a) \(5x - 30 = 0\)

\(5x = 0 + 30\)     

\(5x = 30\)

\(x = 30:5\)

\(x = 6\)      

Vậy phương trình có nghiệm \(x = 6\).

b) \(4 - 3x = 11\)

\( - 3x = 11 - 4\)

\( - 3x =  7\)

\(x = \left( { 7} \right):\left( { - 3} \right)\)

\(x = \dfrac{-7}{3}\)

Vậy phương trình có nghiệm \(x = \dfrac{7}{3}\).

c) \(3x + x + 20 = 0\)               

\(4x + 20 = 0\)

\(4x = 0 - 20\)

\(4x =  - 20\)

\(x = \left( { - 20} \right):4\)

\(x =  - 5\)   

Vậy phương trình có nghiệm \(x =  - 5\).

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

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

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

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

\(x = \dfrac{{ - 9}}{4}\)

Vậy phương trình có nghiệm \(x = \dfrac{{ - 9}}{4}\).

20/11:15/11:20/27:4/7

20/11; 15/11;20/27; 20/27;4/7

\(\dfrac{20}{11},\dfrac{15}{11},\dfrac{20}{27},\dfrac{4}{7}\)

Ta có : \(P=\dfrac{20}{x^2+y^2}+\dfrac{20}{2xy}+\dfrac{1}{xy}\)

Áp dụng BĐT C.B.S

\(\Rightarrow20\left(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\right)\ge20.\dfrac{4}{\left(x+y\right)^2}\ge20\)

Áp dụng BĐT Cauchy

\(xy\le\dfrac{\left(x+y\right)^2}{4}=1\Rightarrow\dfrac{1}{xy}\ge1\)

Cộng hai BĐT trên lại \(\Rightarrow P\ge21\) => MinP=21 khi x=y=1

`D.(-5)/(-7)`

Do `(-5)/(-7)=5/7>0`

Mà các cái còn lại <0

D nha

\(\dfrac{-5}{-7}=\dfrac{5}{7}>1\\ \dfrac{-11}{12}< 0\\ \dfrac{-20}{23}< 0\\ \dfrac{-27}{360}< 0\)

Phân số lớn nhất là \(\dfrac{-5}{-7}\). Chọn D

\(\dfrac{121}{200}\)

` 11/10 . 11/20 = (11.11)/(10.20)=121/200 `

\(\dfrac{x-1}{50}+\dfrac{x-2}{49}=\dfrac{x-3}{48}+\dfrac{x-4}{47}\)

\(\Rightarrow\dfrac{x-1}{50}-1+\dfrac{x-2}{49}-1=\dfrac{x-3}{48}-1+\dfrac{x-4}{47}-1\)

\(\Rightarrow\dfrac{x-51}{50}+\dfrac{x-51}{49}=\dfrac{x-51}{48}+\dfrac{x-51}{47}\)

\(\Rightarrow\dfrac{x-51}{50}+\dfrac{x-51}{49}-\dfrac{x-51}{48}-\dfrac{x-51}{47}=0\)

\(\Rightarrow\left(x-51\right)\left(\dfrac{1}{50}+\dfrac{1}{49}-\dfrac{1}{48}-\dfrac{1}{47}\right)=0\)

Vì \(\dfrac{1}{50}+\dfrac{1}{49}-\dfrac{1}{48}-\dfrac{1}{47}\ne0\) nên \(x-51=0\Rightarrow x=51\)

\(\dfrac{x+25}{6}+\dfrac{x+20}{11}+\dfrac{x+16}{15}+3=0\)

\(\Rightarrow\dfrac{x+25}{6}+1+\dfrac{x+20}{11}+1+\dfrac{x+16}{15}+1=0\)

\(\Rightarrow\dfrac{x+31}{6}+\dfrac{x+31}{11}+\dfrac{x+31}{15}=0\)

\(\Rightarrow\left(x+31\right)\left(\dfrac{1}{6}+\dfrac{1}{11}+\dfrac{1}{15}\right)=0\)

Vì \(\dfrac{1}{6}+\dfrac{1}{11}+\dfrac{1}{15}\ne0\) nên \(x+31=0\Rightarrow x=-31\)

\(\dfrac{x-15}{6}+\dfrac{x-10}{11}=\dfrac{x-3}{18}+\dfrac{x-7}{14}\)

\(\Rightarrow\dfrac{x-15}{6}-1+\dfrac{x-10}{11}-1=\dfrac{x-3}{18}-1+\dfrac{x-7}{14}-1\)

\(\Rightarrow\dfrac{x-21}{6}+\dfrac{x-21}{11}=\dfrac{x-21}{18}+\dfrac{x-21}{14}\)

\(\Rightarrow\dfrac{x-21}{6}+\dfrac{x-21}{11}-\dfrac{x-21}{18}-\dfrac{x-21}{14}=0\)

\(\Rightarrow\left(x-21\right)\left(\dfrac{1}{6}+\dfrac{1}{11}-\dfrac{1}{18}-\dfrac{1}{14}\right)=0\)

Vì \(\dfrac{1}{6}+\dfrac{1}{11}-\dfrac{1}{18}-\dfrac{1}{14}\ne0\) nên \(x-21=0\Rightarrow x=21\)

.

.

=\(\dfrac{11}{31}\)*\(\left(\dfrac{-2}{17}-\dfrac{-9}{17}\right)\)+\(\dfrac{7}{31}\)*\(\dfrac{-20}{17}\)

=\(\dfrac{11}{31}\)*\(\dfrac{-7}{17}\)+\(\dfrac{7}{31}\)*\(\dfrac{-20}{17}\)

=\(\dfrac{-77}{512}\)+\(\dfrac{-140}{512}\)

=\(\dfrac{-217}{512}\)

nhanh giúp mk với

a) \(\dfrac{3}{11}+\dfrac{14}{11}=\dfrac{17}{11}\)

b) \(\dfrac{1}{16}+\dfrac{3}{4}=\dfrac{1}{16}+\dfrac{12}{16}=\dfrac{13}{16}\)

c) \(\dfrac{2}{20}+\dfrac{7}{10}=\dfrac{2}{20}+\dfrac{14}{20}=\dfrac{16}{20}=\dfrac{4}{5}\)