a . \(\left|\frac{5}{2}x-3,5\right|=\frac{2}{3}\)

\(\Rightarrow\)\(\frac{5}{2}x-3,5=\frac{2}{3}\)hoặc \(-\frac{2}{3}\)

\(\Rightarrow\)\(\frac{5}{2}x=\frac{2}{3}+3,5\)hoặc \(-\frac{2}{3}+3,5\)

\(\Rightarrow\)\(\frac{5}{2}x=\frac{2}{3}+\frac{7}{2}\)hoặc \(-\frac{2}{3}+\frac{7}{2}\)

\(\Rightarrow\)\(\frac{5}{2}x=\frac{25}{6}\)hoặc \(\frac{17}{6}\)

\(\Rightarrow\)\(x=\frac{5}{3}\)hoặc \(\frac{17}{15}\)
Vậy \(x\in\){ \(\frac{5}{3}\); \(\frac{17}{15}\)}

b. \(\left|\frac{4}{6}x-\frac{5}{8}\right|=18+2.\left(-8\right).\)

    \(\left|\frac{4}{6}x-\frac{5}{8}\right|=18+\left(-16\right)\)

    \(\left|\frac{4}{6}x-\frac{5}{8}\right|=2\)

\(\Rightarrow\)\(\frac{4}{6}x-\frac{5}{8}=2\)hoặc \(-2\)

\(\Rightarrow\)\(\frac{4}{6}x=2+\frac{5}{8}\)hoặc \(-2+\frac{5}{8}\)

\(\Rightarrow\)\(\frac{4}{6}x=\frac{21}{8}\)hoặc \(\frac{-11}{8}\)

\(\Rightarrow\)\(x=\frac{63}{16}\)hoặc \(\frac{-33}{16}\)

Vậy \(x\in\){ \(\frac{63}{16}\); \(\frac{-33}{16}\)}

a)\(x=-\frac{4}{7}\Rightarrow\left|x\right|=\left|-\frac{4}{7}\right|=\frac{4}{7}\)

b)\(x=\frac{-3}{-11}=\frac{3}{11}\Rightarrow\left|x\right|=\left|\frac{3}{11}\right|=\frac{3}{11}\)

c)\(x=-0,749\Rightarrow\left|x\right|=\left|-0,749\right|=0,749\)

d)\(x=-34\Rightarrow\left|x\right|=\left|-34\right|=34\)

So sánh:

2016^20+2016^19+7/2016^18+2016^19 và 20^20+20^19+10/20^18+20^19

2016^20+2016^19+7/2016^18+2016^19 > 20^20+20^19+10/20^18+20^19

giải chi tiết hộ mik

a)  Ta có (6x² - 3xy²) + M = x² + y² - 2xy²

<=> M = x² +  y² - 2xy² - (6x² - 3xy²) 

<=> M = x² +  y² - 2xy² - 6x² + 3xy²

<=> M = -5x² + xy² + y²

b) Ta có M - (2xy - 4y²) = 5xy + x² - 7y²

<=> M = 5xy + x² - 7y² + 2xy - 4y² 

<=> M = x² + 7xy - 11y² 

câu nào

\(2\times\left|x-\frac{2}{3}\right|-\frac{1}{5}=0.\)

\(2\times\left|x-\frac{2}{3}\right|=\frac{1}{5}\)

\(\left|x-\frac{2}{3}\right|=\frac{1}{5}\): \(2\)

\(\left|x-\frac{2}{3}\right|=\frac{1}{10}\)

\(\Rightarrow\)\(x-\frac{2}{3}=\frac{1}{10}\)hoặc \(-\frac{1}{10}\)

\(\Rightarrow\)\(x=\frac{1}{10}+\frac{2}{3}\)hoặc \(-\frac{1}{10}+\frac{2}{3}\)

\(\Rightarrow\)\(x=\frac{23}{30}\)hoặc \(\frac{17}{30}\)

Vậy \(x\in\){ \(\frac{23}{30}\); \(\frac{17}{30}\)}

\(2\left|x-\frac{2}{3}\right|-\frac{1}{5}=0\)

\(\Leftrightarrow2\left|x-\frac{2}{3}\right|=\frac{1}{5}\)

\(\Leftrightarrow\left|x-\frac{2}{3}\right|=\frac{1}{10}\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{2}{3}=\frac{1}{10}\\x-\frac{2}{3}=-\frac{1}{10}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{23}{30}\\x=\frac{17}{30}\end{cases}}\)

Vậy ....

- 9,01 - 9,02 - 9,03 - .... - 9,99

= [ ( -9,99 ) + ( -9,01 ) ] . 99 : 2

= -19 . 99 : 2

= -1,881 : 2

= -940,5

\(\frac{5}{4}\)\(-\)\(|x+\frac{1}{3}|\)\(=\)\(\frac{1}{5}\)

\(|x+\frac{1}{3}|\)\(=\)\(\frac{5}{4}\)\(-\)\(\frac{1}{5}\)

\(|x+\frac{1}{3}|\)\(=\)\(\frac{21}{20}\)

TH1 : \(x+\frac{1}{3}\)\(=\)\(\frac{21}{20}\)

\(x=\)\(\frac{21}{20}\)\(-\)\(\frac{1}{3}\)

\(x=\)\(\frac{43}{60}\)

TH2: \(x+\frac{1}{3}\)\(=\)\(\frac{-21}{20}\)

\(x=\)\(\frac{-21}{20}\)\(-\)\(\frac{1}{3}\)

\(x=\)\(\frac{-83}{60}\)

Vậy x = \(\frac{43}{60}\)hoặc x = \(\frac{-83}{60}\)

\(\left(3x+\frac{3}{5}\right).\left(\left|x\right|-\frac{1}{4}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x+\frac{3}{5}=0\\\left|x\right|-\frac{1}{4}=0\end{cases}}\Rightarrow\orbr{\begin{cases}3x=\frac{3}{5}\\\left|x\right|-\frac{1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x = \frac{1}{5}\\x∈ \left\{\frac{-1}{4};\frac{1}{4}\right\}\end{cases}}\)