x² - 3x + 2 = 0

<=> x² - 2x - x + 2 = 0

<=> x(x  - 2) - (x - 2) = 0

<=> (x - 1)(x - 2) = 0

<=> x - 1 = 0 hoặc x - 2 = 0

<=> x = 1 hoặc x = 2

Vậy S = {1; 2}

`x^2 - 3x + 2 = 0`

$\rightrightarrows$ `x^2 - x + 2-2x=0`

$\rightrightarrows$ `x( x-1 ) - 2( x-1 ) = 0`

$\rightrightarrows$ `(x-2)(x-1)=0`

$\rightrightarrows$ \(\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\)

$\rightrightarrows$ \(\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Vậy `x in { 1;2}` 

a) \(=-\dfrac{3}{7}.\left(\dfrac{-2}{18}+\dfrac{-7}{18}+\dfrac{15}{18}\right)=-\dfrac{3}{7}.\dfrac{1}{3}=-\dfrac{1}{7}\)

b) Có các cặp: a=80,b=48 ; a=48,b=80 ; a=112,b=16 ; a=16,b=112

c)\(=\dfrac{2}{3}.x-\dfrac{700}{11}:\left(\dfrac{13}{15}+\dfrac{13}{35}+\dfrac{13}{63}+\dfrac{13}{99}\right)=-5\)

\(\dfrac{2}{3}.x-\dfrac{700}{11}:\dfrac{52}{33}=-5\)

\(\dfrac{2}{3}.x=-\dfrac{590}{13}\)

\(x=-\dfrac{590}{13}:\dfrac{2}{3}=-\dfrac{885}{13}\)

`\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 = \(\dfrac{48}{49}\): \(\dfrac{12}{15}\) = \(\dfrac{60}{49}\)