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16 tháng 6 2023

\(a,\sqrt{3^2}-\sqrt{\left(-7\right)^2}+\sqrt{\left(-1\right)^2}\)

\(=3-7+1\)

\(=-3\)

\(b,-2\sqrt{\left(-2\right)^2}+3\sqrt{\left(-5\right)^2}+\sqrt{3^2}\)

\(=-2.2+3.5+3\)

\(=-4+15+3\)

\(=14\)

\(c,\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(2+\sqrt{2}\right)^2}\)
\(=\left|2-\sqrt{2}\right|+\left|2+\sqrt{2}\right|\)

\(=2-\sqrt{2}+2+\sqrt{2}\)

\(=4\)

\(d,\sqrt{\left(3-\sqrt{2}\right)^2}-\sqrt{\left(1-\sqrt{2}\right)^2}\)

\(=\left|3-\sqrt{2}\right|-\left|1-\sqrt{2}\right|\)

\(=3-\sqrt{2}-\left(-1+\sqrt{2}\right)\)

\(=3-\sqrt{2}+1-\sqrt{2}\)

\(=-2\sqrt{2}+4\)

14 tháng 8 2023

\(b,\sqrt{2}.\sqrt{7+3\sqrt{5}}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{14+6\sqrt{5}}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{\sqrt{5^2}+2.3\sqrt{5}+3^2}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{\left(\sqrt{5}+3\right)^2}-\dfrac{4}{\sqrt{5}-1}\\ =\left|\sqrt{5}+3\right|-\dfrac{4}{\sqrt{5}-1}\\ =\dfrac{\left(\sqrt{5}+3\right)\left(\sqrt{5}-1\right)-4}{\sqrt{5}-1}\\ =\dfrac{2+2\sqrt{5}-4}{\sqrt{5}-1}\\ =\dfrac{-2+2\sqrt{5}}{\sqrt{5}-1}\\ =\dfrac{2\left(-1+\sqrt{5}\right)}{\sqrt{5}-1}\\ =2\)

\(c,\sqrt{27}-6\sqrt{\dfrac{1}{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}\\ =3\sqrt{3}-\dfrac{6}{\sqrt{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}\)

\(=\dfrac{3\sqrt{3}.\sqrt{3}-6+\sqrt{3}-3}{\sqrt{3}}\\ =\dfrac{9-6+\sqrt{3}-3}{\sqrt{3}}\\ =\dfrac{\sqrt{3}}{\sqrt{3}}\\ =1\)

\(d,\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}\\ =\dfrac{\left(9-2\sqrt{3}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{\left(3\sqrt{6}-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}\\ =\dfrac{27\sqrt{6}+18\sqrt{2}-18\sqrt{2}-4\sqrt{6}}{\left(3\sqrt{6}\right)^2-\left(2\sqrt{2}\right)^2}\\ =\dfrac{23\sqrt{6}}{54-8}\\ =\dfrac{23\sqrt{6}}{46}\\ =\dfrac{\sqrt{6}}{2}\)

14 tháng 8 2023

Giải chi tiết từng bước nha

8 tháng 10 2021

\(a,ĐK:x\le\dfrac{5}{3}\\ PT\Leftrightarrow-3x+5=49\\ \Leftrightarrow x=-\dfrac{44}{3}\left(tm\right)\\ b,ĐK:x\ge-12\\ PT\Leftrightarrow\dfrac{1}{2}x+6=2\\ \Leftrightarrow\dfrac{1}{2}x=-4\\ \Leftrightarrow x=-8\left(tm\right)\\ c,ĐK:x\ge-\dfrac{1}{2}\\ PT\Leftrightarrow2x+1=13+4\sqrt{3}\\ \Leftrightarrow x=\dfrac{12+4\sqrt{3}}{2}=6+2\sqrt{3}\left(tm\right)\\ d,PT\Leftrightarrow\left|3x-1\right|=8\Leftrightarrow\left[{}\begin{matrix}3x-1=8\\1-3x=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{7}{3}\end{matrix}\right.\)