a) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)^2=27.\)

\(\Leftrightarrow x^3+27-x\left(x^2-4x+4\right)-27=0.\)

\(\Leftrightarrow x^3-x^3+4x^2-4x=0.\)

\(\Leftrightarrow4x\left(x-1\right)=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=0.\\x-1=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0.\\x=1.\end{matrix}\right.\)

Vậy \(S=\left\{0;1\right\}.\)

a) \(x+\dfrac{4}{9}=\dfrac{5}{27}\)  

    \(x=\dfrac{5}{27}-\dfrac{4}{9}\)

   \(x=-\dfrac{7}{27}\)

b) \(x-\dfrac{4}{11}=\dfrac{7}{33}\)

   \(x=\dfrac{7}{33}+\dfrac{4}{11}\)

   \(x=\dfrac{19}{33}\)

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

  \(\dfrac{8}{5}-x=\dfrac{2}{15}\)

          \(x=\dfrac{8}{5}-\dfrac{2}{15}\)

          \(x=\dfrac{22}{15}\)

d) \(x-\dfrac{3}{4}=\dfrac{1}{2}+\dfrac{2}{6}\)

   \(x-\dfrac{3}{4}=\dfrac{5}{6}\)

   \(x=\dfrac{5}{6}+\dfrac{3}{4}\)

   \(z=\dfrac{19}{12}\)

Ví dụ : 5/2 = 5 phần 2

a) \(\dfrac{4}{5}\left(x-\dfrac{1}{3}\right)-3\dfrac{1}{2}=50\%\)

\(\dfrac{4}{5}x-\dfrac{4}{15}-\dfrac{7}{2}=\dfrac{1}{2}\)

\(\dfrac{4}{5}x=\dfrac{1}{2}+\dfrac{7}{2}+\dfrac{4}{15}\)

\(\dfrac{4}{5}x=\dfrac{64}{15}\)

\(x=\dfrac{64}{15}:\dfrac{4}{5}\\ x=\dfrac{16}{3}\)

b) \(8\dfrac{3}{5}.x-2\dfrac{1}{5}.x=16\%\)

\(\dfrac{43}{5}x-\dfrac{11}{5}x=\dfrac{4}{25}\)

\(\dfrac{32}{5}x=\dfrac{4}{25}\)

\(x=\dfrac{4}{25}:\dfrac{32}{5}\)

\(x=\dfrac{1}{40}\)

2: Tìm x

a) Ta có: x+25=40

nên x=40-25=15

Vậy: x=15

b) Ta có: 198-(x+4)=120

\(\Leftrightarrow x+4=198-120=78\)

hay x=78-4=74

Vậy: x=74

c) Ta có: \(\left(2x-7\right)\cdot3=125\)

\(\Leftrightarrow2x-7=\dfrac{125}{3}\)

\(\Leftrightarrow2x=\dfrac{125}{3}+7=\dfrac{125}{3}+\dfrac{21}{3}=\dfrac{146}{3}\)

\(\Leftrightarrow x=\dfrac{146}{3}:2=\dfrac{146}{6}=\dfrac{73}{3}\)

Vậy: \(x=\dfrac{73}{3}\)

d) Ta có: \(x+16⋮x+1\)

\(\Leftrightarrow x+1+15⋮x+1\)

mà \(x+1⋮x+1\)

nên \(15⋮x+1\)

\(\Leftrightarrow x+1\inƯ\left(15\right)\)

\(\Leftrightarrow x+1\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)

hay \(x\in\left\{0;-2;2;-4;4;-6;14;-16\right\}\)

Vậy: \(x\in\left\{0;-2;2;-4;4;-6;14;-16\right\}\)

\(a,x+25=40\\ \Rightarrow x=40-25\\ \Rightarrow x=15\\ b,198-\left(x+4\right)=120\\ \Rightarrow-\left(x+4\right)=120-198\\ \Rightarrow-\left(x+4\right)=-78\\ \Rightarrow x+4=78\\ \Rightarrow x=78-4\\ \Rightarrow x=74\\ c,\left(2x-7\right).3=125\\ \Rightarrow2x-7=\dfrac{125}{3}\\ \Rightarrow2x=\dfrac{125}{3}+7\\ \Rightarrow2x=\dfrac{146}{3}\\ \Rightarrow x=\dfrac{146}{3}:2\Rightarrow x=\dfrac{73}{3}\\ d,\left(x+16\right)⋮\left(x+1\right)\\ \Rightarrow\left[\left(x+1\right)+15\right]⋮\left(x+1\right)\\ mà:\left(x+1\right)⋮\left(x+1\right)\\ \Rightarrow15⋮\left(x+1\right)\\ \Rightarrow\left(x+1\right)\inƯ\left(15\right)\\ \Rightarrow\left(x+1\right)\in\left\{-15;-1;1;15\right\}\\ \Rightarrow x\in\left\{-16;-2;0;14\right\}\)

Tự kết luận nhé bạn

ở oooo

e: \(\Leftrightarrow3x\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

Bài 1

2/5 - 3/5 : (0,75 + 30%)

= 2/5 - 3/5 : 21/20

= 2/5 - 4/7

= -6/35

‐--------------------

Bài 2

a) x/6 = -11/3

x = -11/3 . 6

x = -22

b) 16/5 của x là 48

x = 48 : 16/5

x = 15

bài1

2/5-3/5:(0,75+30%)=2/5-3/5:(75/100+30/100)

                               =2/5-3/5:105/100

                               =2/5-3/5:21/20

                               = 2/5-3/5. 20/21

                               =2/5-1/1.4/7

                               =2/5-4/7

                               =2/5+-4/7

                               =14/35+-20/35

                               =34/35

Bài 3: 

a: Ta có: 60-3(x-2)=51

\(\Leftrightarrow x-2=3\)

hay x=5

b: Ta có: \(4x-20=25:2^2\)

\(\Leftrightarrow4x=\dfrac{25}{4}+20=\dfrac{105}{4}\)

hay \(x=\dfrac{105}{16}\)

c: Ta có: \(8\cdot6+288:\left(x-3\right)^2=50\)

\(\Leftrightarrow288:\left(x-3\right)^2=50-48=2\)

\(\Leftrightarrow\left(x-3\right)^2=144\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)