\(\frac{x-1}{9}+\frac{x-2}{4}+\frac{x-5}{5}=4\)

\(\Rightarrow\left(\frac{x-1}{9}-1\right)+\left(\frac{x-2}{4}-2\right)+\left(\frac{x-5}{5}-1\right)=4-4\)

\(\frac{x-10}{9}+\frac{x-10}{4}+\frac{x-10}{5}=0\)

\(\left(x-10\right).\left(\frac{1}{9}+\frac{1}{4}+\frac{1}{5}\right)=0\)

mà \(\frac{1}{9}+\frac{1}{4}+\frac{1}{5}\ne0\)

=> x - 10 = 0

=> x = 10

\(\left(3x+5\right)\left(7-2x\right)+6x\left(x+4\right)=0\)

\(\Leftrightarrow 21x-6x^2+35-10x+6x^2+24x=0\)

\(\Leftrightarrow35x+35=0\)

\(\Leftrightarrow35\left(x+1\right)=0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

\(x^3-25x=0\)

\(\Leftrightarrow x\left(x^2-25\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-25=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm5\end{cases}}\)

Vậy ......

À câu này mình từng làm 1 lần rồi nè: https://olm.vn/hoi-dap/question/1274928.html

\(18-2x^2-2ax-6a\)

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

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

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

Tham khảo nhé~

\(x^4-4x^3-7x^2+22x+24\)

\(=\left(x^4-4x^3\right)-\left(7x^2-28x\right)-\left(6x-24\right)\)

\(=x^4.\left(x-4\right)-7x.\left(x-4\right)-6.\left(x-4\right)\)

\(=\left(x-4\right).\left(x^4-7x-6\right)\)

Tham khảo nhé~