\(m_{HCl\left(1\right)}=600.2,5\%=15g\\ m_{HCl\left(2\right)}=400.15\%=60g\\ C\%_{HCl.sau}=\dfrac{75}{1000}\cdot100\%=7,5\%\)

Cảm ơn nhieeuf ạ

1: \(2\cdot\sin x+\left(m-2\right)\cdot cosx=m\)

Để phương trình có nghiệm thì \(2^2+\left(m-2\right)^2\ge m^2\)

=>\(m^2-4m+4+4\ge m^2\)

=>-4m+8>=0

=>-4m>=-8

=>m<=2

2: \(cos^2x+m\cdot\sin2x=5\)

=>\(m\cdot\sin2x+\frac{1+cos2x}{2}=5\)

=>\(m\cdot\sin2x+\frac12\cdot cos2x=5-\frac12=\frac92\)

Để phương trình có nghiệm thì \(m^2+\left(\frac12\right)^2>=\left(\frac92\right)^2\)

=>\(m^2\ge\frac{81}{4}-\frac14=\frac{80}{4}=20\)

=>\(\left[\begin{array}{l}m\ge2\sqrt5\\ m\le-2\sqrt5\end{array}\right.\)

3: \(m\cdot\sin4x+cos^22x-7=0\)

=>\(m\cdot\sin4x+\frac{1+cos4x}{2}=7\)

=>\(m\cdot\sin4x+\frac12\cdot cos4x=7-\frac12=\frac{13}{2}\)

Để phương trình có nghiệm thì \(m^2+\left(\frac12\right)^2\ge\left(\frac{13}{2}\right)^2\)

=>\(m^2\ge\frac{169}{4}-\frac14=\frac{168}{4}=42\)

=>\(\left[\begin{array}{l}m\ge\sqrt{42}\\ m\le-\sqrt{42}\end{array}\right.\)

4: Để phương trình có nghiệm thì \(\left(m-2\right)^2+m^2\ge\left(2m\right)^2\)

=>\(m^2-4m+4+m^2-4m^2\ge0\)

=>\(-2m^2-4m+4\ge0\)

=>\(-2\left(m^2+2m-2\right)\le0\)

=>\(m^2+2m-2\le0\)

=>\(m^2+2m+1\le3\)

=>\(\left(m+1\right)^2\le3\)

=>\(-\sqrt3\le m+1\le\sqrt3\)

=>\(-\sqrt3-1\le m\le\sqrt3-1\)

BTKL: \(a=\Sigma m_{O_2}=m_{hh\text{ oxit}}-m_{hh\text { KL}}=14,2-7,8=6,4(g)\)

cho em hỏi xíu là \(m_{hh}\) là gì vậy ạ?

Ta có : \(T=1+2^2+2^4+2^6+....+2^{100}\)

\(\Rightarrow2^2\cdot T=2^2+2^4+2^6+2^8+....+2^{102}\)

\(\Rightarrow4T-T=\left(2^2+2^4+2^6+...+2^{102}\right)-\left(1+2^2+2^4+...+2^{100}\right)\)

\(\Rightarrow3T=2^{102}-1\Rightarrow T=\frac{2^{102}-1}{3}\)

Đề bài yêu cầu gì?

Dạ giống v nè 

ĐKXĐ: x<>1 và y<>0

\(\left\{{}\begin{matrix}\dfrac{1}{x-1}+\dfrac{1}{y}=-1\\\dfrac{3}{x-1}-\dfrac{2}{y}=7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{2}{x-1}+\dfrac{2}{y}=-2\\\dfrac{3}{x-1}-\dfrac{2}{y}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x-1}=5\\\dfrac{1}{x-1}+\dfrac{1}{y}=-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x-1=1\\\dfrac{1}{y}=-1-1=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-\dfrac{1}{2}\end{matrix}\right.\left(nhận\right)\)