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a) \(\sqrt{\left(-5\right)^2}+\sqrt{5^2}-\sqrt{\left(-3\right)^2}-\sqrt{3^2}-\left(\sqrt{7}\right)^2=\sqrt{25}+\sqrt{25}-\sqrt{9}-\sqrt{9}\)
\(=5+5-3-3\)
\(=4\)
c) \(\sqrt{\left(-10\right)^2}+10.\left(-\sqrt{5}\right)^2=\sqrt{100}+10.5\)
\(=10+10.5\)
\(=10+50\)
\(=60\)
Học tốt nha^^
a) \(\sqrt{\left(-5\right)^2}+\sqrt{5^2}-\sqrt{\left(-3\right)^2}-\sqrt{3^2}\)
\(=5+5-3-3\)
\(=4\)
b) \(\left(\sqrt{4^2}+\sqrt{\left(-4\right)^2}\right).\sqrt{4^{-3}}-\sqrt{3^{-4}}\)
\(=\left(4+4\right).\frac{1}{8}-\frac{1}{9}\)
\(=8.\frac{1}{8}-\frac{1}{9}\)
\(=1-\frac{1}{9}\)
\(=\frac{8}{9}\)
\(\frac{3}{4}+\frac{1}{4}:\left(-\frac{2}{3}\right)-\left(-5\right)\)
\(=\frac{3}{4}+\frac{1}{4}.\left(-\frac{3}{2}\right)+5\)
\(=\frac{3}{4}-\frac{3}{8}+5\)
\(=\frac{3}{8}+5=\frac{43}{8}\)
\(12.\left(\frac{2}{5}-\frac{5}{6}\right)^2=12.\left(-\frac{13}{30}\right)^2=12.\frac{169}{900}=\frac{169}{75}\)
\(\left(-2\right)^2+\sqrt{36}-\sqrt{9}+\sqrt{25}=4+6-3+5=12\)
\(\left(9\frac{3}{4}:3.4.2\frac{7}{34}\right):\left(-1\frac{9}{16}\right)=\left(\frac{39}{4}:3.4.\frac{75}{34}\right):\left(-\frac{25}{16}\right)=\frac{975}{34}.\left(-\frac{16}{25}\right)=-\frac{312}{17}\)
\(\frac{\sqrt{3^2}+\sqrt{39^2}}{\sqrt{91^2}-\sqrt{\left(-7\right)^2}}=\frac{3+39}{91-7}=\frac{42}{84}=\frac{1}{2}\)
\(a,\cdot\left\{\left[\left(2\sqrt{2}\right)^2:2,4\right]\cdot\left[5,25:\left(\sqrt{7}\right)^2\right]\right\}:\left\{\left[2\dfrac{1}{7}:\dfrac{\left(\sqrt{5}\right)^2}{7}\right]:\left[2^2:\dfrac{\left(2\sqrt{2}\right)^2}{\sqrt{81}}\right]\right\}\\ =\left[\left(8:2,4\right)\cdot\left(5,25:7\right)\right]:\left[\left(\dfrac{15}{7}:\dfrac{5}{7}\right):\left(4:\dfrac{8}{9}\right)\right]\\ =\left(\dfrac{10}{3}\cdot\dfrac{3}{4}\right):\left(3:\dfrac{9}{2}\right)\\ =\dfrac{5}{2}:\dfrac{2}{3}\\ =\dfrac{15}{4}\)
a: \(\dfrac{\left\{\left[\left(2\sqrt{2}\right)^2:2,4\right]\cdot\left[5,25:\left(\sqrt{7}^2\right)\right]\right\}}{\left\{\left[2\dfrac{1}{7}:\dfrac{\left(\sqrt{5}\right)^2}{7}\right]:\left[2^2:\dfrac{\left(2\sqrt{2}\right)^2}{\sqrt{81}}\right]\right\}}\)
\(=\dfrac{\dfrac{8}{2,4}\cdot\dfrac{5,25}{7}}{\left(\dfrac{15}{7}:\dfrac{5}{7}\right):\left(4:\dfrac{8}{9}\right)}\)
\(=\dfrac{\dfrac{10}{3}\cdot\dfrac{3}{4}}{3:\left(4\cdot\dfrac{9}{8}\right)}\)
\(=\dfrac{\dfrac{10}{4}}{3:\left(\dfrac{9}{2}\right)}=\dfrac{5}{2}:\left(3\cdot\dfrac{2}{9}\right)=\dfrac{5}{2}:\dfrac{2}{3}=\dfrac{15}{4}\)
b: \(\sqrt{\left(x-\sqrt{2}\right)^2}=\left|x-\sqrt{2}\right|>=0\forall x\)
\(\sqrt{\left(y+\sqrt{2}\right)^2}=\left|y+\sqrt{2}\right|>=0\forall y\)
\(\left|x+y+z\right|>=0\forall x,y,z\)
Do đó: \(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+\left|x+y+z\right|>=0\forall x,y,z\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-\sqrt{2}=0\\y+\sqrt{2}=0\\x+y+z=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\sqrt{2}\\y=-\sqrt{2}\\z=0\end{matrix}\right.\)
\(\sqrt{\left(3-\sqrt{5}\right)^2}=\left|3-\sqrt{5}\right|=3-\sqrt{5}\)
\(\sqrt{\left(\sqrt{3-\sqrt{4}}\right)^2}=\left|\sqrt{3-\sqrt{4}}\right|=\sqrt{3-\sqrt{4}}\)
\(\sqrt{\left(7-\sqrt{34}\right)^2}=\left|7-\sqrt{34}\right|=7-\sqrt{34}\)
cảm ơn bạn linh Nguyễn