Gấp rưỡi = 3/2

Sau khi đổ 5 lít từ thùng thứ nhất sang thùng thứ 2 thì tổng số dầu không thay đổi.

Ta có sơ đồ khi đó:

Thùng 1 |-----|-----|            (Tổng: 80 lít)

Thùng 2 |-----|-----|-----|

Tổng số phần bằng nhau là:

2 + 3 = 5 (phần)

Số dầu thùng 1 khi đó là:

80 : 5 x 2 = 32 (lít)

Số dầu thùng 1 lúc đầu là:

32 + 5 = 37 (lít)

Số dầu thùng 2 lúc đầu là:

80 - 37 = 43 (lít)

ĐS: 37 lít

       43 lít

\(x-2\sqrt{x}=0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)

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

Vậy x=0 hoặc x=4 là giá trị cần tìm

11) \(\dfrac{5}{7}\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{4}{7}\right)+\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{4}{7}\right):\dfrac{7}{5}\)

= \(\dfrac{5}{7}\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{4}{7}\right)+\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{4}{7}\right)\cdot\dfrac{5}{7}\)

= \(\dfrac{5}{7}\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{4}{7}+\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{4}{7}\right)\)

= \(\dfrac{5}{7}\cdot0\)

=0

12) \(\dfrac{43}{5}\left(\dfrac{17}{3}-\dfrac{16}{9}+2\right)-\dfrac{43}{5}\left(\dfrac{17}{3}-\dfrac{16}{9}\right)\)

= \(\dfrac{43}{5}\left(\dfrac{17}{3}-\dfrac{16}{9}+2-\dfrac{17}{3}+\dfrac{16}{9}\right)\)

= \(\dfrac{43}{5}\cdot2=\dfrac{43}{10}\)

10( x - 4 ) - 8( x + 2 ) = 12

10x - 40 - 8x + 16 = 12 

=> x( 10 - 8 ) - 40 + 16 = 12

           2x - 40 + 16 = 12

            2x - 40 = -4

                   2x = -4 + 40 = 36

                     x = 36 : 2 = 18 

a , \(\left(\dfrac{-2}{3}+1\dfrac{1}{4}-\dfrac{1}{6}\right):\dfrac{-24}{10}\)

=\(\left(\dfrac{-2}{3}+\dfrac{5}{4}-\dfrac{1}{6}\right):\dfrac{-12}{5}\)

=\(\left(\dfrac{-8}{12}+\dfrac{15}{12}-\dfrac{2}{12}\right)\cdot\dfrac{-5}{12}\)

=\(\dfrac{5}{12}\cdot\dfrac{-5}{12}=\dfrac{-25}{144}\)

b , \(\dfrac{13}{15}\cdot0,25\cdot3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right)1\dfrac{23}{24}\)

=\(\dfrac{13}{15}\cdot\dfrac{1}{4}\cdot3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right)\cdot\dfrac{57}{24}\)

=\(\dfrac{13}{20}-\dfrac{47}{60}\cdot\dfrac{57}{24}\)

=\(\dfrac{13}{20}-\dfrac{893}{480}=\dfrac{312}{480}-\dfrac{893}{480}=\dfrac{-581}{480}\)

c , \(\left(\dfrac{12}{32}+\dfrac{5}{-20}-\dfrac{10}{24}\right):\dfrac{2}{3}\)

=\(\left(\dfrac{180}{480}-\dfrac{120}{480}-\dfrac{200}{480}\right)\cdot\dfrac{3}{2}\)

= \(\dfrac{-7}{24}\cdot\dfrac{3}{2}=\dfrac{-7}{16}\)

d , \(4\dfrac{1}{2}:\left(2,5-3\dfrac{3}{4}\right)+\left(-\dfrac{1}{2}\right)\)

=\(\dfrac{9}{2}:\left(\dfrac{5}{2}-\dfrac{15}{4}\right)-\dfrac{1}{2}\)

=\(\dfrac{9}{2}:\dfrac{-5}{4}-\dfrac{1}{2}=\dfrac{9}{2}\cdot\dfrac{-4}{5}-\dfrac{1}{2}=\dfrac{-18}{5}-\dfrac{1}{2}=\dfrac{-41}{10}\)

e , \(\dfrac{-5}{2}:\left(\dfrac{3}{4}-\dfrac{1}{2}\right)=\dfrac{-5}{2}\left(\dfrac{3}{4}-\dfrac{2}{4}\right)\)

=\(\dfrac{-5}{2}:\dfrac{1}{4}=\dfrac{-5}{2}\cdot4=-10\)

Ta có : 1+\(\dfrac{1}{x+2}\) = \(\dfrac{12}{8-x^3}\) (đkxđ x\(\ne\pm2\) )

\(\Leftrightarrow\) \(\dfrac{1}{x+2}\) = \(\dfrac{12}{8-x^3}-1\)

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

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

\(\Leftrightarrow8-x^3=\left(x+2\right)\left(x^3+4\right)\)

\(\Leftrightarrow8-x^3=x^4+4x+2x^3+8\)

\(\Leftrightarrow-x^3-x^4-4x-2x^3=8-8\)

\(\Leftrightarrow-x^4-3x^3-4x=0\)

\(\Leftrightarrow-x\left(x^3+3x^2+4\right)=0\)

\(\Rightarrow-x=0\)\(\Rightarrow x=0\) (TM x\(\ne\pm2\))

Ta có : 2013\(^{2014}\) = 2013\(^{2013+1}\) = 2013\(^{2013}\)*2013

= 4052169\(^{2013}\)

Mà 4052169>2014 \(\Rightarrow\) 2013\(^{2014}\)>2014\(^{2013}\)