a) \(-\dfrac{3}{7}-x=-\dfrac{1}{2}\\ x=-\dfrac{3}{7}-\left(-\dfrac{1}{2}\right)\\ x=\dfrac{-3}{7}+\dfrac{1}{2}=\dfrac{-6}{14}+\dfrac{7}{14}=\dfrac{1}{14}\)

b) \(x-\dfrac{4}{5}=\dfrac{1}{2}-\dfrac{1}{3}\\ x-\dfrac{4}{5}=\dfrac{3}{6}-\dfrac{2}{6}=\dfrac{1}{6}\\ x=\dfrac{1}{6}+\dfrac{4}{5}=\dfrac{5}{30}+\dfrac{24}{30}\\ x=\dfrac{29}{30}\)

c) \(-x-\dfrac{3}{4}=-\dfrac{8}{11}\\ -x=-\dfrac{8}{11}+\dfrac{3}{4}\\ -x=-\dfrac{32}{44}+\dfrac{33}{44}=\dfrac{1}{44}\\ x=-\dfrac{1}{44}\)

d) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\\ \dfrac{11}{12}-\dfrac{2}{5}-x=\dfrac{2}{3}\\ x=\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}\\ x=\dfrac{55}{60}-\dfrac{24}{60}-\dfrac{40}{60}\\ x=-\dfrac{9}{60}\)

a) \(\dfrac{-3}{7}-x=\dfrac{-1}{2}\)

\(\Rightarrow x=\dfrac{1}{2}-\dfrac{3}{7}\)

\(\Rightarrow x=\dfrac{1}{14}\)

Vậy \(x=\dfrac{1}{14}\)

b) \(x-\dfrac{4}{5}=\dfrac{1}{2}-\dfrac{1}{3}\)

\(\Rightarrow x-\dfrac{4}{5}=\dfrac{1}{6}\)

\(\Rightarrow x=\dfrac{1}{6}+\dfrac{4}{5}\)

\(\Rightarrow x=\dfrac{29}{30}\)

Vậy \(x=\dfrac{29}{30}\)

c) \(-x-\dfrac{3}{4}=\dfrac{-8}{11}\)

\(\Rightarrow-x=\dfrac{3}{4}-\dfrac{8}{11}\)

\(\Rightarrow-x=\dfrac{1}{44}\)

\(\Rightarrow x-\dfrac{1}{44}\)

Vậy \(x=-\dfrac{1}{44}\)

d) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\Rightarrow\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)

\(\Rightarrow\dfrac{2}{5}+x=\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{1}{4}-\dfrac{2}{5}\)

\(\Rightarrow x=\dfrac{-3}{20}\)

Vậy \(x=\dfrac{-3}{20}\)

Sossssssss

Mình sửa đề nhé: (x^3+y^3):(x+y)

\(a.\left(\dfrac{-4}{5}+\dfrac{3}{7}-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{2}{7}\right)\\ =\dfrac{-4}{5}+\dfrac{2}{7}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{2}{7}\\ =\left(-\dfrac{4}{5}-\dfrac{1}{5}\right)+\left(\dfrac{2}{7}+\dfrac{2}{7}\right)+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)\\ =-1+\dfrac{4}{7}+0=-\dfrac{3}{7}\)

\(b.\left(7-\dfrac{3}{4}+\dfrac{1}{3}\right)-\left(6+\dfrac{5}{4}-\dfrac{4}{3}\right)-\left(5-\dfrac{7}{4}+\dfrac{5}{3}\right)\\ =7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\\ =\left(7-6-5\right)+\left(\dfrac{7}{4}-\dfrac{3}{4}-\dfrac{5}{4}\right)+\left(\dfrac{1}{3}+\dfrac{4}{3}-\dfrac{5}{3}\right)\\=1+\dfrac{-1}{4}+0=\dfrac{3}{4}\)

a) 

\(\left(-\dfrac{4}{5}+\dfrac{3}{7}-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{2}{7}\right)\\ =\left(-\dfrac{4}{5}-\dfrac{1}{5}\right)+\left(\dfrac{3}{7}+\dfrac{2}{7}\right)+\left(-\dfrac{1}{2}+\dfrac{1}{2}\right)\\ =-\dfrac{5}{5}+\dfrac{5}{7}+0\\ =-1+\dfrac{5}{7}\\ =-\dfrac{2}{7}\)

b) 

\(\left(7-\dfrac{3}{4}+\dfrac{1}{2}\right)-\left(6+\dfrac{5}{4}-\dfrac{4}{3}\right)-\left(5-\dfrac{7}{4}+\dfrac{5}{3}\right)\\ =7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\\ =\left(7-6-5\right)+\left(-\dfrac{3}{4}-\dfrac{5}{4}+\dfrac{7}{4}\right)+\left(\dfrac{1}{3}+\dfrac{4}{3}-\dfrac{5}{3}\right)\\ =\left(-4\right)+\left(\dfrac{-1}{4}\right)+0\\ =-\dfrac{17}{4}\)

c) 

