+2222222222

B=1+2-3-4+5+6-7-8+9+10-11-12+...-299-300+301+302

Có tất cả:(300-1):1+1=300(số)

Nếu ta nhóm 4 số hạng vào một nhóm thì ta được:300:4=75(cặp)

B=(1+2-3-4)+(5+6-7-8)+(9+10-11-12)+...+(287+288-299-300)+(301+302)

B=(-4)+(-4)+(-4)+...+(-4)+(301+302)

B=75.(-4)+603

B=(-300)+603

B=303

Hình như bạn viết sai đề rồi phải là 300 chứ sao lại là 330 cái chỗ 299-330 đấy

B=303

k mk mk k lại cho

chép đúng đề mà,là 330 chứ k phải 300 đâu

a, |3x - 6| + 4x + 3 = 12

|3x - 6| = 12 - 3 - 4x

|3x - 6| = 9 - 4x

TH1: 3x - 6 = 9 - 4x ⇔ 3x + 4x = 9 + 6 ⇔ 7x = 15 ⇔ x = 15/7 

TH2: 3x - 6 = 4x - 9 ⇔ 3x - 4x = -9 + 6 ⇔ -x = -3 ⇔ x = 3. 

b, |6x - 12| + 3x - 1 = 43

|6x - 12| = 43 + 1 - 3x

|6x - 12| = 44 - 3x

TH1: 6x - 12 = 44 - 3x ⇔ 6x + 3x = 44 + 12 ⇔ 9x = 56 ⇔ x = 6.

TH2: 6x - 12 = 3x - 44 ⇔ 6x - 3x = -44 + 12 ⇔ 3x = -32 ⇔ x = -32/3.

c, |8 - 4x| + 7x - 5 = 16 - 2x

|8 - 4x| = (16 + 5) + (-2x - 7x)

|8 - 4x| = 21 - 9x

TH1: 8 - 4x = 21 - 9x ⇔ -4x + 9x = 21 - 8 ⇔ 5x = 13 ⇔ x = 13/5

TH2: 8 - 4x = 9x - 21 ⇔ -4x - 9x = -21 - 8 ⇔ -15x = -29 ⇔ x = 29/15

Giải:

a) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)  

     \(\dfrac{-5}{6}-x=\dfrac{1}{4}\)

               \(x=\dfrac{-5}{6}-\dfrac{1}{4}\) 

               \(x=\dfrac{-13}{12}\) 

b) \(2.\left(x-\dfrac{1}{3}\right)=\left(\dfrac{1}{3}\right)^2+\dfrac{5}{9}\) 

    \(2.\left(x-\dfrac{1}{3}\right)=\dfrac{1}{9}+\dfrac{5}{9}\) 

    \(2.\left(x-\dfrac{1}{3}\right)=\dfrac{2}{3}\)  

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

             \(x-\dfrac{1}{3}=\dfrac{1}{3}\) 

                    \(x=\dfrac{1}{3}+\dfrac{1}{3}\) 

                    \(x=\dfrac{2}{3}\) 

c) \(\left|2x-\dfrac{3}{4}\right|-\dfrac{3}{8}=\dfrac{1}{8}\) 

           \(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{8}+\dfrac{3}{8}\) 

            \(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{2}\) 

\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{1}{2}\\2x-\dfrac{3}{4}=\dfrac{-1}{2}\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{1}{8}\end{matrix}\right.\) 

d) \(\dfrac{2}{3}x+\dfrac{1}{6}x=3\dfrac{5}{8}\) 

\(x.\left(\dfrac{2}{3}+\dfrac{1}{6}\right)=\dfrac{29}{8}\) 

            \(x.\dfrac{5}{6}=\dfrac{29}{8}\) 

                \(x=\dfrac{29}{8}:\dfrac{5}{6}\) 

                \(x=\dfrac{87}{20}\)

Tìm các cặp số x nguyên thỏa mãn: (12x - 1)(6x - 1)(4x - 1)(3x - 1) = 330 

<=> (12x - 1)2(6x - 1)3(4x - 1)4(3x - 1) = 330.24 
<=> (12x - 1)(12x - 2)(12x - 3)(12x - 4) = 7920 
<=> [ (12x - 1)(12x - 4)] [ (12x - 3)(12x - 2) ] - 7920 = 0 
<=> (144x² - 60x + 4)(144x² - 60x + 6) - 7920 = 0 

Đặt (144x² - 60x + 4) = t 
=> t(t + 2) - 7920 = 0 
=> t² + 2t - 7920 = 0 

∆' = 1² + 7920 = 7921 => √∆' = 89 

=> t1 = - 90 hay t2 = 88 

Khi t = - 90 
=> (144x² - 60x + 4) = -90 
=> 144x² - 60x + 94 = 0 
=> 72x² - 30x + 47 = 0 
∆' = (-15)² - 47.72 = - 3159 => (loại) 

Khi t = 88 
=> (144x² - 60x + 4) = 88 
=> 144x² - 60x - 84 = 0 
=> 36x² - 15x + 21 = 0 
∆ = (-15)² + 4.36.21 = 3249 => √∆ = 57 
=> x = 1 hay x = -42/2.36 (loại vì x là số nguyên) 

