Tổng số tạ hàng đoàn xe đã chở được là:

   92 \(\times\) 4 + 64 \(\times\) 3 = 560 (tạ hàng)

Trung bình mỗi tốp xe chở được số tạ hàng là:

     560 : 2 = 280 (tạ)

Tổng số xe tham gia chở hàng là:

            4 + 3 = 7(xe)

Trung bình mỗi xe chở được số tạ hàng là:

           560 : 7 = 80 (tạ)

Đáp số: a,280 tạ

             b, 80 tạ 

ai giúp mình với

  12647 + 12587 + (2532 : 9) 

= 25234 + \(\dfrac{25432}{9}\)

= \(\dfrac{25234\times9}{9}\) + \(\dfrac{25432}{9}\)

= \(\dfrac{227106}{9}\) + \(\dfrac{25432}{9}\)

= \(\dfrac{252538}{9}\)

\(\dfrac{n}{n+1}\) 

Gọi ước chung lớn nhất của n và n + 1 là d

Ta có : \(\left\{{}\begin{matrix}n⋮d\\n+1⋮d\end{matrix}\right.\)

         ⇔\(\left\{{}\begin{matrix}n⋮d\\n+1-n⋮d\end{matrix}\right.\)

         ⇒ 1 ⋮ d ⇒ d = 1

Vậy Ước chung lớn nhất của n và n + 1 là 1

Hay phân số \(\dfrac{n}{n+1}\) là phân số tối giản

Dễ mà 

1.

a,x+11/30=19/20

=> x=19/20-11/30

=> x=7/12

b,\(x:3\dfrac{1}{5}=1\dfrac{1}{2}\)

\(\Rightarrow x:\dfrac{16}{5}=\dfrac{3}{2}\)

\(\Rightarrow x=\dfrac{3}{2}\cdot\dfrac{16}{5}\)

\(\Rightarrow x=\dfrac{24}{5}\)

c, \(7\cdot x-3\cdot x=3,2\)

\(\Rightarrow x\left(7-3\right)=3,2\)

\(\Rightarrow4x=3,2\)

\(\Rightarrow x=3,2:4\)

\(\Rightarrow x=0,8\)

d, \(x\cdot13-x:\dfrac{1}{4}=36\)

\(\Rightarrow x\cdot13-x\cdot4=36\)

\(\Rightarrow x\left(13-4\right)=36\)

\(\Rightarrow12x=36\)

\(\Rightarrow x=36:12\)

\(\Rightarrow x=3\)

e,\(8,75\cdot x+1,25\cdot x=20\)

\(\Rightarrow x\left(8,75+1,25\right)=20\)

\(\Rightarrow10x=20\)

\(\Rightarrow x=20:10\)

\(\Rightarrow x=2\)

2. 

a. \(\dfrac{3}{5}+25-\dfrac{1}{5}\)

\(=\left(\dfrac{3}{5}-\dfrac{1}{5}\right)+25\)

\(=\dfrac{2}{5}+25\)

\(=\dfrac{127}{5}=25,4\)

b. \(\dfrac{6}{7}:\dfrac{1}{2}\cdot\dfrac{3}{4}-\dfrac{5}{8}\)

\(=\dfrac{6}{7}\cdot2\cdot\dfrac{3}{4}-\dfrac{5}{8}\)

\(=\dfrac{12}{7}\cdot\dfrac{3}{4}-\dfrac{5}{8}\)

\(=\dfrac{9}{7}-\dfrac{5}{8}\)

\(=\dfrac{37}{56}\)

c,\(\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}\right)\cdot\left(1-\dfrac{1}{5}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\)

\(=\dfrac{1\cdot2\cdot3\cdot4}{2\cdot3\cdot4\cdot5}\)

\(=\dfrac{1}{5}\)

d. \(\dfrac{5932+6001\cdot5931}{5932\cdot6001-69}\)

\(=1\)

a. \(8x\left(x-2007\right)-2x+4034=0\)

\(\Rightarrow\left(x-2017\right)\left(4x-1\right)\)

\(\Rightarrow\left[{}\begin{matrix}x-2017=0\\4x-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2017\\4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)

Vậy x=2017 hoặc x=1/4

b.\(\dfrac{x}{2}+\dfrac{x^2}{8}=0\)

\(\Rightarrow\dfrac{x}{2}\left(1+\dfrac{x}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{2}=0\\1+\dfrac{x}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{x}{4}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

Vậy x=0 hoặc x=-4

c.\(4-x=2\left(x-4\right)^2\)

\(\Rightarrow\left(4-x\right)-2\left(x-4\right)^2=0\)

\(\Rightarrow\left(4-x\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4-x=0\\2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{2}\end{matrix}\right.\)

Vậy x=4 hoặc x=7/2

d.\(\left(x^2+1\right)\left(x-2\right)+2x=4\)