\(\left(0,25+\dfrac{7}{9}-\dfrac{1}{7}\right)-\left(0,75-\dfrac{2}{9}-\dfrac{1}{7}\right)\\ =0,25+\dfrac{7}{9}-\dfrac{1}{7}-0,75+\dfrac{2}{9}+\dfrac{1}{7}\\ =\left(0,25-0,75\right)+\left(\dfrac{7}{9}+\dfrac{2}{9}\right)+\left(-\dfrac{1}{7}+\dfrac{1}{7}\right)\\ =-\dfrac{1}{2}+\dfrac{9}{9}+0\\ =-\dfrac{1}{2}+1\\ =\dfrac{1}{2}\)

d) 

\(\dfrac{\dfrac{2}{7}+\dfrac{1}{3}-\dfrac{2}{9}}{\dfrac{3}{7}+\dfrac{1}{2}-\dfrac{1}{3}}\\ =\dfrac{\dfrac{2}{7}+\dfrac{2}{6}-\dfrac{2}{9}}{\dfrac{3}{7}+\dfrac{3}{6}-\dfrac{3}{9}}\\ =\dfrac{2\left(\dfrac{1}{7}+\dfrac{1}{6}-\dfrac{1}{9}\right)}{3\left(\dfrac{1}{7}+\dfrac{1}{6}-\dfrac{1}{9}\right)}\\ =\dfrac{2}{3}\)

a) \(\dfrac{-3}{21}+\dfrac{-2}{7}+\dfrac{1}{3}=\dfrac{-3}{21}+\dfrac{-6}{21}+\dfrac{7}{21}=-\dfrac{2}{21}\)

b) \(\dfrac{-13}{15}+\dfrac{5}{-18}+\dfrac{4}{9}=\dfrac{-78}{90}+\dfrac{-25}{90}+\dfrac{40}{90}=\dfrac{63}{90}=\dfrac{7}{10}\)

c) \(\dfrac{-2}{5}-\left(\dfrac{-3}{11}\right)=\dfrac{-2}{5}+\dfrac{3}{11}=\dfrac{-22}{55}+\dfrac{15}{55}=\dfrac{-7}{55}\)

d) \(\left(-4\right)-\left(\dfrac{-4}{5}\right)-\dfrac{2}{3}=\left(-4\right)+\dfrac{4}{5}-\dfrac{2}{3}=\dfrac{-60}{15}+\dfrac{12}{15}-\dfrac{10}{15}=\dfrac{-58}{15}\)

e) \(\dfrac{3}{5}+\left(\dfrac{-4}{3}\right)-\dfrac{-3}{4}=\dfrac{3}{5}+\left(\dfrac{-4}{3}\right)+\dfrac{3}{4}=\dfrac{36}{60}+\dfrac{-80}{60}+\dfrac{45}{60}=\dfrac{1}{60}\)

g) \(\dfrac{5}{8}-\left(-\dfrac{2}{5}\right)-\dfrac{3}{10}=\dfrac{5}{8}+\dfrac{2}{5}-\dfrac{3}{10}=\dfrac{25}{40}+\dfrac{16}{40}-\dfrac{12}{40}=\dfrac{29}{40}\)

h) \(\dfrac{3}{4}-\left(-\dfrac{5}{3}\right)+\left(\dfrac{1}{12}+\dfrac{2}{9}\right)=\dfrac{3}{4}+\dfrac{5}{3}+\left(\dfrac{3}{36}+\dfrac{8}{36}\right)=\dfrac{27}{36}+\dfrac{60}{36}+\dfrac{11}{36}=\dfrac{98}{36}=\dfrac{49}{18}\)

a) \(\dfrac{-3}{21}+\dfrac{-2}{7}+\dfrac{1}{3}=\dfrac{-3}{21}-\dfrac{6}{21}+\dfrac{7}{21}\\ =\dfrac{-3-6+7}{21}=-\dfrac{2}{21}\)

b) \(\dfrac{-13}{15}+\dfrac{5}{-18}+\dfrac{4}{9}=\dfrac{-78}{90}-\dfrac{25}{90}+\dfrac{40}{90}\\ =\dfrac{-78-25+40}{90}=\dfrac{-63}{90}=-\dfrac{7}{10}\)

c) \(\dfrac{-2}{5}-\left(\dfrac{-3}{11}\right)=-\dfrac{2}{5}+\dfrac{3}{11}\\ =-\dfrac{22}{55}+\dfrac{15}{55}=-\dfrac{7}{55}\)

