\(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)
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28 tháng 4 2021
(x+1/5)2+17/25=26/25
(x+1/5)2 =26/25-17/25
(x+1/5)2 =9/25
⇒(x+1/5)2=(3/5)2 hoặc (x+1/5)2=(-3/5)2
x+1/5=3/5 hoặc x+1/5=-3/5
x=2/5 hoặc x=-4/5
Chúc bạn học tốt!
28 tháng 4 2021
Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=\dfrac{-3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=\dfrac{-4}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{2}{5};\dfrac{-4}{5}\right\}\)
TA
3 tháng 3 2018
Các bạn ơi giúp mk với 
các bạn ơi mk sắp phải đi học rồi giúp mk với
NT
1
19 tháng 7 2017
\(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{26}{25}-\dfrac{17}{25}=\dfrac{9}{25}\) \(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=\dfrac{-3}{5}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{5}-\dfrac{1}{5}\\x=\dfrac{-3}{5}-\dfrac{1}{5}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\x=\dfrac{-4}{5}\end{matrix}\right.\)
vậy \(x=\dfrac{2}{5};x-\dfrac{-4}{5}\)
8 tháng 7 2021
1) Ta có: \(\left(-\dfrac{2}{3}\right)^2\cdot\dfrac{-9}{8}-25\%\cdot\dfrac{-16}{5}\)
\(=\dfrac{4}{9}\cdot\dfrac{-9}{8}-\dfrac{1}{4}\cdot\dfrac{-16}{5}\)
\(=\dfrac{-1}{2}+\dfrac{4}{5}\)
\(=\dfrac{-5}{10}+\dfrac{8}{10}=\dfrac{3}{10}\)
2) Ta có: \(-1\dfrac{2}{5}\cdot75\%+\dfrac{-7}{5}\cdot25\%\)
\(=\dfrac{-7}{5}\cdot\dfrac{3}{4}+\dfrac{-7}{5}\cdot\dfrac{1}{4}\)
\(=\dfrac{-7}{5}\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=-\dfrac{7}{5}\)
3) Ta có: \(-2\dfrac{3}{7}\cdot\left(-125\%\right)+\dfrac{-17}{7}\cdot25\%\)
\(=\dfrac{-17}{7}\cdot\dfrac{-5}{4}+\dfrac{-17}{7}\cdot\dfrac{1}{4}\)
\(=\dfrac{-17}{7}\cdot\left(\dfrac{-5}{4}+\dfrac{1}{4}\right)\)
\(=\dfrac{17}{7}\)
4) Ta có: \(\left(-2\right)^3\cdot\left(\dfrac{3}{4}\cdot0.25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)
\(=\left(-8\right)\cdot\left(\dfrac{3}{4}\cdot\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=\left(-8\right)\cdot\dfrac{3}{16}:\dfrac{54-28}{24}\)
\(=\dfrac{-3}{2}\cdot\dfrac{24}{26}\)
\(=\dfrac{-72}{52}=\dfrac{-18}{13}\)
18 tháng 10 2021
\(\dfrac{\dfrac{4}{5}:\left(\dfrac{4}{5}\cdot\dfrac{5}{4}\right)}{\dfrac{16}{25}-\dfrac{1}{25}}+\dfrac{\left(\dfrac{27}{25}-\dfrac{2}{25}\right):\dfrac{4}{7}}{\left(\dfrac{59}{9}-\dfrac{13}{4}\right)\cdot\dfrac{36}{17}}+\left(\dfrac{6}{5}\cdot\dfrac{1}{2}\right):\dfrac{4}{5}\)
\(=\dfrac{4}{5}:\dfrac{3}{5}+\dfrac{7}{4}:7+\dfrac{3}{5}:\dfrac{4}{5}\)
\(=\dfrac{4}{3}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\dfrac{7}{3}\)
O
1 tháng 4 2018
\(M=\dfrac{1,6:\left(1\dfrac{3}{5}.1,25\right)}{0,64-\dfrac{1}{25}}+\dfrac{\left(1,08-\dfrac{2}{25}\right):\dfrac{4}{7}}{\left(5\dfrac{5}{9}-2\dfrac{1}{4}\right).2\dfrac{1}{4}}+0,6.0,5:\dfrac{2}{5}\)
\(M=\dfrac{1,6:\left(\dfrac{8}{5}.1,25\right)}{0,64-\dfrac{1}{25}}+\dfrac{\left(1,08-\dfrac{2}{25}\right):\dfrac{4}{7}}{\left(\dfrac{50}{9}-\dfrac{9}{4}\right).\dfrac{35}{17}}+0,6.0,5:\dfrac{2}{5}\)
\(M=\dfrac{1,6:2}{0,64-\dfrac{1}{25}}+\dfrac{1:\dfrac{4}{7}}{\dfrac{119}{36}.\dfrac{35}{17}}+0,6.0,5:\dfrac{2}{5}\)
\(M=\dfrac{0,8}{0,6}+\dfrac{1,75}{\dfrac{245}{36}}+0,6.0,5:\dfrac{2}{5}\)
\(M=\dfrac{4}{3}+\dfrac{9}{35}+0,6.0,5:\dfrac{2}{5}\)
\(M=\dfrac{167}{105}+0,6.0,5:\dfrac{2}{5}\)
\(M=\dfrac{167}{105}+\dfrac{3}{10}:\dfrac{2}{5}\)
\(M=\dfrac{167}{105}+\dfrac{3}{4}\)
\(M=\dfrac{983}{420}.\)
PN
9 tháng 7 2018
\(=\dfrac{0,8:1}{0,6}+\dfrac{1:\dfrac{4}{7}}{3\dfrac{11}{36}.2\dfrac{2}{17}}+0,6:\dfrac{4}{5}\)
\(=\dfrac{0,4}{0,3}+\dfrac{\dfrac{7}{4}}{7}+0,75\)
\(=1\dfrac{0,4}{0,3}\)
27 tháng 9 2021
\(E=\dfrac{98:\left(\dfrac{4}{5}\cdot\dfrac{5}{4}\right)}{\dfrac{16}{25}-\dfrac{1}{25}}+\dfrac{\left(\dfrac{27}{25}-\dfrac{2}{25}\right)\cdot\dfrac{7}{4}}{\left(\dfrac{59}{9}-\dfrac{13}{4}\right)\cdot\dfrac{36}{17}}\\ E=\dfrac{98}{\dfrac{3}{5}}+\dfrac{\dfrac{7}{4}}{\dfrac{119}{36}\cdot\dfrac{36}{17}}\\ E=\dfrac{490}{3}+\dfrac{\dfrac{7}{4}}{7}=\dfrac{490}{3}+\dfrac{1}{4}=\dfrac{1963}{12}\)
27 tháng 9 2021
bạn ơi chỗ kia mik nhìn hơi loạn tí bạn giải thích giúp mik với
\(\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)
\(\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
\(\left(x+\dfrac{1}{5}\right)^2=\dfrac{26}{25}+\dfrac{17}{25}\\ \left(x+\dfrac{1}{5}\right)^2=\dfrac{43}{25}\\ x+\dfrac{1}{5}=\dfrac{\sqrt{45}}{5}\\ x=\dfrac{\sqrt{45}}{5}-\dfrac{1}{5}\\ x=\dfrac{-1+\sqrt{43}}{5}\)