`\hat{A}/\hat{B}=3/15=>\hat{B}=15/3\hat{A}`

Xét `\triangle ABC` có:

  `\hat{A}+\hat{B}+\hat{C}=180^o`

 `=>\hat{A}+15/3\hat{A}+4\hat{A}=180^o`

 `=>10\hat{A}=180^o =>\hat{A}=18^o`

      `=>{(\hat{B}=15/3 .18^o =90^o ),(\hat{C}=4.18^o =72^o ):}`

| x +1 | + |y +2| + 3

|x  - 1| ≥ 0

|y +2| ≥ 0

|x - 1| + | y +2 | + 3 ≥ 3

dấu = xảy ra ⇔ x = 1 và y = -2

c= 2017
a + b + 2017 = 2018
a - b +  2017  = 2019.

...........
f(2 ) = 4 a + 2 b + 2017 

Cách làm bài 4 b giống 4 a là bỏ trị tuyệt đối , bạn có 2 số

đáp số bài 4 b 
x = 3/20 hay x = - 7/20

\(\dfrac{2x}{3}\) = \(\dfrac{3y}{4}\) = \(\dfrac{4z}{5}\) và x + y + z = 4

\(\dfrac{12x}{18}\) = \(\dfrac{12y}{16}\) = \(\dfrac{12y}{15}\)= \(\dfrac{12x+12y+12z}{18+16+15}\)= \(\dfrac{12\left(x+y+z\right)}{49}\)= \(\dfrac{12.4}{49}\) =\(\dfrac{48}{49}\)

x = \(\dfrac{48}{49}\): \(\dfrac{12}{18}\)= \(\dfrac{72}{49}\), y = \(\dfrac{48}{49}\): \(\dfrac{12}{16}\) = \(\dfrac{64}{49}\), z = \(\dfrac{48}{49}\): \(\dfrac{12}{15}\) = \(\dfrac{60}{49}\)

1\(\dfrac{2}{3}\): x = - \(\dfrac{4}{3}\)

\(\dfrac{5}{3}\): X = \(\dfrac{-4}{3}\)

X =          \(\dfrac{5}{3}\)   : \(\dfrac{-4}{3}\) 

X = \(\dfrac{5}{3}\) . \(\dfrac{-3}{4}\)

X = \(\dfrac{-5}{4}\)

-\(\dfrac{2}{7}\) . X = \(\dfrac{-1}{4}\)

X = \(\dfrac{-1}{4}\) : \(\dfrac{-2}{7}\)

X = \(\dfrac{1}{4}\) . \(\dfrac{7}{2}\)

X =  \(\dfrac{7}{8}\)