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\(\begin{array}{l}a)\sqrt {1600} = 40;\\b)\sqrt {0,16} = 0,4;\\c)\sqrt {2\frac{1}{4}} = \sqrt {\frac{9}{4}} = \frac{3}{2}\end{array}\)
\(\text{a)}\sqrt{1600}=40\)
\(\text{b)}\sqrt{0,16}=0,4\)
\(\text{c)}\sqrt{2\dfrac{1}{4}}=\sqrt{\dfrac{9}{4}}=\dfrac{3}{2}\)
\(\begin{array}{l}a)2.\sqrt 6 .( - \sqrt 6 )\\ = - 2.\sqrt 6 .\sqrt 6 \\ = - 2.{(\sqrt 6 )^2}\\ = - 2.6\\ = - 12\\b)\sqrt {1,44} - 2.{(\sqrt {0,6} )^2}\\ = 1,2 - 2.0,6\\ = 1,2 - 1,2\\ = 0\\c)0,1.{(\sqrt 7 )^2} + \sqrt {1,69} \\ = 0,1.7 + 1,3 \\= 0,7 + 1,3 \\= 2\\d)( - 0,1).{(\sqrt {120} )^2} - \frac{1}{4}.{(\sqrt {20} )^2} \\= ( - 0,1).120 - \frac{1}{4}.20\\ = - 12 - 5\\ = - (12 + 5)\\ = - 17\end{array}\)
a: \(=-2\sqrt{6}\cdot\sqrt{6}=-2\cdot\sqrt{6\cdot6}=-2\cdot6=-12\)
b: \(=1.2-2\cdot0.6=1.2-1.2=0\)
c: \(=0.1\cdot7+1.3=0.7+1.3=2\)
d: \(=-0.1\cdot120-\dfrac{1}{4}\cdot20=-12-5=-17\)
\(\begin{array}{l}a)\frac{x}{{ - 3}} = \frac{7}{{0,75}}\\ \Rightarrow x.0,75 = ( - 3).7\\ \Rightarrow x = \frac{{( - 3).7}}{{0,75}} = - 28\end{array}\)
Vậy x = 28
\(\begin{array}{l}b) - 0,52:x = \sqrt {1,96} :( - 1,5)\\ - 0,52:x = 1,4:( - 1,5)\\ x = \dfrac{(-0,52).(-1,5)}{1,4}\\x = \frac{39}{{70}}\end{array}\)
Vậy x = \(\frac{39}{{70}}\)
\(\begin{array}{l}c)x:\sqrt 5 = \sqrt 5 :x\\ \Leftrightarrow \frac{x}{{\sqrt 5 }} = \frac{{\sqrt 5 }}{x}\\ \Rightarrow x.x = \sqrt 5 .\sqrt 5 \\ \Leftrightarrow {x^2} = 5\\ \Leftrightarrow \left[ {_{x = - \sqrt 5 }^{x = \sqrt 5 }} \right.\end{array}\)
Vậy x \( \in \{ \sqrt 5 ; - \sqrt 5 \} \)
Chú ý:
Nếu \({x^2} = a(a > 0)\) thì x = \(\sqrt a \) hoặc x = -\(\sqrt a \)
a: \(\dfrac{x}{-3}=\dfrac{7}{0.75}=\dfrac{28}{3}\)
=>\(x=\dfrac{28\left(-3\right)}{3}=-28\)
b: \(-\dfrac{0.52}{x}=\dfrac{\sqrt{1.96}}{-1.5}=\dfrac{1.4}{-1.5}\)
=>\(x=0.52\cdot\dfrac{1.5}{1.4}=\dfrac{39}{70}\)
c: \(\dfrac{x}{\sqrt{5}}=\dfrac{\sqrt{5}}{x}\)
=>\(x^2=5\)
=>\(x=\pm\sqrt{5}\)
\(\begin{array}{l}a)\sqrt x - 16 = 0\\\sqrt x = 16\\x = {16^2}\\x = 256\end{array}\)
Vậy x = 256
\(\begin{array}{l}b)2\sqrt x = 1,5\\\sqrt x = 1,5:2\\\sqrt x = 0.75\\x = {(0,75)^2}\\x = 0,5625\end{array}\)
Vậy x = 0,5625
\(\begin{array}{l}c)\sqrt {x + 4} - 0,6 = 2,4\\\sqrt {x + 4} = 2,4 + 0,6\\\sqrt {x + 4} = 3\\x + 4 = 9\\x = 5\end{array}\)
Vậy x = 5
a) \(\sqrt{0,09}-\sqrt{0,64}=\frac{-1}{2}=-0,5\)
b) \(0,1\cdot\sqrt{225}-\sqrt{\frac{1}{4}}=0,1\cdot15-\frac{1}{2}=1\)
c) \(\sqrt{0,36}\cdot\sqrt{\frac{25}{16}+\frac{1}{4}}=\frac{3\sqrt{29}}{20}\)
d) đề baì có sai ko ban?
Bạn Duy giúp mik vs
Tính:
a) \(\sqrt{0,36}+\sqrt{0,49}\)
b) \(\sqrt{\frac{4}{9}}-\sqrt{\frac{25}{36}}\)
a)\(\sqrt{0,36}\)+\(\sqrt{0,49}\)=0,6+0,7=1,3
b)\(\sqrt{\frac{4}{9}}\)-\(\sqrt{\frac{25}{36}}\)=2/3-5/6=4/6-5/6=-1/6
a) \(\sqrt{0,36}+\sqrt{0,49}=\sqrt{\left(0,6\right)^2}+\sqrt{\left(0,7\right)^2}=0,6+0,7=1,3\)
b) \(\sqrt{\frac{4}{9}}-\sqrt{\frac{25}{36}}=\sqrt{\left(\frac{2}{3}\right)^2}-\sqrt{\left(\frac{5}{6}\right)^2}=\frac{2}{3}-\frac{5}{6}=-\frac{1}{6}\)
a)\(\sqrt{0,09}\)+2.\(\sqrt{0,25}\)=0,3+2.0,5
=0,3+1
=1,3
b)0,5.\(\sqrt{100}\)-\(\sqrt{\frac{4}{25}}\)=0,5.10-0,4
=5-0,4
=4,6
c)(\(\sqrt{1\frac{9}{16}}\) -\(\sqrt{\frac{9}{16}}\)):5=(1,25-0,75):5
=0,5:5
=0,1
d)3.\(\sqrt{1\frac{17}{64}}\) -2.\(\sqrt{0,0625}\)=1,125-2.0,25
=1,125-0,5
=0,625
\(A=\left(8+2\cdot3-7\cdot\dfrac{13}{10}+3\cdot\dfrac{5}{4}\right):\left(\dfrac{5\sqrt{6}}{3}\right)^2\\ A=\left(14-\dfrac{91}{10}+\dfrac{15}{4}\right):\dfrac{50}{3}\\ A=\dfrac{173}{20}\cdot\dfrac{3}{50}=\dfrac{519}{1000}\)
\(\begin{array}{l}a)\sqrt {0,49} + \sqrt {0,64} = 0,7 + 0,8 = 1,5;\\b)\sqrt {0,36} - \sqrt {0,81} = 0,6 - 0,9 = - 0,3;\\c)8.\sqrt 9 - \sqrt {64} = 8.3 - 8 = 24 - 8 = 16;\\d)0,1.\sqrt {400} + 0,2.\sqrt {1600} = 0,1.20 + 0,2.40 = 2 + 8 = 10\end{array}\)