$\textbf{a)}$

$\dfrac{x-1}{2}+\dfrac{x-2}{5}=\dfrac14+\dfrac{x-7}{10}$

$\Leftrightarrow\dfrac{5(x-1)+2(x-2)}{10}=\dfrac14+\dfrac{x-7}{10}$

$\Leftrightarrow\dfrac{7x-9}{10}=\dfrac14+\dfrac{x-7}{10}$

$\Leftrightarrow14x-18=5+2x-14$

$\Leftrightarrow12x=9$

$\Leftrightarrow x=\dfrac34.$

$\textbf{b)}$

$\dfrac{3-2}{2x-3}=\dfrac25+\dfrac1{2x-3}-\dfrac32$

$\Leftrightarrow1-\dfrac2{2x-3}=-\dfrac{11}{10}+\dfrac1{2x-3}$

$\Leftrightarrow\dfrac{21}{10}=\dfrac3{2x-3}$

$\Leftrightarrow21(2x-3)=30$

$\Leftrightarrow42x=93$

$\Leftrightarrow x=\dfrac{31}{14}.$

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

Bài này tương tự câu 1 ở đây

Ta có:

\(\left(\right. \frac{13 \frac{2}{9} - 15 \frac{2}{3}}{18 \frac{3}{7} - 17 \frac{1}{4}} \cdot \frac{30^{2} - 5^{4}}{25 - 12 \cdot 5^{2}} \left.\right) \cdot x = \frac{\frac{2}{11} + \frac{3}{13} + \frac{4}{15} + \frac{5}{17}}{4 \frac{1}{11} + \frac{5}{13} + \frac{9}{15} + \frac{13}{17}}\)

Bước 1: Đổi hỗn số về phân số

• \(13 \frac{2}{9} = \frac{119}{9}\),
• \(15 \frac{2}{3} = \frac{47}{3}\),
• \(18 \frac{3}{7} = \frac{129}{7}\),
• \(17 \frac{1}{4} = \frac{69}{4}\)

Bước 2: Tính toán từng phần

Ta có:

\(\frac{119}{9} - \frac{47}{3} = \frac{119 - 141}{9} = \frac{- 22}{9}\) \(\frac{129}{7} - \frac{69}{4} = \frac{516 - 483}{28} = \frac{33}{28}\) \(30^{2} - 5^{4} = 900 - 625 = 275\) \(25 - 12 \cdot 25 = 25 - 300 = - 275\)

Khi đó:

\(\left(\right. \frac{- 22}{9} \div \frac{33}{28} \cdot \frac{275}{- 275} \left.\right) = \left(\right. \frac{- 22}{9} \cdot \frac{28}{33} \cdot \left(\right. - 1 \left.\right) \left.\right) = \frac{616}{297}\)

Bước 3: Tính vế phải

Tử số:

\(\frac{2}{11} + \frac{3}{13} + \frac{4}{15} + \frac{5}{17} = \frac{35494}{36465}\)

Mẫu số:

\(4 \frac{1}{11} + \frac{5}{13} + \frac{9}{15} + \frac{13}{17} = \frac{149645}{36465}\)

→ Vế phải:

\(\frac{35494}{36465} \div \frac{149645}{36465} = \frac{35494}{149645}\)

Bước 4: Giải phương trình

\(\frac{616}{297} \cdot x = \frac{35494}{149645} \Rightarrow x = \frac{35494}{149645} \cdot \frac{297}{616} = \frac{813}{7118}\)

Vậy:

\(\boxed{x = \frac{813}{7118}}\)

hỉu không =]]]

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

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

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

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

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

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

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

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

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

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) 

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

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

( 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 $\boxed{x\in\{-4,\ 2,\ 3\}.}$

a) 

( 4x - 9 ) ( 2,5 + (-7/3) . x ) = 0

\(\Rightarrow\orbr{\begin{cases}4x-9=0\\2,5+\frac{-7}{3}x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{9}{4}\\x=\frac{15}{14}\end{cases}}\)

P/s: đợi xíu làm câu b

b) \(\frac{1}{x\left(x+1\right)}\cdot\frac{1}{\left(x+1\right)\left(x+2\right)}\cdot\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2015}\)

\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2015}\)

\(\frac{-1}{x+3}=\frac{1}{2015}\)

\(\Leftrightarrow x+3=-2015\)

\(\Leftrightarrow x=-2018\)

Vậy,.........

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

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

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

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

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

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