{3}+2\sqrt{5}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}\)
\(=\frac{2\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}{3\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}=\frac{2}{3}\)
V
2
MT
5 tháng 7 2015
\(\sqrt{75}-3\sqrt{27}-6\sqrt{\frac{1}{3}}+2\sqrt{\frac{4}{3}}\)
=\(\sqrt{25}\sqrt{3}-3\sqrt{9}\sqrt{3}-6\sqrt{\frac{1}{3}}+2\sqrt{4}\sqrt{\frac{1}{3}}\)
=\(5\sqrt{3}-9\sqrt{3}-6\sqrt{\frac{1}{3}}+4\sqrt{\frac{1}{3}}\)
=\(-4\sqrt{3}-2\sqrt{\frac{1}{3}}\)
=\(-4\sqrt{3}-\frac{2}{\sqrt{3}}\)
=\(-4\sqrt{3}-\frac{2\sqrt{3}}{3}\)
=\(\frac{-12\sqrt{3}}{3}-\frac{2\sqrt{3}}{3}=\frac{-14\sqrt{3}}{3}\)
YY
5 tháng 7 2015
\(\sqrt{75}-3\sqrt{27}-6\sqrt{\frac{1}{3}}+2\sqrt{\frac{4}{3}}=\sqrt{3.25}-9\sqrt{3}-2\sqrt{3}+\frac{4\sqrt{3}}{3}=-4\sqrt{3}-2\sqrt{3}+\frac{4\sqrt{3}}{3}=-6\sqrt{3}+\frac{4\sqrt{3}}{3}=-\sqrt{\frac{14\sqrt{3}}{3}}\)
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)
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\\ =\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}^2-\sqrt{5}^2\right)\\ =-\left(2-5\right)\\ =-\left(-3\right)\\ =3\)
\(M=2\sqrt{3^2.3}-6\frac{\sqrt{2^2.3}}{3}+\frac{3}{5}\sqrt{5^2.3}\)
\(M=6\sqrt{3}-4\sqrt{3}+3\sqrt{3}=5\sqrt{3}\)
\(P=\frac{2}{x-1}\sqrt{\frac{\left(x-1\right)^2}{\left(2x\right)^2}}=\frac{2}{x-1}.\frac{\left|x-1\right|}{2x}=\frac{-2\left(x-1\right)}{\left(x-1\right).2x}=-\frac{1}{x}\)