\(\text{(5x-1)(2x-1/3) =0}\)

\(\Rightarrow\hept{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}\Rightarrow\hept{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}\Rightarrow\hept{\begin{cases}x=0.2\\x=0.17\end{cases}}}}\)

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=>\(\orbr{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}}\)=>\(\orbr{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}}\)=>\(\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)

\(\text{1 số khi nhân 0 thì bằng 0}\)

\(\text{vậy ta có 2 trường hợp có thể xảy ra:}\)

\(1.5x-1=0\Rightarrow x=\frac{1}{5}=0.2\)

\(2.2x-\frac{1}{3}=0\Rightarrow x=\frac{1}{6}\)

Cái này chi tiết lắm rồi,nếu k hỉu nữa thì bn hơi bị....k giỏi toán cho lắm

\( \left(5x-2\right)\left(x-3\right)=0\)

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

\(\left(x^2-9\right)^2-\left(x-3\right)^2=0\)

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

\(\Leftrightarrow\left(x-3\right)^2\left[\left(x+3\right)^2-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-3\right)^2=0\\\left(x+3\right)^2=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=1\\x+3=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\\x=-4\end{matrix}\right.\)

\(\left(x^2-9\right)^2-\left(x-3\right)^2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\\x=-4\end{matrix}\right.\)

\(\left(2x-1\right)^2+\left(x-3\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(2x-1+x-3\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(3x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\3x=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{4}{3}\end{matrix}\right.\)

Vậy \(S=\left\{\dfrac{1}{2};\dfrac{4}{3}\right\}\)

a)\(2x^2\)+\(3\left(x^2-1\right)\)=\(5x\left(x+1\right)\)

\(2x^2\)+\(3x^2\)\(-3\)=\(5x^2+5x\)

\(5x^2-5x^2-5x=3\)

\(-5x=3\)

\(x=\frac{-3}{5}\)

tự ghi dấu suy ra ở đằng trước nhé

b) Vì \(2x\left(5-3x\right)=2x\left(3x-5\right)-3\left(x-7\right)=3\)

nên chỉ cần giải: \(6x^2-10x-3x+21=3\)

\(\Leftrightarrow6x^2-13x+21=3\)

\(\Leftrightarrow6x^2-13x+18=0\)

\(\Rightarrow\)pt vô nghiệm

Ta có: \(x^3-5x^2+6x=0\)

\(\Leftrightarrow x\left(x^2-5x+6\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x-3\right)=0\)

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

Vậy: S={0;2;3}

\(x^3-5x^2+6x=0\)

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

\(\Leftrightarrow x^2\left(x-2\right)-3x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x^2-3x\right)\left(x-2\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x-3\right)=0\)

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

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

\(S=\left\{0,2,3\right\}\)

x^3 -5x^2 + 6x=0

\(< =>x\left(x^2-5x+6\right)=0\)

\(< =>x=0;x^2-5x+6=0\)

+/ \(x^2-5x+6=0\)

\(< =>x^2-2x-3x+6=0\)

\(< =>\left(x^2-2x\right)-\left(3x-6\right)=0\)

\(< =>x\left(x-2\right)-3\left(x-2\right)=0\)

\(< =>\left(x-2\right)\left(x-3\right)=0\)

=> \(x-2=0;x-3=0\)

=> \(x=2;3\)

Vayayj tập nghiệm của......

\(4x^2+4x+1+4x+2-2x^2-x\le0\)

\(\Leftrightarrow2x^2+7x+3\le0\Leftrightarrow\left(2x+1\right)\left(x+3\right)\le0\)

TH1 : \(\left\{{}\begin{matrix}2x+1\ge0\\x+3\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{1}{2}\\x\le-3\end{matrix}\right.\)<=> -1/2 =< x =< -3 

TH2 : \(\left\{{}\begin{matrix}2x+1\le0\\x+3\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le-\dfrac{1}{2}\\x\ge-3\end{matrix}\right.\)( vô lí ) 

1:

a: =>(|x|+4)(|x|-1)=0

=>|x|-1=0

=>x=1; x=-1

b: =>x^2-4>=0

=>x>=2 hoặc x<=-2

d: =>|2x+5|=2x-5

=>x>=5/2 và (2x+5-2x+5)(2x+5+2x-5)=0

=>x=0(loại)