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TT
1
19 tháng 6 2017
\(A=...\)
\(=3\sqrt{7}-2.5\sqrt{7}+6\sqrt{7}-\dfrac{1}{7}.2\sqrt{7}\)
\(=\left(3-2.5+6-\dfrac{1}{7}.2\right)\sqrt{7}\)
\(=-\dfrac{9\sqrt{7}}{7}\)
1 tháng 10 2021
a: Ta có: \(2\sqrt{28}+2\sqrt{63}-3\sqrt{175}+\sqrt{112}-\sqrt{20}\)
\(=4\sqrt{7}+6\sqrt{7}-15\sqrt{7}+4\sqrt{7}-2\sqrt{5}\)
\(=-\sqrt{7}-2\sqrt{5}\)
ND
17 tháng 7 2018
\(a.6\sqrt{3}-2\sqrt{12}+5\sqrt{300}-7\sqrt{243}=6\sqrt{3}-4\sqrt{3}+50\sqrt{3}-63\sqrt{3}=\left(6-4+50-63\right)\sqrt{3}=-11\sqrt{3}\)
\(b.\sqrt{28}+3\sqrt{63}-6\sqrt{175}-\dfrac{1}{5}\sqrt{252}=2\sqrt{7}+9\sqrt{7}-30\sqrt{7}-\dfrac{6}{5}\sqrt{7}=\left(2+9-30-\dfrac{6}{5}\right)\sqrt{7}=-20,2\sqrt{7}\)\(c.5\sqrt{44}-2\sqrt{275}-3\sqrt{176}=10\sqrt{11}-10\sqrt{11}-12\sqrt{11}=-12\sqrt{11}\)
\(d.2\sqrt{75}-\sqrt{12}+2\sqrt{147}-7\sqrt{103}=10\sqrt{3}-2\sqrt{3}+14\sqrt{3}-7\sqrt{103}=22\sqrt{3}-7\sqrt{103}\)
17 tháng 7 2018
\(a.6\sqrt{3}-2\sqrt{12}+5\sqrt{300}-7\sqrt{243}=6\sqrt{3}-4\sqrt{3}+50\sqrt{3}-63\sqrt{3}=-11\sqrt{3}\)
\(b.\sqrt{28}+3\sqrt{63}-6\sqrt{175}-\dfrac{1}{5}\sqrt{252}=2\sqrt{7}+9\sqrt{7}-30\sqrt{7}-\dfrac{6}{5}\sqrt{7}=-\dfrac{101}{5}\sqrt{7}\)
\(c.5\sqrt{44}-2\sqrt{275}-3\sqrt{176}=20\sqrt{11}-10\sqrt{11}-12\sqrt{11}=-2\sqrt{11}\)
\(d.2\sqrt{75}-\sqrt{12}+2\sqrt{147}-7\sqrt{103}=10\sqrt{3}-2\sqrt{3}+14\sqrt{3}-7\sqrt{103}=22\sqrt{3}-7\sqrt{103}\)
PA
1
AT
7 tháng 7 2021
a) \(A=\sqrt{28}-\sqrt{63}+\dfrac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=2\sqrt{7}-3\sqrt{7}+\dfrac{\sqrt{7}\left(\sqrt{7}+1\right)}{\sqrt{7}}-\left|\sqrt{7}+1\right|\)
\(=-\sqrt{7}+\sqrt{7}+1-\sqrt{7}-1=-\sqrt{7}\)
\(B=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\right)\dfrac{4\sqrt{x}+12}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-3+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{4\left(\sqrt{x}+3\right)}{\sqrt{x}}=\dfrac{2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{4\left(\sqrt{x}+3\right)}{\sqrt{x}}\)
\(=\dfrac{8}{\sqrt{x}-3}\)
b) \(A>B\Rightarrow-\sqrt{7}>\dfrac{8}{\sqrt{x}-3}\Rightarrow\dfrac{8}{\sqrt{x}-3}+\sqrt{7}< 0\)
\(\Rightarrow\dfrac{\sqrt{7x}+8-3\sqrt{7}}{\sqrt{x}-3}< 0\)
Ta có: \(\left\{{}\begin{matrix}8=\sqrt{64}\\3\sqrt{7}=\sqrt{63}\end{matrix}\right.\Rightarrow8-3\sqrt{7}>0\Rightarrow8-3\sqrt{7}+\sqrt{7x}>0\)
\(\Rightarrow\sqrt{x}-3< 0\Rightarrow\sqrt{x}< 3\Rightarrow x< 9\Rightarrow0< x< 9\)
12 tháng 9 2023
a) \(A=\sqrt{28}-\sqrt{63}+\dfrac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\sqrt{2^2\cdot7}-\sqrt{3^2\cdot7}+\dfrac{\sqrt{7}\cdot\left(\sqrt{7}+1\right)}{\sqrt{7}}-\left|\sqrt{7}+1\right|\)
\(=2\sqrt{7}-3\sqrt{7}+\sqrt{7}+1-\sqrt{7}-1\)
\(=-\sqrt{7}\)
\(B=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\right)\cdot\dfrac{4\sqrt{x}+12}{\sqrt{x}}\)
\(=\left[\dfrac{\sqrt{x}-3+\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]\cdot\dfrac{4\sqrt{x}+12}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{4\left(\sqrt{x}+3\right)}{\sqrt{x}}\)
\(=\dfrac{2\cdot4}{\sqrt{x}-3}\)
\(=\dfrac{8}{\sqrt{x}-3}\)
b) \(A>B\) khi
\(\dfrac{8}{\sqrt{x}-3}< -\sqrt{7}\)
\(\Leftrightarrow8< -\sqrt{7x}+3\sqrt{7}\)
\(\Leftrightarrow x< \dfrac{\left(3\sqrt{7}-8\right)^2}{7}\)
39
1
\(A=\sqrt{9.7}-2\sqrt{25.7}+\sqrt{9.7.4}-\frac{1}{7}\sqrt{4.7}\)
\(=3\sqrt{7}-10\sqrt{7}+6\sqrt{7}-\frac{2}{7}\sqrt{7}\)
\(=\frac{-9}{7}\sqrt{7}\)
Nếu đúng tk nhé
a = \(\sqrt{63}-2\sqrt{175}+\sqrt{252}-\frac{1}{7}\sqrt{28}\)
= \(\sqrt{\frac{4}{7}}\left(1,5-5+3-1\right)\)
= \(-1,5\sqrt{\frac{4}{7}}\)