Câu 4:

PTHH: \(Zn+S\underrightarrow{t^o}ZnS\)

Ta có: \(\left\{{}\begin{matrix}n_{Zn}=\dfrac{0,65}{65}=0,01\left(mol\right)\\n_S=\dfrac{0,384}{32}=0,012\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\) Lưi huỳnh còn dư, Kẽm p/ứ hết

\(\Rightarrow\left\{{}\begin{matrix}n_{ZnS}=0,01\left(mol\right)\\n_{S\left(dư\right)}=0,002\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ZnS}=0,01\cdot97=0,97\left(g\right)\\m_{S\left(dư\right)}=0,002\cdot32=0,064\left(g\right)\end{matrix}\right.\)

Câu 4:  (Bonus)

Ta có: \(n_{FeS}=\dfrac{17,6}{88}=0,2\left(mol\right)\)

Bảo toàn nguyên tố Lưu huỳnh: \(n_{FeS}=n_{PbS}=0,2\left(mol\right)\)

\(\Rightarrow m_{PbS}=0,2\cdot239=47,8\left(g\right)\)

c: Ta có: \(\left(\dfrac{1}{2}\right)^{2x+1}=\dfrac{1}{8}\)

\(\Leftrightarrow2x+1=3\)

\(\Leftrightarrow2x=2\)

hay x=1

d: Ta có: \(\left(-\dfrac{1}{3}\right)^{x+3}=\dfrac{1}{81}\)

\(\Leftrightarrow x+3=4\)

hay x=1

\(A=\dfrac{2^{13}\cdot3^7}{2^{15}\cdot3^2\cdot9^2}=\dfrac{2^{13}\cdot3^7}{2^{15}\cdot3^6}=\dfrac{3}{4}\)

\(C=27\cdot\left(-\dfrac{3}{2}\right)^{-5}\cdot\left(-\dfrac{2}{5}\right)^{-4}:\left(\dfrac{2}{125}\right)^{-1}\)

\(=27\cdot\dfrac{-32}{243}\cdot\dfrac{625}{16}\cdot\dfrac{2}{125}\)

\(=\dfrac{-32}{9}\cdot\dfrac{1}{8}\cdot5\)

\(=-\dfrac{20}{9}\)

Giúp mình với ạ !! Mình cần gấp lắm . 

Câu hỏi đâu vậy bạn

1.

a, \(\sqrt{3}sin2x+2cos^2x=0\)

\(\Leftrightarrow\sqrt{3}sin2x+2cos^2x-1=-1\)

\(\Leftrightarrow\sqrt{3}sin2x+cos2x=-1\)

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sin2x+\dfrac{1}{2}cos2x=-\dfrac{1}{2}\)

\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=-\dfrac{1}{2}\)

\(\Leftrightarrow2x-\dfrac{\pi}{3}=\dfrac{2\pi}{3}+k2\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\)

1.

c, ĐK: \(x\ne k2\pi\)

\(\dfrac{2sin\left(x+\dfrac{\pi}{6}\right)-cos2x}{cosx-1}\)

\(\Leftrightarrow\sqrt{3}sinx+cosx-cos2x=cosx-1\)

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx-\dfrac{1}{2}cos2x=-\dfrac{1}{2}\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{6}\right)=-\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=-\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{6}=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\left(l\right)\\x=\dfrac{4\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{4\pi}{3}+k2\pi\)

2.

ĐK: \(n\ge2;n\in N\)

\(A_n^2+C_{n+1}^3=10\left(n-1\right)\)

\(\Leftrightarrow\dfrac{n!}{\left(n-2\right)!}+\dfrac{\left(n+1\right)!}{3!.\left(n-2\right)!}=10\left(n-1\right)\)

\(\Leftrightarrow n\left(n-1\right)+\dfrac{\left(n+1\right)n\left(n-1\right)}{6}=10\left(n-1\right)\)

\(\Leftrightarrow6n^2-6n+n^3-n=60n-60\)

\(\Leftrightarrow n^3+6n^2-67n+60=0\)

\(\Leftrightarrow\left[{}\begin{matrix}n=-12\left(l\right)\\n=5\left(tm\right)\\n=1\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow n=5\)

a: \(M=\left|x\right|+x\)

\(=\left[{}\begin{matrix}x+x=2x\left(x\ge0\right)\\-x+x=0\left(x< 0\right)\end{matrix}\right.\)

b: \(N=\left|x\right|:x=\pm1\)

t cx vừa hỏi câu này xong:))