A

A

A

a: =5/6(1/8+2/3)=5/6*19/24=95/144

b: =3/4(7/5-1/2)=27/40

c: =35/24*2/3=35/36

địt con mẹ mày lên

\(\dfrac{1}{3}\)-\(\dfrac{2}{15}\)+\(\dfrac{14}{15}\)=\(\dfrac{3}{15}\)-\(\dfrac{2}{15}\)+\(\dfrac{14}{15}\)

=\(\dfrac{13}{15}\)

\(\dfrac{3}{5}\)+4-\(\dfrac{6}{7}\)=\(\dfrac{3}{5}\)+\(\dfrac{20}{5}\)-\(\dfrac{6}{7}\)=\(\dfrac{23}{5}\)-\(\dfrac{6}{7}\)=\(\dfrac{161}{35}\)-\(\dfrac{30}{35}\)=\(\dfrac{131}{35}\)

đầy đủ phần ngoặc hay gì đó

\(\dfrac{11}{2}\): \(\dfrac{1}{4}\) \(\times\) \(\dfrac{5}{3}\)

= \(\dfrac{11}{2}\) \(\times\) \(\dfrac{4}{1}\) \(\times\) \(\dfrac{5}{3}\)

= 22 \(\times\) \(\dfrac{5}{3}\)

= \(\dfrac{110}{3}\)

\(\dfrac{5}{2}-\dfrac{1}{4}+\dfrac{5}{3}\)

= \(\dfrac{30}{12}-\dfrac{3}{12}+\dfrac{20}{12}\)

= \(\dfrac{7}{12}\) 

\(\dfrac{14}{5}\times\dfrac{2}{3}\)+ 5

= \(\dfrac{28}{15}\) + 5

= \(\dfrac{28}{15}\) + \(\dfrac{75}{15}\)

= \(\dfrac{103}{15}\)

a) C1

 \(\dfrac{1}{8}\times\dfrac{5}{6}+\dfrac{5}{6}\times\dfrac{2}{3}=\dfrac{5}{6}\times\left(\dfrac{1}{8}+\dfrac{2}{3}\right)=\dfrac{5}{6}\times\dfrac{19}{24}=\dfrac{95}{144}\)

   C2

\(\dfrac{1}{8}\times\dfrac{5}{6}+\dfrac{5}{6}\times\dfrac{2}{3}=\dfrac{5}{48}+\dfrac{10}{18}=\dfrac{95}{144}\)

b) C1

  \(\dfrac{7}{5}\times\dfrac{3}{4}-\dfrac{1}{2}\times\dfrac{3}{4}=\dfrac{3}{4}\times\left(\dfrac{7}{5}-\dfrac{1}{2}\right)=\dfrac{3}{4}\times\dfrac{9}{10}=\dfrac{27}{40}\)

    C2 

   \(\dfrac{7}{5}\times\dfrac{3}{4}-\dfrac{1}{2}\times\dfrac{3}{4}=\dfrac{21}{20}-\dfrac{3}{8}=\dfrac{27}{40}\)

c) C1

 \(\left(\dfrac{5}{6}+\dfrac{5}{8}\right)\times\dfrac{2}{3}=\dfrac{35}{24}\times\dfrac{2}{3}=\dfrac{35}{36}\)

  C2 

  \(\left(\dfrac{5}{6}+\dfrac{5}{8}\right)\times\dfrac{2}{3}=\left(\dfrac{5}{6}\times\dfrac{2}{3}\right)+\left(\dfrac{5}{8}\times\dfrac{2}{3}\right)=\dfrac{5}{9}+\dfrac{5}{12}=\dfrac{35}{36}\)

Đề bài mình viết thiếu là CM biểu thức sau không phụ thuộc vào x ( nghĩa là kết quả phải ra số tự nhiên không có x ) 

\(A=\left(2x+1\right)\left(x-1\right)-2x\left(x+2\right)-5\left(-x+3\right)+4\)

\(=2x^2-2x+x-1-2x^2-4x+5x-15+4\)

\(=-12\left(đpcm\right)\)

\(B=\left(4x+3\right)\left(2x-5\right)-\left(8x+1\right)\left(x+3\right)+13\left(3x+1\right)+2\)

\(=8x^2-20x+6x-15-\left(8x^2+24x+x+3\right)+39x+13+2\)

\(=-3\left(đpcm\right)\)