\(\Rightarrow\frac{3}{4}x+5-\frac{2}{3}x+4-\frac{1}{6}x-1=\frac{1}{3}x+4-\frac{1}{3}+3\)+3

\(\Rightarrow\left(\frac{3}{4}x-\frac{2}{3}x-\frac{1}{6}x\right)+\left(5+4-1\right)=\frac{1}{3}x+\left(4-\frac{1}{3}+3\right)\)

=>\(\frac{-1}{12}x+8=\frac{1}{3}x+\frac{20}{3}\)\(\Rightarrow\frac{-1}{12}x+8-\frac{1}{3}x=\frac{20}{3}\)

\(\Rightarrow\left(\frac{-1}{12}-\frac{1}{3}\right)x+8=\frac{20}{3}\)

\(\Rightarrow\frac{-5}{12}x+8=\frac{20}{3}\Rightarrow\frac{-5}{12}x=\frac{20}{3}-8\)

\(\Rightarrow\frac{-5}{12}x=\frac{-4}{3}\Rightarrow x=\frac{-4}{3}:\frac{-5}{12}=\frac{16}{5}\)

b)

\(x-2.\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)

\(x-2\cdot\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)

\(x-2=\frac{16}{9}:\left(\frac{1}{3}-\frac{1}{9}\right)\)

\(x-2=8\)

=> x = 10

a) 

\(A=\frac{1}{2}.\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{2013}{2014}\cdot\frac{2014}{2015}\cdot\frac{2015}{2016}\)

\(A=\frac{1}{2016}\)

Bài làm:

Ta có: \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.....\frac{30}{62}.\frac{31}{64}=2^x\)

\(\Leftrightarrow\frac{1.2.3.....30.31}{2.2.2.3.2.4.....2.31.2.32}=2^x\)

\(\Leftrightarrow\frac{1}{2^{31}.2^5}=2^x\)

\(\Leftrightarrow\frac{1}{2^{36}}=2^x\)

\(\Rightarrow x=-36\)

mk cần cả giải thích

giúp mk vs!!!

( 1/7 . x - 2/7 ) . ( -1.5 . x + 3/5 ) . ( 1/ 3 . x + 4/3) + 0

 <=> +) 1/7 . x - 2/7 = 0                                    +)    (- 1 / 5) . x +3/5 = 0                              +)  1/ 3 . x + 4/ 3 = 0

                    x = 2                                                                  x = 3                                                         x = 4

                                                    Vậy x = 2 : x = 3 ; x=4

$\left(\dfrac17x-\dfrac27\right)\left(-\dfrac15x+\dfrac35\right)\left(\dfrac13x+\dfrac43\right)=0$

$\Leftrightarrow\dfrac{x-2}{7}\cdot\dfrac{3-x}{5}\cdot\dfrac{x+4}{3}=0$

$\Leftrightarrow x-2=0\ \text{hoặc}\ 3-x=0\ \text{hoặc}\ x+4=0$

$\Leftrightarrow x=2\ \text{hoặc}\ x=3\ \text{hoặc}\ x=-4.$

Vậy $\boxed{x\in\{-4,\ 2,\ 3\}.}$

Thiếu điều kiện xy = 1; x+y khác 0 nhá bn

Bài này tương tự câu 1 ở đây

Đặt $A=\dfrac1{(x+y)^3}\left(\dfrac1{x^3}+\dfrac1{y^3}\right)+\dfrac3{(x+y)^4}\left(\dfrac1{x^2}+\dfrac1{y^2}\right)+\dfrac6{(x+y)^5}\left(\dfrac1x+\dfrac1y\right).$

$=\dfrac{(x+y)^2(x^3+y^3)+3xy(x+y)(x^2+y^2)+6x^2y^2}{x^3y^3(x+y)^5}.$

$=\dfrac{(x+y)^2(x+y)(x^2-xy+y^2)+3xy(x+y)(x^2+y^2)+6x^2y^2}{x^3y^3(x+y)^5}.$

$=\dfrac{(x+y)\left[(x+y)^2(x^2-xy+y^2)+3xy(x^2+y^2)\right]+6x^2y^2}{x^3y^3(x+y)^5}.$

$=\dfrac{(x+y)\left[x^4+x^3y+x^2y^2+xy^3+y^4+3x^3y+3xy^3\right]+6x^2y^2}{x^3y^3(x+y)^5}.$

$=\dfrac{(x+y)\left(x^4+4x^3y+x^2y^2+4xy^3+y^4\right)+6x^2y^2}{x^3y^3(x+y)^5}.$

$=\dfrac{(x+y)^4-6x^2y^2}{x^3y^3(x+y)^4}+\dfrac{6x^2y^2}{x^3y^3(x+y)^5}.$

$=\dfrac{(x+y)^5}{x^3y^3(x+y)^5}.$

$=\dfrac1{x^3y^3}.$