\(\dfrac{2x^2-x}{x-1}+\dfrac{x+1}{1-x}+\dfrac{2-x^2}{x-1}.\left(x\ne1\right).\)

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

hi

`30+2,8:(4/25 - 9/15 -1/6)`

`=30+2,8 :( -11/25-1/6)`

`=30+14/5 :(-19/150)`

`=30-60/13`

`=330/13`

\(30+2,8:\left(\dfrac{4}{25}-\dfrac{9}{15}-\dfrac{1}{6}\right)=30+\dfrac{14}{5}:\dfrac{-91}{150}=30+\dfrac{-60}{13}=\dfrac{330}{13}\)

b: \(=\dfrac{x+5+x+x-5}{x\left(x+5\right)}=\dfrac{3x}{x\left(x+5\right)}=\dfrac{3}{x+5}\)

\(a,=-3x^3+x^2+9x^2-3x-12x+4=-3x^3+10x^2-15x+4\\ b,=\dfrac{x+5+x+x-5}{x\left(x+5\right)}=\dfrac{3x}{x\left(x+5\right)}=\dfrac{3}{x+5}\)

\(\dfrac{\dfrac{2}{3}+0,25-0,6}{\dfrac{2}{3}-0,25+0,6}:\dfrac{\dfrac{2}{5}-\dfrac{1}{6}+\dfrac{3}{7}}{\dfrac{2}{5}+\dfrac{1}{6}-\dfrac{3}{7}}\)

\(=\dfrac{\dfrac{19}{60}}{\dfrac{61}{60}}:\dfrac{\dfrac{139}{210}}{\dfrac{29}{210}}=\dfrac{19}{61}:\dfrac{139}{29}=\dfrac{551}{8479}\)

Mình không chắc lắm!! Chúc bạn học tốt!!!

\(\dfrac{5x+5y}{3x-3y}:\dfrac{5x}{x^2-y^2}.\)

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

\(=\dfrac{x+y}{3}.\dfrac{x+y}{x}.\)

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

\(\dfrac{1}{38}>\dfrac{1}{40}>\dfrac{1}{42}>...>\dfrac{1}{50}\)

=>\(\dfrac{1}{38}+\dfrac{1}{40}+\dfrac{1}{42}+\dfrac{1}{44}+\dfrac{1}{46}+\dfrac{1}{48}+\dfrac{1}{50}< 7\cdot\dfrac{1}{38}=\dfrac{7}{38}< 1\)

Vậy tổng trên bé hơn 1

A=-1-3-5-...-2017

=-(1+3+5+...+2017)

Xét tổng B=1+3+5+...+2017

Tổng B có:(2017-1):2+1=1009(số hạng)

Tổng B=\(\dfrac{\left(2017+1\right)\cdot1009}{2}=1009\cdot1009=1018081\)

=>A=-B=-1018081

\(\dfrac{1}{38}+\dfrac{1}{40}+\dfrac{1}{42}+...+\dfrac{1}{50}\) có: \(\left(50-38\right):2+1\)= \(7\) (số hạng)

Ta có: \(\dfrac{1}{38}< \dfrac{1}{7};\dfrac{1}{40}< \dfrac{1}{7};\dfrac{1}{42}< \dfrac{1}{7};...;\dfrac{1}{50}< \dfrac{1}{7}\)

=> \(\dfrac{1}{38}+\dfrac{1}{40}+\dfrac{1}{42}+...+\dfrac{1}{50}< \dfrac{1}{7}+\dfrac{1}{7}+...+\dfrac{1}{7}\)( 7 số hạng)

=> \(\dfrac{1}{38}+\dfrac{1}{40}+\dfrac{1}{42}+...+\dfrac{1}{50}< \dfrac{7}{7}=1\)

Vậy: \(\dfrac{1}{38}+\dfrac{1}{40}+\dfrac{1}{42}+...+\dfrac{1}{50}< 1\)

A= (-1-3-5-7-...-2017)

A= 1+3+5+7+...+2017

A có: (2017-1):2+1=1009 (số hạng)

Tổng A = \(\dfrac{\left(2017+1\right).1009}{2}=1018081\)

A=1018081

a,

\(\dfrac{4^6\cdot9^5+6^9\cdot120}{-8^4\cdot3^{12}-6^{11}}=\dfrac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{-2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{-2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{-2^{11}\cdot3^{11}\left(2\cdot3-1\right)}=\dfrac{2^{13}\cdot3^{11}}{-2^{11}\cdot3^{11}\left(2\cdot3+1\right)}=\dfrac{2^2}{7}=\dfrac{4}{7}\)

b,

\(\dfrac{1}{1-\dfrac{1}{1-2-1}}+\dfrac{1}{1+\dfrac{1}{1+2-1}}=\dfrac{1}{1-\dfrac{1}{-2}}+\dfrac{1}{1+\dfrac{1}{2}}=\dfrac{1}{1+\dfrac{1}{2}}+\dfrac{1}{1+\dfrac{1}{2}}=\dfrac{2}{\dfrac{3}{2}}=\dfrac{4}{3}\)