\(\Rightarrow\left(x-2\right)\left(x^2+3\right)=0\)

Nxet: (x²+3)>0 với mọi x

=> x-2=0 <=>x=2

Vậy x=2

a, 8\(x\).(\(x-2007\)) - 2\(x\) + 4034 = 0

     4\(x\)(\(x\) - 2007) - \(x\) + 2017 = 0

     4\(x^2\) - 8028\(x\) - \(x\) + 2017 = 0

     4\(x^2\) - 8029\(x\) + 2017 = 0

     4(\(x^2\) - 2. \(\dfrac{8029}{8}\) \(x\) +( \(\dfrac{8029}{8}\))²) - (\(\dfrac{8029}{4}\))²  + 2017 = 0

    4.(\(x\) + \(\dfrac{8029}{8}\))² = (\(\dfrac{8029}{4}\))² - 2017

       \(\left[{}\begin{matrix}x=-\dfrac{8029}{8}+\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\\x=-\dfrac{8029}{8}-\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\end{matrix}\right.\) 

Giúp mk với ai trả lời nhanh mk cho một đúng nhanh lên giúp mk mk đang cần

Gọi thương của phép chia là k

Theo đề bài

\(113k+4=108k+39\)

\(\Leftrightarrow5k=35\Leftrightarrow k=7\)

Số bị chia là

\(113.7+4=795\)

Số chia là

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

\(=x^3+x^2+\dfrac{x}{3}+\dfrac{1}{27}\)

2) \(\left(2x+y^2\right)^3=\left(2x\right)^3+3.\left(2x\right)^2.y^2+3.2x.\left(y^2\right)^2+\left(y^2\right)^3\)

\(=8x^3+12x^2y^2+6xy^4+y^6\)

3) \(\left(\dfrac{1}{2}x^2+\dfrac{1}{3}y\right)^3=\left(\dfrac{1}{2}x^2\right)^3+3.\left(\dfrac{1}{2}x^2\right)^2.\dfrac{1}{3}y+3.\dfrac{1}{2}x^2.\left(\dfrac{1}{3}y\right)^2+\left(\dfrac{1}{3}y\right)^3\)

\(=\dfrac{1}{8}x^6+\dfrac{1}{4}x^4y+\dfrac{1}{6}x^2y^2+\dfrac{1}{27}y^3\)

4) \(\left(3x^2-2y\right)^3=\left(3x^2\right)^3-3.\left(3x^2\right)^2.2y+3.3x^2.\left(2y\right)^2-\left(2y\right)^3\)

\(=27x^6-54x^4y+36x^2y^2-8y^3\)

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

\(=\dfrac{8}{27}x^6-\dfrac{1}{3}x^4y+\dfrac{1}{2}x^2y^2-\dfrac{1}{8}y^3\)

6) \(\left(2x+\dfrac{1}{2}\right)^3=\left(2x\right)^3+3.\left(2x\right)^2.\dfrac{1}{2}+3.2x.\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3\)

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

7) \(\left(x-3\right)^3=x^3-3.x^2.3+3.x.3^2-3^3\)

\(=x^3-9x^2+27x-27\)

8) \(\left(x+1\right)\left(x^2-x+1\right)\)

\(=\left(x+1\right)\left(x^2-x.1+1^2\right)\)

\(=x^3+1^3\)

\(=x+1\)

9) \(\left(x-3\right)\left(x^2+3x+9\right)\)

\(=\left(x-3\right)\left(x^2+x.3+3^2\right)\)

\(=x^3-3^3\)

\(=x^3-27\)

10) \(\left(x-2\right)\left(x^2+2x+4\right)\)

\(=\left(x-2\right)\left(x^2+x.2+2^2\right)\)

\(=x^3-2^3\)

\(=x^3-8\)

11) \(\left(x+4\right)\left(x^2-4x+16\right)\)

\(=\left(x+4\right)\left(x^2-x.4+4^2\right)\)

\(=x^3+4^3\)

\(=x^3+64\)

12) \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)

\(=\left(x-3y\right)\left[x^2+x.3y+\left(3y\right)^2\right]\)

\(=x^3-\left(3y\right)^3\)

\(=x^3-27y^3\)

13) \(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)\)

\(=\left(x^2-\dfrac{1}{3}\right)\left[\left(x^2\right)^2+x^2.\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2\right]\)

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

\(=x^6-\dfrac{1}{27}\)

14) \(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)\)

\(=\left(\dfrac{1}{3}x+2y\right)\left[\left(\dfrac{1}{3}x\right)^2-\dfrac{1}{3}x.2y+\left(2y\right)^2\right]\)

\(=\left(\dfrac{1}{3}x\right)^3+\left(2y\right)^3\)

\(=\dfrac{1}{27}x^3+8y^3\)

39.2+14.3+5.4=140