giúp mình bài này với mình gấp lắm

a) \(\dfrac{6}{13}:\left(\dfrac{1}{2}-x\right)=\dfrac{15}{39}\)

\(\dfrac{1}{2}-x=\dfrac{6}{13}:\dfrac{15}{39}\)

\(\dfrac{1}{2}-x=\dfrac{6}{5}\)

\(x=\dfrac{1}{2}-\dfrac{6}{5}\)

\(x=-\dfrac{7}{10}\)

b) \(3\times\left(\dfrac{x}{4}+\dfrac{x}{28}+\dfrac{x}{70}+\dfrac{x}{130}\right)=\dfrac{60}{13}\)

\(3\times x\times\left(\dfrac{1}{4}+\dfrac{1}{28}+\dfrac{1}{70}+\dfrac{1}{130}\right)=\dfrac{60}{13}\)

\(x\times\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+\dfrac{3}{7\times13}\right)=\dfrac{60}{13}\)

\(x\times\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}\right)=\dfrac{60}{13}\)

\(x\times\left(1-\dfrac{1}{13}\right)=\dfrac{60}{13}\)

\(x\times\dfrac{12}{13}=\dfrac{60}{13}\)

\(x=\dfrac{60}{13}:\dfrac{12}{13}\)

\(x=5\)

Bài 1:

Ta có: \(4-2\left(x+1\right)=2\)

\(\Leftrightarrow2\left(x+1\right)=2\)

\(\Leftrightarrow x+1=1\)

hay x=0

Bài 2: 

Ta có: \(\left|2x-3\right|-1=2\)

\(\Leftrightarrow\left|2x-3\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)

chưa biết

a) Ta có: \(\dfrac{1}{7}+x=-\dfrac{2}{3}\)

\(\Leftrightarrow x=-\dfrac{2}{3}-\dfrac{1}{7}=\dfrac{-14}{21}-\dfrac{3}{21}\)

hay \(x=-\dfrac{17}{21}\)

Vậy: \(x=-\dfrac{17}{21}\)

b) Ta có: \(\dfrac{-2}{3}:x=\dfrac{-5}{6}\)

\(\Leftrightarrow x=\dfrac{-2}{3}:\dfrac{-5}{6}=\dfrac{-2}{3}\cdot\dfrac{6}{-5}=\dfrac{-12}{-15}=\dfrac{4}{5}\)

Vậy: \(x=\dfrac{4}{5}\)

c) Ta có: \(\left(\dfrac{3}{5}-2x\right)\cdot\dfrac{5}{8}=1\)

\(\Leftrightarrow\left(\dfrac{3}{5}-2x\right)=1:\dfrac{5}{8}=\dfrac{8}{5}\)

\(\Leftrightarrow-2x=\dfrac{8}{5}-\dfrac{3}{5}=1\)

hay \(x=-\dfrac{1}{2}\)

Vậy: \(x=-\dfrac{1}{2}\)

d) Ta có: \(\dfrac{3}{4}+\dfrac{2}{5}x=\dfrac{29}{60}\)

\(\Leftrightarrow x\cdot\dfrac{2}{5}=\dfrac{29}{60}-\dfrac{3}{4}=\dfrac{29}{60}-\dfrac{45}{60}=\dfrac{-16}{60}=\dfrac{-4}{15}\)

hay \(x=\dfrac{-4}{15}:\dfrac{2}{5}=\dfrac{-4}{15}\cdot\dfrac{5}{2}=\dfrac{-20}{30}=-\dfrac{2}{3}\)

Vậy: \(x=-\dfrac{2}{3}\)

e) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}=\dfrac{-7}{20}\)

hay \(x=-\dfrac{1}{4}:\dfrac{7}{20}=\dfrac{-1}{4}\cdot\dfrac{20}{7}=\dfrac{-20}{28}=\dfrac{-5}{7}\)

Vậy: \(x=-\dfrac{5}{7}\)

f) Ta có: \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\Leftrightarrow-x+\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}=0\)

\(\Leftrightarrow-x+\dfrac{55}{60}-\dfrac{24}{60}-\dfrac{40}{60}=0\)

\(\Leftrightarrow-x-\dfrac{9}{60}=0\)

\(\Leftrightarrow-x=\dfrac{9}{60}=\dfrac{3}{20}\)

hay \(x=-\dfrac{3}{20}\)

