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 = \(\frac{-83}{60}\)

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

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

Ta có: 

\(\left|3x-1\right|+\left|x+4\right|\ge\left|3x-1+x+4\right|=\left|4x+3\right|\)

Dấu \(=\)khi \(\left(3x-1\right)\left(x+4\right)\ge0\Leftrightarrow\orbr{\begin{cases}x\ge\frac{1}{3}\\x\le-4\end{cases}}\).

Vậy nghiệm của phương trình đã cho là \(x\in(-\infty,-4]\)và \(x\in[\frac{1}{3},+\infty)\).

a

C= |x-1| + |x-5|

Do x-1 + x-5 luôn > 0

=> x-1 + x-5 = 0

=> 2x -6 = 0

=> 2x = 6

=> x = 3

mình ghi nhầm, lớn hơn hoặc bằng 0 nha