\(\left|x-\dfrac{2}{5}\right|-\dfrac{1}{2}=\dfrac{1}{3}.\dfrac{1}{4}-\dfrac{1}{5}\)

\(\Rightarrow\left|x-\dfrac{2}{5}\right|-\dfrac{1}{2}=\dfrac{-7}{60}\)

\(\Rightarrow\left|x-\dfrac{2}{5}\right|=\dfrac{23}{60}\)

\(\Rightarrow x-\dfrac{2}{5}=\dfrac{23}{60}\) hoặc \(x-\dfrac{2}{5}=\dfrac{-23}{60}\)

\(\Rightarrow x=\dfrac{47}{60}\) hoặc \(x=\dfrac{1}{60}\)

Vậy \(x\in\left\{\dfrac{47}{60};\dfrac{1}{60}\right\}\)

4) | x-1/3| -1/3=1/3

\(x+\left|\dfrac{1}{2}-\dfrac{1}{3}\right|=\left|\dfrac{-2}{3}-\dfrac{1}{4}\right|\)

\(x+\left|\dfrac{1}{6}\right|=\left|\dfrac{-11}{12}\right|\)

\(x+\dfrac{1}{6}=\dfrac{11}{12}\)

\(x=\dfrac{11}{12}-\dfrac{1}{6}\)

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

Vậy ...

1) \(2\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\)

\(\Leftrightarrow2\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}+\dfrac{3}{2}\)

\(\Leftrightarrow2\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}\)

\(\Leftrightarrow\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}:2\)

\(\Leftrightarrow\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{8}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=-\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=-\dfrac{7}{8}+\dfrac{1}{3}\\\dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=-\dfrac{13}{24}\\\dfrac{1}{2}x=\dfrac{29}{24}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\left(-\dfrac{13}{24}\right):\dfrac{1}{2}\\x=\dfrac{29}{24}:\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{13}{12}\\x=\dfrac{29}{12}\end{matrix}\right.\)

2) \(\dfrac{3}{4}-2\left|2x-\dfrac{2}{3}\right|=2\)

\(\Leftrightarrow2\left|2x-\dfrac{2}{3}\right|=\dfrac{3}{4}-2\)

\(\Leftrightarrow2\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{8}\)

\(\Leftrightarrow\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{8}:2\)

\(\Leftrightarrow\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{16}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{2}{3}=\dfrac{-5}{16}\\2x-\dfrac{2}{3}=\dfrac{5}{16}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{-5}{16}+\dfrac{2}{3}\\2x=\dfrac{5}{16}+\dfrac{2}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{17}{48}\\2x=\dfrac{47}{48}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{48}:2\\x=\dfrac{47}{48}:2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{96}\\x=\dfrac{47}{96}\end{matrix}\right.\)

1) . \(\dfrac{1}{2}-\left|\dfrac{1}{5}-\dfrac{1}{4}\right|+\left(-\dfrac{1}{3}\right)^2\\ =\dfrac{1}{2}-\left(\dfrac{1}{4}-\dfrac{1}{5}\right)+\dfrac{1}{9}\\ =\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{9}\)

\(=\dfrac{61}{180}\)

2) . \(\dfrac{1}{3}+\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{-2}{3}\right)^2\\ =\dfrac{1}{3}+\dfrac{4}{3}\cdot\dfrac{1}{6}+\dfrac{4}{9}\\ =\dfrac{1}{3}+\dfrac{2}{9}+\dfrac{4}{9}\\ =1\)

Các bạn ơi giúp mk với các bạn ơi mk sắp phải đi học rồi giúp mk với

các bạn có thương mk ko

\(N=\dfrac{2^{19}\cdot3^9-3\cdot3^8\cdot5\cdot2^{18}}{3^9\cdot2^9\cdot2^{10}-2^{20}\cdot3^{10}}\)

\(=\dfrac{2^{18}\cdot3^9\cdot\left(2-5\right)}{3^9\cdot2^{19}\left(1-2\cdot3\right)}=\dfrac{1}{2}\cdot\dfrac{-3}{-5}=\dfrac{3}{10}\)

\(P=\dfrac{8\cdot10+8\cdot24+8\cdot560}{6\cdot45+6\cdot108+6\cdot120\cdot21}=\dfrac{8\left(10+24+560\right)}{6\left(45+108+120\cdot21\right)}=\dfrac{4}{3}\cdot\dfrac{2}{9}=\dfrac{8}{27}\)

a. 5² + (x+3) = 5²

=> x + 3    = 5² - 5²

=>  x + 3   =  0

=>  x  = -3

b. 2³ + (x-3²) = 5³ - 4³

=> 8 + (x-9) =  125 - 64

=> x - 9 = 125 - 64 - 8

=> x - 9 =  53

=> x    =  53 + 9

=> x    =   62

a; \(\dfrac{6}{x}\) < \(\dfrac{x}{7}\) < \(\dfrac{8}{x}\)

    vì \(x\) \(\in\) N* ta có: 6.7 < \(x.x\) < 7.8

                             42 < \(x^2\) < 56

                            \(x^2\) = 49 

                           \(x\) = \(\pm\) 7

Vì \(x\) \(\in\) N*; \(x\) = 7

b;  \(\dfrac{x}{11}\) < \(\dfrac{12}{x}\) < \(\dfrac{x}{9}\)

   9.12<   \(x^2\) < 11.12 

    108 < \(x^2\) < 132

            \(x^2\) = 121

            \(\left[{}\begin{matrix}x=-11\\x=11\end{matrix}\right.\)

    Vì \(x\in\) N*

   \(x\)  = 11

1. ta có: \(\sqrt{\dfrac{4}{9}-\sqrt{\dfrac{25}{36}}}=\sqrt{\dfrac{4}{9}-\dfrac{5}{6}}=\sqrt{-\dfrac{7}{18}}\)

Mà \(-\dfrac{7}{18}\) là số âm \(\Rightarrow\) Bài toán không có kết quả.

2. Ta có:

\(\left(x-1\right)^2=\dfrac{9}{16}\)

\(\Rightarrow\left(x-1\right)^2=\left(\dfrac{3}{4}\right)^2\)

\(\Rightarrow x-1=\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{3}{4}+1\)

\(\Rightarrow x=1\dfrac{3}{4}\)

Vậy \(x=1\dfrac{3}{4}\)

Câu 2 không phải toán lớp 6 mà bạn.

Ta có: \(x=\sqrt{x}\)

\(\Rightarrow x=1\)

Vậy \(x=1\)

Bạn Trần Đăng Nhất làm thiếu nha:

\(x=\sqrt{x}=>x^2=\left(\sqrt{x}\right)^2\)

\(=>x^2=x=>x^2-x=0\)

\(=>x\left(x-1\right)=0\)

\(=>\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Vậy có 2 giá trị của x là 0 và 1..

CHÚC BẠN HỌC TỐT.....