a, \(9x+3x\left(2x^2+x-3\right)=9x+6x^3+3x^2-9x\)

b, \(\left(3x-1\right)^2-9x\left(x+1\right)=9x^2-6x+1-9x^2-9x=1-15x\)

c, \(\left(x-1\right)^2-x\left(x+1\right)=x^2-2x+1-x^2-x=1-3x\)

Ko hiểu làm sao mà giúp được

\(\dfrac{1}{2}-3x+\left|x-1\right|=0\\ \Rightarrow3x+\left|x-1\right|=\dfrac{1}{2}-0\\ \Rightarrow3x+\left|x-1\right|=\dfrac{1}{2}\\ \Rightarrow\left|x-1\right|=\dfrac{1}{2}-3x\\ \Rightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{2}-3x\\x-1=-\dfrac{1}{2}+3x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x+3x=\dfrac{1}{2}+1\\x-3x=-\dfrac{1}{2}+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}4x=\dfrac{3}{2}\\2x=\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{8}\\x=\dfrac{1}{4}\end{matrix}\right.\)

__

\(\dfrac{1}{2}\left|2x-1\right|+\left|2x-1\right|=x+1\\ \Rightarrow\left|2x-1\right|\cdot\left(\dfrac{1}{2}+1\right)=x+1\\ \Rightarrow\left|2x-1\right|\cdot\dfrac{3}{2}=x+1\\ \Rightarrow\left|2x-1\right|=x+1:\dfrac{3}{2}\\ \Rightarrow\left|2x-1\right|=x+\dfrac{2}{3}\\ \Rightarrow\left[{}\begin{matrix}2x-1=x+\dfrac{2}{3}\\2x-1=-x-\dfrac{2}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-x=\dfrac{2}{3}+1\\2x+x=-\dfrac{2}{3}+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\3x=\dfrac{1}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{1}{9}\end{matrix}\right.\)

\(\left(2x+x^2\right)\left(x^2-3x+2\right)=0\Leftrightarrow x\left(x+2\right)\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=1\\x=2\end{matrix}\right.\\ A=\left\{-2;0;1;2\right\}\)

\(3\le x^3\le27\Leftrightarrow x\in\left\{2;3\right\}\\ B=\left\{2;3\right\}\)

\(\Leftrightarrow A\cup B=\left\{-2;0;1;2;3\right\}\)

a) Ta có: \(2x\left(x+3\right)-3\left(x^2+1\right)=x+1-x\left(x-2\right)\)

\(\Leftrightarrow2x^2+6x-3x^2-3=x+1-x^2+2x\)

\(\Leftrightarrow-x^2+6x-3=-x^2+3x+1\)

\(\Leftrightarrow-x^2+6x-3+x^2-3x-1=0\)

\(\Leftrightarrow3x-4=0\)

\(\Leftrightarrow3x=4\)

\(\Leftrightarrow x=\frac{4}{3}\)

Vậy: \(S=\left\{\frac{4}{3}\right\}\)

b) Ta có: \(3x\left(x-2\right)-x\left(1+3x\right)=14\)

\(\Leftrightarrow3x^2-6x-x-3x^2-14=0\)

\(\Leftrightarrow-7x-14=0\)

\(\Leftrightarrow-7x=14\)

hay x=-2

Vậy: S={-2}

c) Ta có: \(2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(\Leftrightarrow2x^2-10x-3x-2x^2=26\)

\(\Leftrightarrow-13x=26\)

hay x=-2

Vậy: S={-2}

a: =>(3x+1)(3x-1)-(3x+1)(2x-3)=0

=>(3x+1)(3x-1-2x+3)=0

=>(3x+1)(x+2)=0

=>x=-1/3 hoặc x=-2

b: =>(3x+1)(6x+2)-(3x+1)(x-2)=0

=>(3x+1)(6x+2-x+2)=0

=>(3x+1)(5x+4)=0

=>x=-1/3 hoặc x=-4/5

a,

(x²-x+1)(x+1)-x³+3x=15

x³-x²+x+x²-x+1-x³+3x=15

x³-x³-x²+x²+x-x+3x+1=15

3x+1=15

3x=15-1

3x=14

x=14/3

b,

(x+3)(x-2)+3x=\(\frac{4}{x+\frac{3}{4}}\)

x²-2x+3x-6+3x=\(\frac{4}{x+\frac{3}{4}}\)

x²-2x+3x+3x-6=\(\frac{4}{x+\frac{3}{4}}\)

