It is believed that he won the prize in the contest yesterday.

He is believed to have won the prize in the contest yesterday.

a) \(M_X=M_{Br2}=160\) (đvC)

b) CT của hợp chất : X2O3

Ta có : \(2X+16.3=160\)

=> X=56

Vậy X là Fe

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

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

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

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

\(g,ĐK:x\ge0\\ PT\Leftrightarrow10\sqrt{x}+8\sqrt{x}-11\sqrt{x}=21\\ \Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\left(tm\right)\\ h,ĐK:x\ge0\\ PT\Leftrightarrow6\sqrt{3x}+2\sqrt{3x}-3\sqrt{3x}=15\\ \Leftrightarrow\sqrt{3x}=5\Leftrightarrow3x=25\Leftrightarrow x=\dfrac{25}{3}\left(tm\right)\\ i,ĐK:x\ge0\\ PT\Leftrightarrow12\sqrt{x}-21-2\sqrt{x}+10=6\sqrt{x}-12\\ \Leftrightarrow4\sqrt{x}=-1\Leftrightarrow\sqrt{x}=-\dfrac{1}{4}\Leftrightarrow x\in\varnothing\\ j,ĐK:x\ge2\\ PT\Leftrightarrow6\sqrt{x-2}-15\cdot\dfrac{1}{5}\sqrt{x-2}=20+4\sqrt{x-2}\\ \Leftrightarrow\sqrt{x-2}=-20\Leftrightarrow x\in\varnothing\)

\(k,ĐK:x\ge3\\ PT\Leftrightarrow6\sqrt{x-3}-\dfrac{1}{5}\cdot5\sqrt{x-3}-\dfrac{1}{7}\cdot7\sqrt{x-3}=20\\ \Leftrightarrow4\sqrt{x-3}=20\Leftrightarrow\sqrt{x-3}=5\\ \Leftrightarrow x-3=25\Leftrightarrow x=28\left(tm\right)\\ l,ĐK:x\ge5\\ PT\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\\ \Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)

Các bạn chỉ cần giúp mik câu c ạ

\(C=\dfrac{x^3}{x^2-4}-\dfrac{x}{x-2}+\dfrac{2}{x+2}\)

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

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

\(=\dfrac{x^3-x^2-4}{x^2-4}\)

a,\(C=\dfrac{x^3}{x^2-4}-\dfrac{x}{x-2}-\dfrac{2}{x+2}\)

\(\Rightarrow C=\dfrac{x^3}{\left(x-2\right)\left(x+2\right)}-\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow C=\dfrac{x^3}{\left(x-2\right)\left(x+2\right)}-\dfrac{x^2+2x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow C=\dfrac{x^3-x^2-2x-2x+4}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow C=\dfrac{x^3-x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow C=\dfrac{x^2\left(x-1\right)-4\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow C=\dfrac{\left(x^2-4\right)\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow C=\dfrac{\left(x-2\right)\left(x+2\right)\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow C=x-1\)

b, C=0\(\Rightarrow x-1=0\Rightarrow x=1\)

c, Để C nhận giá trị dương thì \(x-1\ge0\Rightarrow x\ge1\)