Question

[Nguyễn Thanh Huyền]

cần giúp gấp vs ạ huhu ( đề bài : tìm x biêt)

Answer 1

[Nguyễn Lê Phước Thịnh]

k: \(\left(4x-16\right)\left(-72+9x\right)=0\)

=>\(4\cdot\left(x-4\right)\cdot9\left(x-8\right)=0\)

=>\(36\left(x-4\right)\left(x-8\right)=0\)

=>\(\left(x-4\right)\left(x-8\right)=0\)

=>\(\left[{}\begin{matrix}x-4=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=8\end{matrix}\right.\)

m: \(\left(20+5x\right)\left(4x-8\right)=0\)

=>\(5\cdot\left(x+4\right)\cdot4\left(x-2\right)=0\)

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

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

n: \(\left(-4x+48\right)\left(2x-24\right)=0\)

=>\(-4\left(x-12\right)\cdot2\left(x-12\right)=0\)

=>\(\left(x-12\right)^2=0\)

=>x-12=0

=>x=12

o: \(\left(4x+16\right)\left(-2x+20\right)\left(-40+x\right)=0\)

=>\(4\cdot\left(x+4\right)\cdot\left(-2\right)\left(x-10\right)\left(x-40\right)=0\)

=>\(\left(x+4\right)\left(x-10\right)\left(x-40\right)=0\)

=>\(\left[{}\begin{matrix}x+4=0\\x-10=0\\x-40=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=10\\x=40\end{matrix}\right.\)

p: \(\left(-5x+40\right)\left(-x+2023\right)\left(2x-2\right)=0\)

=>\(-5\left(x-8\right)\cdot\left(-1\right)\cdot\left(x-2023\right)\cdot2\left(x-1\right)=0\)

=>\(\left(x-8\right)\left(x-2023\right)\left(x-1\right)=0\)

=>\(\left[{}\begin{matrix}x-8=0\\x-1=0\\x-2023=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=1\\x=2023\end{matrix}\right.\)

q: \(2024x\left(4x-8\right)\left(5+5x\right)=0\)

=>\(x\cdot4\left(x-2\right)\cdot5\left(x+1\right)=0\)

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

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

r: \(-4x\left(3x+9\right)\left(2x-16\right)=0\)

=>\(-4x\cdot3\left(x+3\right)\cdot2\left(x-8\right)=0\)

=>\(x\left(x+3\right)\left(x-8\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x+3=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=8\end{matrix}\right.\)

s: \(\left(-100+5x\right)\left(2x-10\right)\left(6x+6\right)=0\)

=>\(5\cdot\left(x-20\right)\cdot2\left(x-5\right)\cdot6\left(x+1\right)=0\)

=>\(\left(x-20\right)\left(x-5\right)\left(x+1\right)=0\)

=>\(\left[{}\begin{matrix}x-20=0\\x-5=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=20\\x=5\\x=-1\end{matrix}\right.\)

t: \(\left(-2x+4\right)\left(2x+16\right)\cdot\left(7-x\right)=0\)

=>\(-2\left(x-2\right)\cdot2\left(x+8\right)\cdot\left(-1\right)\cdot\left(x-7\right)=0\)

=>\(\left(x-2\right)\left(x+8\right)\left(x-7\right)=0\)

=>\(\left[{}\begin{matrix}x-2=0\\x-7=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-8\\x=7\end{matrix}\right.\)