Vậy: \(x=-\dfrac{3}{20}\)

g) Ta có: \(\left|x+\dfrac{1}{3}\right|-4=\dfrac{-1}{2}\)

\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{-1}{2}+4=\dfrac{-1}{2}+\dfrac{8}{2}=\dfrac{7}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{7}{2}\\x+\dfrac{1}{3}=-\dfrac{7}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{21}{6}-\dfrac{2}{6}=\dfrac{19}{6}\\x=-\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{-21}{6}-\dfrac{2}{6}=\dfrac{-23}{6}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{19}{6};-\dfrac{23}{6}\right\}\)

cảm ơn cậu cutee gì đó ơi nha

a) \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)

\(\Leftrightarrow x^3-27+x\left(4-x^2\right)=1\)

\(\Leftrightarrow x^3-27+4x-x^3=1\)

\(\Leftrightarrow4x-27=1\Leftrightarrow4x=28\Leftrightarrow x=7\)

b) \(\left(x-2\right)^3-\left(x-2\right)\left(x^2+2x+4\right)+6\left(x-2\right)\left(x+2\right)=60\)

\(\Leftrightarrow x^3-6x^2+12x+4-x^3-8+6\left(x^2-4\right)=60\)

\(\Leftrightarrow-6x^2+12x-4+6x^2-24=60\)

\(\Leftrightarrow12x-28=60\Leftrightarrow x=\frac{22}{3}\)

a) (x - 3)(x^2 + 3x + 9) + x(x + 2)(2 - x) = 1

x^3 + 3x^2 + 9x - 3x^2 - 9x - 27 + 2x^2 - x^3 + 4x - 2x^2 = 1

4x - 27 = 1

4x = 28

x = 7

b) (x - 2)^3 - (x - 2)(x^2 + 2x + 4) + 6(x - 2)(x + 2) = 60

x^3 - 4x^2 + 4x - 2x^2 + 8x - 8 - x(x^2 + 2x + 4) + 2(x^2 + 2x + 4) + 6x - 24 = 60

x^3 + 12x - 32 - x^2 - 2x^2 - 4x + 2x^2 + 4x + 8 = 60

12x - 24 = 60

12x = 60 + 24

12x = 84

x = 7

câu a ko rõ ràng

Ta có \(x+5⋮x+2\)

=> x + 2 + 3 \(⋮\)x + 2

Vì x + 2  \(⋮\)x + 2

=> 3  \(⋮\)x + 2

=> x + 2 \(\in\)Ư(3)

=> x + 2  \(\in\) {1 ; 3 - 1 ; - 3}

=> x  \(\in\){-1 ; 1 ; - 3 ; - 5}

b) (x - 2)(x + 3) = 0

 \(\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

=> x  \(\in\){2 ; - 3}

c) (2x + 60)(9 - x²) = 0

=> \(\orbr{\begin{cases}2x+60=0\\9-x^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-30\\x=\pm3\end{cases}}\)

=> x  \(\in\){- 30 ; 3 ; - 3}

d) Vì \(x;y\inℤ\Rightarrow\hept{\begin{cases}x-2\inℤ\\y+1\inℤ\end{cases}}\)

Ta có 3 = 1.3 = (-1).(-3)

Lập bảng xét các trường hợp 

Vậy các cặp số (x ; y) thỏa mãn là (3 ; 2) ; (1 ; - 4) ; (5 ; 0) ; (- 1; - 2)

a, 720:[118-(2x -10) ]=60

           [118-(2x-10)]=720:60

           118-(2x-10)=12

                   2x-10=118-12

                    2x-10=106

                     2x     =106+10

                        2x  =116

                         x=116:2

                          x=58

Bài 1:a) \(\left(x-3\right)^3=x^3-9x^2+27x-27\)

b)\(\left(\frac{1}{5}x-1\right)\left(\frac{1}{5}x+1\right)=\frac{1}{25}x^2-1\)

Bài 3:

a)x(x-6) + 10x - 60 =0

\(\Rightarrow x^2-6x+10x-60=0\)

\(\Rightarrow x^2+4x+60=0\)

\(\Rightarrow\left(x+2\right)^2+54=0\)

Vì \(\left(x+2\right)^2+54\ge54\forall x\)

\(\Rightarrow\)không có giá trị nào của x thỏa mãn.