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27 tháng 3 2022

`Answer:`

a. \(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)

\(\Leftrightarrow\left|2x-\frac{3}{4}\right|=\frac{17}{2}+\frac{7}{4}\)

\(\Leftrightarrow\left|2x-\frac{3}{4}\right|=\frac{41}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-\frac{3}{4}=\frac{41}{4}\\2x-\frac{3}{4}=-\frac{41}{4}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=\frac{41}{4}+\frac{3}{4}\\2x=-\frac{41}{4}+\frac{3}{4}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=11\\2x=-\frac{19}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=11:2\\x=-\frac{19}{2}:2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{11}{2}\\x=-\frac{19}{4}\end{cases}}\)

b. \(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)

\(\Leftrightarrow\left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\)

\(\Leftrightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)

\(\Leftrightarrow\left(x+\frac{1}{5}\right)=\left(\frac{3}{5}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{5}=\frac{3}{5}\\x+\frac{1}{5}=-\frac{3}{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}-\frac{1}{5}\\x=-\frac{3}{5}-\frac{1}{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{5}\\x=-\frac{4}{5}\end{cases}}\)

c. \(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)

\(\Leftrightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)

\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}-\left(-\frac{24}{27}\right)\)

\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)

\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)

\(\Leftrightarrow3x-\frac{7}{9}=-\frac{2}{3}\)

\(\Leftrightarrow3x=-\frac{2}{3}+\frac{7}{9}\)

\(\Leftrightarrow3x=\frac{1}{9}\)

\(\Leftrightarrow x=\frac{1}{9}:3\)

\(\Leftrightarrow x=\frac{1}{27}\)

11 tháng 7 2019

a) (x - 1/2)2 = 4

<=> (x - 1/2)2 = 22

<=> x - 1/2 = -2; 2

<=> x - 1/2 = 2 hoặc x - 1/2 = -2

       x = 2 + 1/2         x = -2 + 1/2

       x = 5/2               x = -3/2

=> x = 5/2 hoặc x = -3/2

b) 10/1/2 - (x + 1/3)2 = 1/1/2

<=> -(x + 1/3)2 = 1/1/2 - 10/1/2

<=> -(x + 1/3)2 = 1/2 - 5

<=> -(x + 1/3)2 = -5.2 + 1/2

<=> -(x + 1/3)2 = -9/2

<=> (x + 1/3)2 = 9/2

<=> x + 1/3 = \(\sqrt{\frac{9}{2}}\)  hoặc x + 1/3 = \(-\sqrt{\frac{9}{2}}\)

       x = \(\frac{3\sqrt{2}}{2}\) - 1/3         x = \(-\frac{3\sqrt{2}}{2}\) -1/3

=> x = \(\frac{3\sqrt{2}}{2}\) - 1/3 hoặc x = \(-\frac{3\sqrt{2}}{2}\) -1/3

c) (x - 1/5)2 + 17/25 = 26/25

<=> (x - 1/5)2 = 26/25 - 17/25

<=> (x - 1/5)2 = (3/5)2

<=> x - 1/5 = -3/5; 3/5

<=> x - 1/5 = 3/5 hoặc x - 1/5 = -3/5

       x = 3/5 + 1/5         x = -3/5 + 1/5

       x = 4/5                  x = -2/5

=> x = 4/5 hoặc x = -2/5

20 tháng 3 2016

x + 1/5 = 26/25 - 17/25                                                                         x       = 9/25 - 1/5                                                                           x       =      4/5                                                                                                                                                                                                                                                                                                                           

19 tháng 4 2019

nhầm:mux3=mũ3

10 tháng 4 2016

a)\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)

=\(\left(x+\frac{1}{5}\right)^2=\frac{9}{25}=\frac{3^2}{5^2}\)

=\(x+\frac{1}{5}=\frac{3}{5}\)

\(x=\frac{2}{5}\)

b)\(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=\frac{24}{27}\)

=\(x=-\frac{35}{27}\)

30 tháng 4 2015

a) \(x=\frac{1}{27}\)

b) \(x=\frac{2}{5}\)

19 tháng 3 2016

1a. x=-0,8
b)-1va 5/27-(3x-7/9)3=-24/27 mik ko hỉu đề
2.n= 6

d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)

\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)

hay \(x=\dfrac{25}{372}\)

Vậy: \(x=\dfrac{25}{372}\)

e) 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: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)

f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)

\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)

\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)

\(\Leftrightarrow3x=\dfrac{1}{9}\)

hay \(x=\dfrac{1}{27}\)

g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)

\(\Leftrightarrow\dfrac{4}{3}x=2\)

hay \(x=\dfrac{3}{2}\)

Vậy: \(x=\dfrac{3}{2}\)

h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)

  Vậy ...

i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)

  Vậy ...