(\(\sqrt{\dfrac{1}{4}}\) - 1,2) : 1\(\dfrac{1}{20}\) - (- \(\dfrac{5}{2}\))2 + \(\left|1,25-\dfrac{3}{4}\right|\)
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20 tháng 12 2022
\(=\left(\dfrac{1}{2}-\dfrac{6}{5}\right):\dfrac{21}{20}-\dfrac{25}{4}+\dfrac{1}{2}=\dfrac{-7}{10}\cdot\dfrac{20}{21}-\dfrac{23}{4}\)
\(=\dfrac{-2}{3}-\dfrac{23}{4}=\dfrac{-8-69}{12}=-\dfrac{77}{12}\)
25 tháng 6 2023
g: \(=\left(-\sqrt{5}-2\right)\left(\sqrt{5}-2\right)\)
=-(căn 5+2)(căn 5-2)
=-(5-4)=-1
h: \(=\left(\dfrac{4}{3}\sqrt{3}+\sqrt{2}+\dfrac{\sqrt{30}}{3}\right)\left(\dfrac{\sqrt{30}}{5}+\sqrt{2}-\dfrac{4}{5}\sqrt{5}\right)\)
=4/5*căn 10+4/3*căn 6-16/15*căn 15+2/5*căn 15+2-4/5*căn 10+30/15+2/3*căn 15-4/3*căn 6
=4
AN
26 tháng 12 2017
Tính hợp lí nếu có thể:
a,\(\left(1,2-\sqrt{\dfrac{1}{4}}\right):1\dfrac{1}{20}+\left|\dfrac{3}{4}-1,25\right|-\left(-\dfrac{3}{2}\right)^2\)
\(=\left(\dfrac{12}{10}-\dfrac{1}{2}\right):\dfrac{21}{20}+\left|\dfrac{3}{4}-\dfrac{5}{4}\right|-\dfrac{9}{4}\)
\(=\dfrac{7}{10}\cdot\dfrac{20}{21}+\left|-\dfrac{1}{2}\right|-\dfrac{9}{4}\)
\(=\dfrac{2}{3}+\dfrac{1}{2}-\dfrac{9}{4}=\dfrac{8+6-27}{12}=-\dfrac{13}{12}\)
5 tháng 10 2021
a: Ta có: \(\left(4\sqrt{2}-\dfrac{11}{2}\sqrt{8}-\dfrac{1}{3}\sqrt{288}+\sqrt{50}\right)\cdot\left(\dfrac{1}{2}\sqrt{2}\right)\)
\(=\dfrac{1}{2}\sqrt{2}\cdot\left(4\sqrt{2}-11\sqrt{2}-4\sqrt{2}+5\sqrt{2}\right)\)
\(=\dfrac{1}{2}\sqrt{2}\cdot6\sqrt{2}=3\)
G
8 tháng 8 2023
a) ĐK: \(x\ge0\)
PT \(\Leftrightarrow\sqrt{4x}\left(\dfrac{3}{4}-1-\dfrac{1}{4}\right)+5=0\)
\(\Leftrightarrow2\sqrt{x}.\left(-\dfrac{1}{2}\right)+5=0\)
\(\Leftrightarrow x=25\) (thỏa)
Vậy \(x=25\)
b) Đk: \(x\le3\)
PT \(\Leftrightarrow\sqrt{3-x}-\sqrt{9\left(3-x\right)}+\dfrac{5}{4}\sqrt{16\left(3-x\right)}=6\)
\(\Leftrightarrow\sqrt{3-x}\left(1-\sqrt{9}+\dfrac{5}{4}.\sqrt{16}\right)=6\)
\(\Leftrightarrow\sqrt{3-x}=2\Leftrightarrow x=-1\) (thỏa)
Vậy \(x=-1\)
8 tháng 8 2023
2:
a:
Sửa đề: \(P=\left(\dfrac{2}{\sqrt{1+a}}+\sqrt{1-a}\right):\left(\dfrac{2}{\sqrt{1-a^2}}+1\right)\)
\(P=\dfrac{2+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}:\dfrac{2+\sqrt{1-a^2}}{\sqrt{1-a^2}}\)
\(=\dfrac{2+\sqrt{1-a^2}}{\sqrt{1+a}}\cdot\dfrac{\sqrt{1-a^2}}{2+\sqrt{1-a^2}}=\sqrt{\dfrac{1-a^2}{1+a}}\)
\(=\sqrt{1-a}\)
b: Khi a=24/49 thì \(P=\sqrt{1-\dfrac{24}{49}}=\sqrt{\dfrac{25}{49}}=\dfrac{5}{7}\)
c: P=2
=>1-a=4
=>a=-3
20 tháng 12 2021
Bài 2:
a: \(=\sqrt{2}-\dfrac{2}{5}\sqrt{2}+2\sqrt{2}+2\sqrt{2}=\dfrac{23}{5}\sqrt{2}\)
MC
21 tháng 7 2018
\(a)\left(2\dfrac{5}{6}+1\dfrac{4}{9}\right):\left(10\dfrac{1}{12}-9\dfrac{1}{2}\right)\)
\(=\left(\dfrac{17}{6}+\dfrac{13}{9}\right):\left(10\dfrac{1}{12}-9\dfrac{6}{12}\right)\)
\(=\left(\dfrac{153}{54}+\dfrac{78}{54}\right):\left(1\dfrac{-5}{12}\right)\)
\(=\dfrac{231}{54}:\dfrac{7}{12}\)
\(=\dfrac{198}{27}\)
MC
21 tháng 7 2018
\(b)\dfrac{0,8\left(\dfrac{4}{5}:1,25\right)}{0,64-\dfrac{1}{25}}\)
\(=\dfrac{0,8\left(0,8:1,25\right)}{0,64-0,04}\)
\(=\dfrac{0,8.0,64}{0,6}\)
\(=\dfrac{0,512}{0,6}\)\(=\dfrac{64}{75}\)
\(=\left(\dfrac{1}{2}-\dfrac{6}{5}\right):\dfrac{21}{20}-\dfrac{25}{4}+\dfrac{1}{2}\)
\(=\dfrac{-7}{10}\cdot\dfrac{20}{21}-\dfrac{25}{4}+\dfrac{2}{4}\)
\(=\dfrac{-2}{3}-\dfrac{23}{4}=\dfrac{-8-69}{12}=\dfrac{-77}{12}\)
\(\left(\sqrt{\dfrac{1}{4}}-1,2\right):1\dfrac{1}{20}-\left(-\dfrac{5}{2}\right)^2+\left|1,25-\dfrac{3}{4}\right|\)
\(=-\dfrac{7}{10}:\dfrac{21}{20}-\dfrac{25}{4}+\left|\dfrac{1}{2}\right|\)
\(=-\dfrac{7}{10}.\dfrac{20}{21}-\dfrac{25}{4}+\dfrac{1}{2}\)
\(=-\dfrac{2}{3}-\dfrac{25}{4}+\dfrac{1}{2}\)
\(=-\dfrac{77}{12}\)