\(3\left(x-2\right)-4x+5=2\left(2x+1\right)-18\)

\(\Leftrightarrow3x-6-4x+5=4x+2-18\)

\(\Leftrightarrow-x-1=4x-16\)

\(\Leftrightarrow-x-1-4x+16=0\)

\(\Leftrightarrow-5x+15=0\)

\(\Leftrightarrow-5x=-15\)

\(\Leftrightarrow x=3\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{3\right\}\)

\(\left(x+1\right)^2+\left(2x-4\right)\left(x+1\right)=0\)

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

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

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

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{\pm1\right\}\)

\(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-3x-4x+12\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left[\left(x^2-3x\right)-\left(4x-12\right)\right]=0\)

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

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

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

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{\dfrac{1}{3};3;4\right\}\)

\(\left(3x-2\right)\left(4x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\4x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{-5}{4}\end{matrix}\right.\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{\dfrac{2}{3};\dfrac{-5}{4}\right\}\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(1mol\)    \(2mol\)     \(1mol\)      \(1mol\)

\(0,05mol\)  \(0,1mol\)  \(0,05mol\)   \(0,05mol\)

\(n_{Zn}=\dfrac{m}{M}=\dfrac{13}{65}=0,2\left(mol\right)\)

\(N_{HCl}=\dfrac{m}{M}=\dfrac{3,65}{36,5}=0,1\left(mol\right)\)

\(\text{Ta thấy Zn dư,HCl phản ứng hết}\)

\(m_{ZnCl_2}=n.M=0,05.136=6,8\left(g\right)\)

\(m_{H_2}=n.M=0,05.2=0,1\left(g\right)\)

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

\(5x-10=0\)

\(\Leftrightarrow5x=10\)

\(\Leftrightarrow x=2\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{2\right\}\)

\(\text{Ta có:}\dfrac{AB}{CD}=\dfrac{5}{4}\left(gt\right)\)

\(\text{Mà CD=20cm(gt)}\)

\(\Rightarrow\dfrac{AB}{20}=\dfrac{5}{4}\Rightarrow AB=\dfrac{20.5}{4}=25\left(cm\right)\)

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

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

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{0;\dfrac{1}{2};-3\right\}\)

\(\dfrac{5x}{2x\left(x+1\right)}+1=\dfrac{-6}{x+1}\text{ĐKXĐ:}x\ne0;x\ne-1\)

\(\Leftrightarrow\dfrac{5x}{2x\left(x+1\right)}+\dfrac{2x\left(x+1\right)}{2x\left(x+1\right)}=\dfrac{-6.2x}{2x\left(x+1\right)}\)

\(\Rightarrow5x+2x^2+2x=-12x\)

\(\Leftrightarrow2x^2+7x=-12x\)

\(\Leftrightarrow2x^2+7x+12x=0\)

\(\Leftrightarrow2x^2+19x=0\)

\(\Leftrightarrow x\left(2x+19\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x+19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\2x=-19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(\text{loại}\right)\\x=\dfrac{-19}{2}\left(\text{nhận}\right)\end{matrix}\right.\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{\dfrac{-19}{2}\right\}\)