Cái bài này nó chưa cho điểm F là điểm nào á nha bạn

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: \(n_{FeS}=n_{PbS}=0,2\left(mol\right)\)

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

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:))

Câu hỏi đâu em?

câu hổi đâu

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

1.

a, \(sin2x-\sqrt{3}cos2x=-1\)

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

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

\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{3}\right)=sin\left(-\dfrac{\pi}{6}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=-\dfrac{\pi}{6}+k2\pi\\2x-\dfrac{\pi}{3}=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+k\pi\\x=\dfrac{3\pi}{4}+k\pi\end{matrix}\right.\)

Do tổng các hệ số thứ 1,2,3 là 46 nên ta có:\(C_n^0+C_n^1+C_n^2=46\)

\(\Leftrightarrow1+\dfrac{n!}{1!\left(n-1\right)!}+\dfrac{n!}{2!\left(n-2\right)!}=46\)

\(\Leftrightarrow1+n+\dfrac{\left(n-1\right)n}{2}=46\)

\(\Leftrightarrow n^2+n-90=0\)

\(\Leftrightarrow\left[{}\begin{matrix}n=9\\n=-10\left(loai\right)\end{matrix}\right.\)

Khai triển biểu thức: \(\left(x+\dfrac{1}{x}\right)^9\)

Hạng tử thứ k+1 trong biểu thức trên

\(\left(x+\dfrac{1}{x}\right)^9=C_9^{k+1}+\left(x^2\right)^{10-k}.\left(\dfrac{1}{x}\right)^{k+1}\)

đến đây mình chịu rùi hjhj b nào làm được giúp b kia với

1.

b, \(cos^2\left(x+\dfrac{\pi}{3}\right)+cos^2\left(x+\dfrac{2\pi}{3}\right)=\dfrac{1}{2}\left(sinx+1\right)\)

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

\(\Leftrightarrow2cos^2\left(x+\dfrac{\pi}{3}\right)-1+2cos^2\left(x-\dfrac{\pi}{3}\right)-1=sinx-1\)

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

\(\Leftrightarrow2cos2x.cos\dfrac{2\pi}{3}=sinx-1\)

\(\Leftrightarrow-cos2x=sinx-1\)

\(\Leftrightarrow2sin^2x-sinx=0\)

\(\Leftrightarrow sinx\left(2sinx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=\dfrac{1}{2}\end{matrix}\right.\)

Đến đây dễ rồi, tự làm tiếp.