Nhân vô rồi chuyển dấu lên và nhóm nhân -1ra ngoài rồi trg ngoặc là dãy có quy luật giải dãy đó r nhân phá ngoặc

Tờ làm luôn, ko ghi đề nữa nhé

\(A=\frac{\frac{24}{12}-\frac{4}{12}+\frac{3}{12}}{\frac{24}{12}+\frac{2}{12}-\frac{3}{12}}\)

\(A=\frac{\frac{23}{12}}{\frac{23}{12}}=1\)

Vậy A=1

\(A=\frac{2-\frac{1}{3}+\frac{1}{4}}{2+\frac{1}{6}-\frac{1}{4}}\)\(=\frac{2-\frac{2}{6}+\frac{2}{8}}{2+\frac{2}{12}-\frac{2}{8}}\)\(=\frac{2\left(1-\frac{1}{6}+\frac{1}{8}\right)}{-2\left(1-\frac{1}{12}+\frac{1}{8}\right)}\)\(=-1\)

31/12

51/24 

nha

I don't no

\(\frac{4}{3}.\left(\frac{1}{6}-\frac{1}{2}\right)\le x\le\frac{2}{3}.\left(\frac{-1}{6}+\frac{3}{4}\right)\)

\(\frac{4}{3}.\frac{-1}{3}\le x\le\frac{2}{3}.\frac{7}{12}\)

\(\frac{-4}{9}\le x\le\frac{7}{18}\)

\(\frac{-8}{18}\le x\le\frac{7}{18}\)

\(\Rightarrow\)X \(\in\) {\(\frac{-7}{18};\frac{-6}{18};\frac{-5}{18};\frac{-4}{18};\frac{-3}{18};\frac{-2}{18};\frac{-1}{18};0;\frac{1}{18};\frac{2}{18};\frac{3}{18};\frac{4}{18};\frac{5}{18};\frac{6}{18}\)}

\(M=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{62}+\frac{1}{63}\)

\(M=1+\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)+\left(\frac{1}{8}+\frac{1}{9}+...+\frac{1}{15}\right)+\left(\frac{1}{16}+\frac{1}{17}+...+\frac{1}{31}\right)+\left(\frac{1}{32}+\frac{1}{33}+...+\frac{1}{63}\right)\)

\(M< 1+\frac{1}{2}.2+\frac{1}{4}.4+\frac{1}{8}.8+\frac{1}{16}.16+\frac{1}{32}.32\)

\(M< 1+1+1+1+1+1\)

\(M< 1.6=6\left(đpcm\right)\)

\(\left(-2\right)^3.\left(\frac{3}{4}-0,25\right):\left(2\frac{1}{4}-1\frac{1}{6}\right)\)

\(=\left(-8\right).\left(\frac{3}{4}-\frac{1}{4}\right):\left(\frac{9}{4}-\frac{7}{6}\right)\)

\(=\left(-8\right).\frac{1}{2}:\frac{13}{12}\)

\(=\left(-4\right).\frac{12}{13}\)

\(=\frac{-48}{13}\)

a) 3/4+ 1/4:x = 2/5

1/4:x = 3/4-2/5

1/4:x= 7/20

x= 7/20:1/4

x= 7/5

b) chưa học

c) 15/8-1/8: (x/4 - 0,5) = 5/4

1/8: (x/4 -1/2)= 15/8-5/4

1/8:( x/4 -1/2) =  5/8

x/4 - 1/2 = 1/8:5/8

x/4 -1/2= 1/5

x/4= 1/5+1/2

x/4 = 7/7

x/4= 7/7× 4/4

x/4= 28/28

4/4=28/28

phần c ko chắc chắn

đúng k nhé

x+1/x^2+x+1 -(x-1)/x^2+x+1=3/x(x^4+x^2+1)

đkxđ x khác 0

[(x+1)(x^2-x+1)-(x-1)(x^2+x+1)] /(x^2+x+1)(x^2-x+1)=3/x(x^4+x^2+1)

[(x^3+1)-(x^3-1)]/x^4+x^2+1=3/x(x^4+x^2+1)

nhân 2 vế pt cho x(x^4+x^2+1) ta được 

x(x^3+1-x^3+1)=3

<=> 2x=3

<=>x=3/2 (thỏa)

S={3/2}

Đặt \(x^2+x+1=a\ne0vàx^2-x+1=b\ne0\)

\(\Rightarrow b-a=-2xvàb+a=2x^2+2\)

    và điều kiện \(x\ne0\)

thì  \(x\left(x^4+x^2+1\right)=xab\)

\(\Rightarrow PT\Leftrightarrow\frac{x+1}{a}-\frac{x-1}{b}=\frac{3}{xab}\)

              \(\Leftrightarrow\frac{bx\left(x+1\right)-ax\left(x-1\right)}{xab}=\frac{3}{xab}\)

             \(\Leftrightarrow bx^2+bx-ax^2+ax=3\)

             \(\Leftrightarrow x^2\left(b-a\right)+x\left(b+a\right)-3=0\)

             \(\Leftrightarrow2x-3=0\)

             \(\Leftrightarrow x=\frac{3}{2}\)(tm)

Vậy \(x=\frac{2}{3}\) là nghiệm của pt