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\(\sqrt{16}=4;\dfrac{2}{3}=0,\left(6\right);\Omega=3,14;-\sqrt{5}\simeq-2,24\)
\(-5,6< -2,23< 0\)
=>\(-5,6< -\sqrt{5}< 0\)(1)
\(0< \dfrac{2}{3}< 3,14< 4\)
=>\(0< \dfrac{2}{3}< \Omega< \sqrt{16}\)(2)
Từ (1) và (2) suy ra \(-5,6< -\sqrt{5}< 0< \dfrac{2}{3}< \Omega< \sqrt{16}\)
\(\left(-\dfrac{2}{5}\right)^2\cdot\left|\dfrac{1}{3}-\dfrac{3}{5}\right|-\dfrac{2}{5}\cdot\sqrt{\dfrac{1}{25}}+\dfrac{4}{3}\)
\(=\dfrac{4}{25}\cdot\dfrac{4}{15}-\dfrac{2}{5}\cdot\dfrac{1}{5}+\dfrac{4}{3}\)
\(=\dfrac{16}{375}-\dfrac{2}{25}+\dfrac{4}{3}\)
\(=\dfrac{16}{375}-\dfrac{30}{375}+\dfrac{500}{375}\)
\(=\dfrac{486}{375}=\dfrac{162}{125}\)
a) Ta có:
\(6 = \sqrt {36} ; - 1,7 = - \sqrt {2,89} \)
Vì 0 < 2,89 < 3 nên 0> \( - \sqrt {2,89} > - \sqrt 3 \) hay 0 > -1,7 > \( - \sqrt 3 \)
Vì 0 < 35 < 36 < 47 nên \(0 < \sqrt {35} < \sqrt {36} < \sqrt {47} \) hay 0 < \(\sqrt {35} < 6 < \sqrt {47} \)
Vậy các số theo thứ tự tăng dần là: \( - \sqrt 3 ; - 1,7;0;\sqrt {35} ;6;\sqrt {47} \)
b) Ta có:
\(\sqrt {5\frac{1}{6}} = \sqrt {5,1(6)} ; - \sqrt {2\frac{1}{3}} = - \sqrt {2,(3)} \); -1,5 = \( - \sqrt {2,25} \)
Vì 0 < 2,25 < 2,3 < 2,(3) nên 0> \( - \sqrt {2,25} > - \sqrt {2,3} > - \sqrt {2,(3)} \) hay 0 > -1,5 > \( - \sqrt {2,3} > - \sqrt {2\frac{1}{3}} \)
Vì 5,3 > 5,1(6) > 0 nên \(\sqrt {5,3} > \sqrt {5,1(6)} \)> 0 hay \(\sqrt {5,3} > \sqrt {5\frac{1}{6}} > 0\)
Vậy các số theo thứ tự giảm dần là: \(\sqrt {5,3} ;\sqrt {5\frac{1}{6}} ;0\); -1,5; \( - \sqrt {2,3} ; - \sqrt {2\frac{1}{3}} \)
a) \(\sqrt{16}+\sqrt{225}.\sqrt{9}=4+15.3=4+45=49\)
b) \(\sqrt{\dfrac{10000}{400}}+\sqrt{\left(-3\right)^2}.\sqrt{6^4}=\dfrac{100}{20}+\sqrt{9}.\sqrt{36^2}=5+3.36=5+108=113\)
\(=\left(\dfrac{3}{2}-3\right).\sqrt{\dfrac{25}{16}}=\left(-\dfrac{3}{2}\right).\dfrac{5}{4}=-\dfrac{15}{8}\)
Câu 1: Thực hiện phép tính :
a) \(2.\left(\dfrac{-2}{3}\right)^2-\dfrac{7}{2}=2.\dfrac{4}{9}-\dfrac{7}{2}\)
\(=\dfrac{8}{9}-\dfrac{7}{2}\)
\(=\dfrac{16}{18}-\dfrac{63}{18}=\dfrac{-47}{18}\)
\(b,5\dfrac{4}{13}.\dfrac{-3}{4}+3\dfrac{9}{13}.\left(-0,75\right)=\dfrac{69}{13}.\dfrac{-3}{4}+\dfrac{48}{13}.\dfrac{-3}{4}\)
\(=\left(\dfrac{69}{13}+\dfrac{48}{13}\right).\dfrac{-3}{4}\)
\(=\dfrac{117}{13}.\dfrac{-3}{4}\)
\(=9.\dfrac{-3}{4}=\dfrac{-27}{4}\)
\(c,\left(-1\right)^{2017}+\left|\dfrac{-1}{13}\right|+\sqrt{\dfrac{144}{169}}=-1+\dfrac{1}{13}+\dfrac{12}{13}\)
\(=-1+\dfrac{13}{13}\)
\(=-1+1=0\)
Câu 3: Tìm x, biết:
a)\(\dfrac{3}{5}-x=25\)
\(x=\dfrac{3}{5}-\dfrac{125}{5}\)
\(x=\dfrac{-122}{5}\)
b)\(\dfrac{2}{3}\left|x-1\right|+\dfrac{1}{4}=\dfrac{5}{3}\)
\(\dfrac{2}{3}\left|x-1\right|=\dfrac{20}{12}-\dfrac{3}{12}\)
\(\dfrac{2}{3}\left|x-1\right|=\dfrac{17}{12}\)
\(\left|x-1\right|=\dfrac{17}{12}:\dfrac{2}{3}\)
\(\left|x-1\right|=\dfrac{17}{12}.\dfrac{3}{2}\)
\(\left|x-1\right|=\dfrac{17}{8}\)
Ta có 2 TH: TH1:\(x-1=\dfrac{17}{8}\) TH2:\(x-1=\dfrac{-17}{8}\) \(x=\dfrac{17}{8}+1\) \(x=\dfrac{-17}{8}+1\) \(x=\dfrac{17}{8}+\dfrac{8}{8}=\dfrac{25}{8}\) \(x=\dfrac{-17}{8}+\dfrac{8}{8}=\dfrac{-9}{8}\) Vậy x∈\(\left\{\dfrac{25}{5};\dfrac{-9}{8}\right\}\)