\(\dfrac{9}{5}+\dfrac{5}{8}=\dfrac{72}{40}+\dfrac{25}{40}=\dfrac{97}{40}\)

\(\dfrac{3}{5}+\dfrac{13}{9}=\dfrac{27}{45}+\dfrac{65}{45}=\dfrac{82}{45}\)

\(\dfrac{2}{3}+\dfrac{11}{2}=\dfrac{4}{6}+\dfrac{33}{6}=\dfrac{37}{6}\)

\(\dfrac{9}{4}+\dfrac{17}{20}=\dfrac{45}{20}+\dfrac{17}{20}=\dfrac{62}{20}=\dfrac{31}{10}\)

\(\dfrac{43}{6}+\dfrac{45}{8}=\dfrac{172}{24}+\dfrac{135}{24}=\dfrac{307}{24}\)

\(\dfrac{5}{9}+\dfrac{7}{15}=\dfrac{25}{45}+\dfrac{21}{45}=\dfrac{46}{45}\)

\(\dfrac{9}{5}+\dfrac{5}{8}\\ =\dfrac{72}{40}+\dfrac{25}{40}\\ =\dfrac{97}{40}\\ ----------\\ \dfrac{3}{5}+\dfrac{13}{9}\\ =\dfrac{27}{45}+\dfrac{65}{45}\\ =\dfrac{92}{45}\\ --------\\ \dfrac{2}{3}+\dfrac{11}{2}\\ =\dfrac{4}{6}+\dfrac{33}{6}\\ =\dfrac{37}{6}\\ --------\\ \dfrac{9}{4}+\dfrac{17}{20}\\ =\dfrac{45}{20}+\dfrac{17}{20}\\ =\dfrac{62}{20}=\dfrac{31}{10}\\ ----------\\ \dfrac{43}{6}+\dfrac{45}{8}\\ =\dfrac{172}{24}+\dfrac{135}{24}\\ =\dfrac{307}{24}\)

\(-----------\\ \dfrac{5}{9}+\dfrac{7}{15}\\ =\dfrac{25}{45}+\dfrac{21}{45}\\ =\dfrac{46}{45}\)

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

\(=\dfrac{\left(x-y\right)\left(x^2-2xy+y^2\right)}{x-y}=\left(x-y\right)^2\)

Soss

\(\dfrac{8}{3}+\dfrac{10}{7}=\dfrac{56}{21}+\dfrac{30}{21}\\ =\dfrac{56+30}{21}=\dfrac{86}{21}\)

\(\dfrac{8}{3}+\dfrac{10}{7}\)

\(=\dfrac{56}{21}+\dfrac{30}{21}\)

\(=\dfrac{56+30}{21}\)

\(=\dfrac{86}{21}\)

\(\dfrac{4}{5}+\dfrac{3}{4}=\dfrac{16}{20}+\dfrac{15}{20}\\ =\dfrac{16+15}{20}=\dfrac{31}{20}\)

\(\dfrac{4}{5}+\dfrac{3}{4}\)

\(=\dfrac{16}{20}+\dfrac{15}{20}\)

\(=\dfrac{16+15}{20}\)

\(=\dfrac{31}{20}\)

a: Trên tia Ox, ta có: OA<OB

nên A nằm giữa O và B

=>OA+AB=OB

=>AB+4=6

=>AB=2(cm)

b: C là trung điểm của OA

=>\(CO=CA=\dfrac{OA}{2}=2\left(cm\right)\)

Vì AO và AB là hai tia đối nhau

nên AC và AB là hai tia đối nhau

=>A nằm giữa hai điểm B và C

Ta có: A nằm giữa B và C

mà AB=AC(=2cm)

nên A là trung điểm của BC

d: Các góc đỉnh D trong hình vẽ là: \(\widehat{ODC};\widehat{ODA};\widehat{ODB};\widehat{CDA};\widehat{CDB};\widehat{ADB}\)

b: \(cos14=sin76;cos37=sin53\)

Vì 47<48<53<76 nên \(sin47< sin48< sin53< sin76\)

=>\(sin47< sin48< cos37< cos14\)

c: \(cot25=tan65;cot38=tan52\)

Vì 52<62<65<73

nên \(tan52< tan62< tan65< tan73\)

=>\(cot38< tan63< cot25< tan73\)

a: Vì 75>45

nên \(tan75>1\)

Vì 40<45

nên \(cot40>1\)

\(cot40=tan50;tan75=tan75\)

mà \(tan50< tan75\)

nên \(1< cot40< tan75\left(1\right)\)

\(cos56=sin34;sin50=sin50\)

mà 34<50

nên \(sin34< sin50< 1\)

=>\(cos56< sin50< 1\left(2\right)\)

Từ (1),(2) suy ra \(cos56< sin50< cot40< tan75\)