Diện tích đáy hình vuông của thùng đó là :

3*3 = 9 (dm²)

Đổi 45 l = 45dm³

Vậy chiều cao của dầu trong thùng là :

45:9 = 5(dm)

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

\(\Leftrightarrow3x-\dfrac{3}{2}-5x-3=-x+\dfrac{1}{5}\)

\(\Rightarrow3x-5x+x=\dfrac{1}{5}+3+\dfrac{3}{2}\)

\(-x=\dfrac{47}{10}\)

=>\(x=-\dfrac{47}{10}\)

Ta có \(\dfrac{1}{12};\dfrac{1}{13};..;\dfrac{1}{19}< \dfrac{1}{11}\)

=> \(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+..+\dfrac{1}{19}< \dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+..+\dfrac{1}{11}\) (9 số hạng )

\(C< 9\cdot\dfrac{1}{11}=\dfrac{9}{11}< 1\) (1)

Vì \(\dfrac{1}{11};\dfrac{1}{12};\dfrac{1}{13};...;\dfrac{1}{19}>0\) => C>0 (2)

Từ (1);(2) có 0<C<1

=> C không phải là số nguyên 

Một giờ vòi thứ 1 chảy được số phần bể là :

1:10 \(=\dfrac{1}{10}\) (bể)

Một giờ vòi thứ 2 chảy đc số phần bể là :

\(1:15=\dfrac{1}{15}\) (bể)

Vì \(\dfrac{1}{10}>\dfrac{1}{15}\) nên trong 1h vòi 1 sẽ chảy nhanh hơn vòi 2 

Trong 1h vòi 1 nhanh hơn vòi 2 số phần bể là 

\(\dfrac{1}{10}-\dfrac{1}{15}=\dfrac{1}{30}\)

\(y\cdot\dfrac{1}{3}+y:\dfrac{3}{5}\) \(+y\cdot\dfrac{1}{2}:\dfrac{3}{4}=8\)

\(y\cdot\dfrac{1}{3}+y\cdot\dfrac{5}{3}+y\cdot\) \(\dfrac{1}{2}\cdot\dfrac{4}{3}=8\)

\(y\cdot\dfrac{1}{3}+y\cdot\dfrac{5}{3}+y\cdot\dfrac{2}{3}=8\)

\(y\left(\dfrac{1}{3}+\dfrac{5}{3}+\dfrac{2}{3}\right)=8\)

\(y\cdot\dfrac{8}{3}=8\)

\(y=8:\dfrac{8}{3}=8\cdot\dfrac{3}{8}=3\)

\(\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{80}+\dfrac{1}{120}+\dfrac{1}{168}+\dfrac{1}{224}\)

\(=\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+\dfrac{1}{8\cdot10}+\dfrac{1}{10\cdot12}+\dfrac{1}{12\cdot14}+\dfrac{1}{14\cdot16}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+\dfrac{2}{8\cdot10}+\dfrac{2}{10\cdot12}+\dfrac{2}{12\cdot14}+\dfrac{2}{14\cdot16}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{14}-\dfrac{1}{16}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{16}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{7}{16}=\dfrac{7}{32}\)

\(\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}+\dfrac{1}{9\cdot11}+...+\dfrac{1}{21\cdot23}+\dfrac{1}{23\cdot25}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+...+\dfrac{2}{21\cdot23}+\dfrac{2}{23\cdot25}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+...+\dfrac{1}{21}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{25}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{5}-\dfrac{1}{25}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{4}{25}=\dfrac{2}{25}\)

Ta có 

\(A=\dfrac{1}{1\cdot2}\cdot\dfrac{4}{2\cdot3}\cdot\dfrac{9}{3\cdot4}\cdot\dfrac{16}{4\cdot5}..\cdot\dfrac{2500}{50\cdot51}\)

\(A=\dfrac{1^2}{1\cdot2}\cdot\dfrac{2^2}{2\cdot3}\cdot\dfrac{3^2}{3\cdot4}\cdot\dfrac{4^2}{4\cdot5}+...+\dfrac{50^2}{50\cdot51}\)

\(A=\dfrac{1^2\cdot2^2\cdot3^2\cdot...\cdot50^2}{1\cdot2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5\cdot..\cdot50\cdot51}\)

\(A=\dfrac{1^2\cdot2^2\cdot3^2\cdot...\cdot50^2}{1\cdot2^2\cdot3^2\cdot4^2\cdot...\cdot50^2\cdot51}\)\(=\dfrac{1}{51}\)

Có \(A=\dfrac{1}{51}=\dfrac{13}{663}< B=\dfrac{13}{150}\)

a, \(M\left(x\right)+N\left(x\right)=\) \(\left(-6x^2-7+2x+5x\right)+\left(12+6x^2-4x-3x\right)\)

\(M\left(x\right)+N\left(x\right)=-6x^2-7+2x+5x+12+6x^2-4x-3x\)

\(M\left(x\right)+N\left(x\right)=\left(-6x^2+6x^2\right)+\left(2x+5x-4x-3x\right)+\left(-7+12\right)\)

\(M\left(x\right)+N\left(x\right)=5\)

b, \(M\left(x\right)-N\left(x\right)=\left(-6x^2-7+2x+5x\right)-\left(12+6x^2-3x-4x\right)\)

\(M\left(x\right)-N\left(x\right)=-6x^2-7+2x+5x-12-6x^2+3x+4x\)

\(M\left(x\right)+N\left(x\right)=\left(-6x^2-6x^2\right)+\left(2x+5x+3x+4x\right)+\left(-7-12\right)\)

\(M\left(x\right)+N\left(x\right)=-12x^2+14x-19\)

c, \(P\left(x\right)=N\left(x\right)+4x3+3x-12\)

\(P\left(x\right)=\left(12+6x^2-4x-3x\right)+12x+3x-12\)

\(P\left(x\right)=12+6x^2-4x-3x-12x+3x-12\)

\(P\left(x\right)=6x^2+\left(-3x-4x+12x+3x\right)+\left(12-12\right)\)

\(P=6x^2+8x\)

Bậc của đa thức:2

Hệ số cao nhất :6

Hệ số tự do :0

So sánh bằng cách tìm phân số trung gian nhé 

Có \(\dfrac{n}{n+3}>\dfrac{n}{n+4}\) 

mà \(\dfrac{n}{n+4}>\dfrac{n-1}{n+4}\) 

=> \(\dfrac{n}{n+3}>\dfrac{n-1}{n+4}\)