Tới đây hết biết , đề có gì sai sai sao ý !

c,

(x²-5)(x+2)+5x=2x²+17

x³+2x²-5x-10+5x=2x²+17

x³+2x²-5x+5x-10=2x²+17

x³+2x²-10=2x²+17

x³-10=17

x³=17+10

x³=27

\(\Rightarrow x=3\)(Vì : 3³=27)

_k_ nhé bn

Nhân ra thôi bạn, có hằng đẳng thức gì đâu !

a) \(\left(x^2-x+1\right)\left(x+1\right)-x^3+3x=15\)

\(\Leftrightarrow\left(x^2-x+1\right)\cdot x+x^2-x+1-x^3+3x=15\)

\(\Leftrightarrow x^3-x^2+x+x^2-x+1-x^3+3x=15\)

\(\Leftrightarrow1+3x=15\Leftrightarrow3x=14\Leftrightarrow x=\frac{14}{3}\)

b) \(\left(x+3\right)\left(x-2\right)+3x=4\cdot\left(x+\frac{3}{4}\right)\)

\(\Leftrightarrow x^2+3x-2x-6+3x=4x+3\)

\(\Leftrightarrow x^2+4x-6=4x+3\)

\(\Leftrightarrow x^2=9\Leftrightarrow\orbr{\begin{cases}x=-3\\x=3\end{cases}}\)

c) \(\left(x^2-5\right)\left(x+2\right)+5x=2x^2+17\)

\(\Leftrightarrow x^3-5x+2x^2-10+5x=2x^2+17\)

\(\Leftrightarrow x^3=27\Leftrightarrow x=3\)

a) \(\dfrac{1}{4}+\dfrac{3}{4}:x=-2\)

\(\dfrac{3}{4}:x=-2-\dfrac{1}{4}=\dfrac{-8}{4}-\dfrac{1}{4}\)

\(\dfrac{3}{4}:x=\dfrac{-9}{4}\)

\(x=\dfrac{3}{4}:\dfrac{-9}{4}=\dfrac{3}{4}.\dfrac{-4}{9}\)

\(x=\dfrac{-1}{3}\)

b) \(\dfrac{3}{4}+2.\left(2x-\dfrac{2}{3}\right)=-2\)

\(2.\left(2x-\dfrac{2}{3}\right)=-2-\dfrac{3}{4}=\dfrac{-8}{4}-\dfrac{3}{4}\)

\(2.\left(2x-\dfrac{2}{3}\right)=\dfrac{-11}{4}\)

\(2x-\dfrac{2}{3}=\dfrac{-11}{4}:2=\dfrac{-11}{4}.\dfrac{1}{2}\)

\(2x-\dfrac{2}{3}=\dfrac{-11}{8}\)

\(2x=\dfrac{-11}{8}+\dfrac{2}{3}=\dfrac{-33}{24}+\dfrac{16}{24}\)

\(2x=\dfrac{-17}{24}\)

\(x=\dfrac{-17}{24}:2=\dfrac{-17}{24}.\dfrac{1}{2}\)

\(x=\dfrac{-17}{48}\)

c) \(\left(\dfrac{1}{2}+5x\right).\left(2x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}+5x=0\\2x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{-1}{2}\\2x=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{10}\\x=\dfrac{3}{2}\end{matrix}\right.\)

a, 1/4 + 3/4 : x = -2

     3/4 : x = -2 - 1/4 

     3/4 : x = -9/4

             x = 3/4 : -9/4

             x = -1/3

b, 3/4 + 2 . ( 2x - 3 ) = 0 

             2 . ( 2x - 3 ) = 0 - 3/4

             2 . ( 2x - 3 ) = -3/4

                     2x - 3 = -3/4 : 2

                     2x - 3 = -3/8

                          2x = -3/8 + 3

                          2x = 21/8

                            x = 21/8 : 2

                            x = 21/16