Đáp số: x = 1

k mk nha

(12x-1)(6x-1)(4x-1)(3x-1)=330 
<=> (36x^2-15x+1) (24x^2-10x+1)=330(1)
Đặt y= 12x^2-5x+1, pt (1) trở thành:
(3y-5)2(y-1)=330
<=>2(3y^2 -4y+5)-330=0
<=>6y^2-8y-320=0
<=>(y-8)(6y+40)=0
<=>y-8=0 hay 6y+40=0
<=>y=8 hay y=-20/3
*Với y=8, ta có:
12x^2-5x+1=8 
<=>12x^2-5x-7=0
<=>(x-1)(12x+7)=0
<=>x-1=0 hay 12x+7=0
<=>x=1 hay x=-7/12
*Với y=-20/3, ta có 
Làm tương tự 
Tập nghiệm S={1;-7/12;0.28;-0.16}

\(\dfrac{5}{x}+1+\dfrac{4}{x}+1=\dfrac{3}{-13}\\ \Rightarrow\dfrac{9}{x}+2=-\dfrac{3}{13}\\ \Rightarrow\dfrac{9}{x}=-\dfrac{59}{13}\\ \Rightarrow x=-\dfrac{207}{59}\)

a. \(\dfrac{5}{x+1}+\dfrac{4}{x+1}=\dfrac{-3}{13}\)

ĐKXĐ: x ≠ -1

⇔ \(\dfrac{65}{13\left(x+1\right)}+\dfrac{52}{13\left(x+1\right)}=\dfrac{-3\left(x+1\right)}{13\left(x+1\right)}\)

⇔ 65 + 52 = -3(x + 1)

⇔ 117 = -3x - 3

⇔ 117 + 3 = -3x

⇔ 120 = -3x 

⇔ x = \(\dfrac{120}{-3}=-40\) (TM)

b. -x + 2 + 2x + 3 + x + \(\dfrac{1}{4}\) + 2x + \(\dfrac{1}{6}\) = \(\dfrac{8}{3}\)

⇔ -x + 2x + x + 2x = \(\dfrac{8}{3}-\dfrac{1}{6}-\dfrac{1}{4}-3-2\)

⇔ 4x = -2,75

⇔ x = \(\dfrac{-2,75}{4}=\dfrac{-11}{16}\)

c. \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+2}\) = \(\dfrac{12}{26}\)

⇔  \(\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{2\left(3x+1\right)}=\dfrac{12}{26}\)

⇔ \(\dfrac{312\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\) + \(\dfrac{520\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\) - \(\dfrac{312\left(2x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\)

= \(\dfrac{48\left(2x+1\right)\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\)

⇔ 312(3x +1) + 520(3x + 1) - 312(2x + 1) = 48(2x + 1)(3x + 1)

⇔ 936x + 312 + 1560x + 520 - 624x - 312 = (96x + 48)(3x + 1)

⇔ 936x + 312 + 1560x + 520 - 624x - 312 = 288x² + 96x + 144x + 48

⇔ 936x + 1560x - 624x - 96x - 144x - 288x² = 48 - 312 - 520 + 312

⇔ 1632x - 288x² = -472

⇔ -288x² + 1632x + 472 = 0 (Tự giải tiếp, dùng phương pháp tách hạng tử)

⇔ x = 5,942459684 \(\approx\) 6

c: Ta có: \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\)

\(\Leftrightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)

\(\Leftrightarrow2x+1=13\)

hay x=6

a) \(\frac{-x}{2}+\frac{2x}{3}+x+\frac{1}{4}+2x+\frac{1}{6}=\frac{3}{8}.\)

\(\frac{-x}{2}+\frac{2x}{3}+3x+\frac{5}{12}=\frac{3}{8}\)

\(x.\left(-\frac{1}{2}+\frac{2}{3}+3\right)+\frac{5}{12}=\frac{3}{8}\)

\(x\cdot\frac{19}{6}=-\frac{1}{24}\)

x = -1/76

b) \(\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{12}{26}\)

\(\frac{3}{2x+1}+\frac{2.5}{2.\left(2x+1\right)}-\frac{2.3}{3.\left(2x+1\right)}=\frac{6}{13}\)

\(\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{6}{13}\)

\(\frac{3+5-2}{2x+1}=\frac{6}{13}\)

\(\frac{6}{2x+1}=\frac{6}{13}\)

=> 2x + 1 = 13

2x = 12

x = 6

\(a,-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+1}{6}=\dfrac{8}{3}\)

\(\Rightarrow-\dfrac{6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{8}{3}\)

\(\Rightarrow\dfrac{-6x+8x+3x+3+4x+2}{12}=\dfrac{8}{3}\)

\(\Rightarrow\dfrac{9x+5}{12}=\dfrac{8}{3}\)

\(\Rightarrow27x+15=96\)

\(\Rightarrow27x=81\)

\(\Rightarrow x=3\left(tm\right)\)

\(b,\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3+5-2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow2x+1=13\)

\(\Rightarrow2x=12\)

\(\Rightarrow x=6\left(tm\right)\)

#Toru

a) \(-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+2}{6}=\dfrac{8}{3}\) 

\(\Rightarrow\dfrac{-6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{4\cdot8}{12}\)

\(\Rightarrow-6x+8x+3x+3+4x+2=32\)

\(\Rightarrow9x+5=32\)

\(\Rightarrow9x=32-5\)

\(\Rightarrow9x=27\)

\(\Rightarrow x=\dfrac{27}{9}\)

\(\Rightarrow x=3\)

b) \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\) (ĐK: \(x\ne-\dfrac{1}{2}\)) 

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow2x+1=13\)

\(\Rightarrow2x=12\)

\(\Rightarrow x=\dfrac{12}{2}\)

\(\Rightarrow x=6\left(tm\right)\)