1. a) Ta có: M  = |x + 15/19| \(\ge\)0 \(\forall\)x

Dấu "=" xảy ra <=> x + 15/19 = 0 <=> x = -15/19

Vậy MinM = 0 <=> x = -15/19

b) Ta có: N = |x  - 4/7| - 1/2 \(\ge\)-1/2 \(\forall\)x

Dấu "=" xảy ra <=> x - 4/7 = 0 <=> x = 4/7

Vậy MinN = -1/2 <=> x = 4/7

2a) Ta có: P = -|5/3 - x|  \(\le\)0 \(\forall\)x

Dấu "=" xảy ra <=> 5/3 - x = 0 <=> x = 5/3

Vậy MaxP = 0 <=> x = 5/3

b) Ta có: Q = 9 - |x - 1/10| \(\le\)9 \(\forall\)x

Dấu "=" xảy ra <=> x - 1/10 = 0 <=> x = 1/10

Vậy MaxQ = 9 <=> x = 1/10

a) 

( 4x - 9 ) ( 2,5 + (-7/3) . x ) = 0

\(\Rightarrow\orbr{\begin{cases}4x-9=0\\2,5+\frac{-7}{3}x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{9}{4}\\x=\frac{15}{14}\end{cases}}\)

P/s: đợi xíu làm câu b

b) \(\frac{1}{x\left(x+1\right)}\cdot\frac{1}{\left(x+1\right)\left(x+2\right)}\cdot\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2015}\)

\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2015}\)

\(\frac{-1}{x+3}=\frac{1}{2015}\)

\(\Leftrightarrow x+3=-2015\)

\(\Leftrightarrow x=-2018\)

Vậy,.........

b)

\(x-2.\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)

\(x-2\cdot\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)

\(x-2=\frac{16}{9}:\left(\frac{1}{3}-\frac{1}{9}\right)\)

\(x-2=8\)

=> x = 10

a) 

\(A=\frac{1}{2}.\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{2013}{2014}\cdot\frac{2014}{2015}\cdot\frac{2015}{2016}\)

\(A=\frac{1}{2016}\)

4a) \(\frac{-2}{3}x=\frac{3}{10}-\frac{1}{5}=\frac{1}{10}\)

\(\Leftrightarrow x=\frac{1}{10}:\frac{-2}{3}=\frac{1}{10}.\frac{3}{-2}=\frac{3}{-20}\)

Vậy x=\(\frac{3}{-20}\)

b) \(\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}\)

\(\Leftrightarrow\left(\frac{2}{3}-\frac{3}{2}\right)x=\frac{5}{12}\)

\(\Leftrightarrow\frac{-5}{6}x=\frac{5}{12}\)

\(\Leftrightarrow x=\frac{5}{12}:\frac{-5}{6}=\frac{5}{12}.\frac{6}{-5}=\frac{1}{-2}\)

Vậy x=\(\frac{1}{-2}\)

g)Sửa đề: \(\left|4x-1\right|=\left(-3\right)^2\)

\(\Leftrightarrow\left|4x-1\right|=9\)

\(\Rightarrow\left[{}\begin{matrix}4x-1=9\\4x-1=\left(-9\right)\end{matrix}\right.\)

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

Vậy \(x\in\left\{\frac{5}{2};-2\right\}\)

i) \(\left(x-1^3\right)=125\)

\(\Leftrightarrow x-1=125\)

\(\Leftrightarrow x=125+1=126\)

Vậy x=126

k) \(\left(x+\frac{1}{2}\right).\left(\frac{2}{3}-2x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{matrix}\right.\)

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

Vậy \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)

xin lỗi câu h tui xin chữa lại là:\(|x+70\%|=2\frac{1}{5}\)

a)\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\Leftrightarrow x\left(x-1\right)^{x+2}\left(x-2\right)=0\)

Do đó \(x\in\left\{0;1;2\right\}\)

b)

\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot...\cdot\frac{31}{64}=2^x\Leftrightarrow\frac{1\cdot2\cdot3\cdot...\cdot31}{4\cdot6\cdot8\cdot...\cdot64}=2^x\Leftrightarrow\frac{31!}{\left(2\cdot2\right)\cdot\left(2\cdot3\right)\cdot\left(2\cdot4\right)\cdot...\cdot\left(2\cdot31\right)\cdot64}=2^x\)

\(\frac{31!}{2^{30}\cdot31!\cdot2^6}=2^x\Leftrightarrow\frac{1}{2^{36}}=2^x\Leftrightarrow2^{-36}=2^x\Rightarrow x=-36\)

Câu b:

\(\frac{21}{8}:\frac{5}{6}+\frac{1}{2}:\frac{5}{6}\)

= \(\frac{63}{20}+\frac{3}{5}\)

= \(\frac{15}{4}\)

\(\left(\frac{21}{8}+\frac{1}{2}\right):\frac{5}{6}\)

\(\frac{25}{8}:\frac{5}{6}\)

\(\frac{25}{8}.\frac{6}{5}\)

\(\frac{30}{8}\)