3( x - 1 ) - 2| x + 3 | (*)

Với x < -3 (*) trở thành 3x - 3 + 2( x + 3 ) = 3x - 3 + 2x + 6 = 5x + 3

Với x >= -3 (*) trở thành 3x - 3 - 2( x + 3 ) = 3x - 3 - 2x - 6 = x - 9

f(x) - g(x) + h(x) = 2x³ - 2x² - 3x + 1 - ( 2x³ + x - 2 ) + 2x² + x + 1

= 2x³ - 2x + 2 - 2x³ - x + 2

= -3x + 4

f(x) - g(x) + h(x) = 0 <=> -3x + 4 = 0 <=> x = 4/3

\(5^{x+4}-3.5^{x+3}=2.5^{11}\)

\(\Leftrightarrow5.5^{x+3}-3.5^{x+3}=2.5^{11}\)

\(\Leftrightarrow2.5^{x+3}=2.5^{11}\)

\(\Leftrightarrow x+3=11\)

\(\Leftrightarrow x=8\)

\(\frac{3}{7}+\frac{1}{14}=\frac{6}{14}+\frac{1}{14}=\frac{7}{14}=\frac{1}{2}\)

\(\frac{3}{7}+\frac{1}{14}=\)\(\frac{6}{14}+\frac{1}{14}\)=\(\frac{7}{14}=\frac{1}{2}\)

\(-2.\left|\frac{2}{5}x-\frac{3}{10}\right|=3\frac{1}{5}\)

\(-2.\left|\frac{2}{5}x-\frac{3}{10}\right|=\frac{16}{5}\)

\(\left|\frac{2}{5}x-\frac{3}{10}\right|=\frac{16}{5}\div\left(-2\right)\)

\(\left|\frac{2}{5}x-\frac{3}{10}\right|=-\frac{8}{5}\)

\(\Rightarrow\)\(\frac{2}{5}x-\frac{3}{10}=\frac{8}{5}\)hoặc \(-\frac{8}{5}\)

Ta xét 2 trường hợp :

Th 1 : 

\(\frac{2}{5}x-\frac{3}{10}=\frac{8}{5}\)

\(\frac{2}{3}x=\frac{8}{5}+\frac{3}{10}\)

\(\frac{2}{3}x=\frac{19}{10}\)

\(x=\frac{19}{4}\)

Th 2 :

\(\frac{2}{5}x-\frac{3}{10}=-\frac{8}{5}\)

\(\frac{2}{3}x=-\frac{8}{5}+\frac{3}{10}\)

\(\frac{2}{3}x=\frac{-13}{10}\)

\(x=\frac{-13}{4}\)

Vậy \(x\in\){ \(\frac{19}{4}\); \(\frac{-13}{4}\)}

Ta có : \(\widehat{K_1}=\widehat{K_3}=60^0\)(do đối đỉnh)

Ta lại có : \(\widehat{K_3}+\widehat{K_2}=180^0\)(do kề bù)

mà \(\widehat{H_1}+\widehat{K_2}+\widehat{K_3}=240^0\)

\(\Rightarrow\widehat{H_1}=60^0\)

\(\Rightarrow\widehat{H_1}=\widehat{K_3}\)

mà 2 góc này là 2 góc ở vị trí đồng vị

\(\Rightarrow a//b\)