Tất cả k dưới đây đều là \(k\in Z\)

6.

\(\Leftrightarrow\sqrt{3}cot\left(3x-\dfrac{\pi}{3}\right)=1\)

\(\Leftrightarrow cot\left(3x-\dfrac{\pi}{3}\right)=\dfrac{1}{\sqrt{3}}\)

\(\Leftrightarrow cot\left(3x-\dfrac{\pi}{3}\right)=cot\left(\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow3x-\dfrac{\pi}{3}=\dfrac{\pi}{3}+k\pi\)

\(\Leftrightarrow3x=\dfrac{2\pi}{3}+k\pi\)

\(\Leftrightarrow x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\) 

7.

\(\Leftrightarrow\sqrt{3}tan\left(3x-15^0\right)=-1\)

\(\Leftrightarrow tan\left(3x-15^0\right)=-\dfrac{1}{\sqrt{3}}\)

\(\Leftrightarrow tan\left(3x-15^0\right)=tan\left(-30^0\right)\)

\(\Leftrightarrow3x-15^0=-30^0+k180^0\)

\(\Leftrightarrow3x=-15^0+k180^0\)

\(\Leftrightarrow x=-3^0+k60^0\)

8.

\(\Leftrightarrow\sqrt{3}cot\left(3x-30^0\right)=1\)

\(\Leftrightarrow cot\left(3x-30^0\right)=\dfrac{1}{\sqrt{3}}\)

\(\Leftrightarrow cot\left(3x-30^0\right)=cot\left(60^0\right)\)

\(\Leftrightarrow3x-30^0=60^0+k180^0\)

\(\Leftrightarrow3x=90^0+k180^0\)

\(\Leftrightarrow x=30^0+k60^0\)

a) Cl₂ + 2NaOH --> NaClO + NaCl + H₂O

Chất oxh: Cl₂, chất khử: Cl₂

b) \(n_{Cl_2}=\dfrac{17,92}{22,4}=0,8\left(mol\right);n_{NaOH}=0,5.4=2\left(mol\right)\)

PTHH: Cl₂ + 2NaOH --> NaClO + NaCl + H₂O

_____0,8---->1,6--------->0,8---->0,8

=> \(\left\{{}\begin{matrix}C_{M\left(NaCl\right)}=\dfrac{0,8}{0,5}=1,6M\\C_{M\left(NaClO\right)}=\dfrac{0,8}{0,5}=1,6M\\C_{M\left(NaOH\right)}=\dfrac{2-1,6}{0,5}=0,8M\end{matrix}\right.\)

Bài 1.

a)Gia tốc vật: \(a=\dfrac{F}{m}=\dfrac{0,2}{0,1}=2\)m/s²

Vận tốc vật: \(v^2-v^2_0=2aS\)

\(\Rightarrow v=\sqrt{2aS}=\sqrt{2\cdot2\cdot10}=\sqrt{40}\approx6,3\)m/s

b)Với \(v'=10\)m/s thì gia tốc vật là:

\(a'=\dfrac{v'^2-v^2_0}{2S}=\dfrac{10^2-0}{2\cdot10}=5\)m/s²

Lực kéo lúc này: \(F'=m\cdot a'=0,1\cdot5=0,5N\)

Công của lực kéo F là:

\(A=F'\cdot s=0,5\cdot10=5J\)

https://hoidap247.com/cau-hoi/558074

tham khảo ạ

nH2SO4=0,045mol    nKOH=0,03

PT: X + H2SO4 ==> XSO4 + H2 (1)

      H2SO4 + 2KOH ==> K2SO4 + 2H2O (2)

Từ 1 --> nH2SO4 dư = 0,015

--> nX=nH2SO4 pứ= 0,045-0,015=0,03

MX=1,2:0,03=40 --> X là Ca

a) 

Fe không phản ứng với H₂SO₄ đặc nguội

\(n_{SO_2}=\dfrac{1,176}{22,4}=0,0525\left(mol\right)\)

PTHH: Cu + 2H₂SO₄ --> CuSO₄ + SO₂ + 2H₂O

         0,0525<-0,105<----0,0525<-0,0525

\(\%m_{Cu}=\dfrac{0,0525.64}{27}.100\%=12,44\%\)

\(\%m_{Fe}=100\%-12,44\%=87,56\%\)

b) \(C_{M\left(dd.H_2SO_4\right)}=\dfrac{0,105}{0,8}=0,13125M\)

c) n_NaOH = 1,25.0,5 = 0,625 (mol)

PTHH: 2NaOH + CuSO₄ --> Cu(OH)₂ + Na₂SO₄

Xét tỉ lệ: \(\dfrac{0,625}{2}>\dfrac{0,0525}{1}\) => NaOH dư, CuSO₄ hết

PTHH: 2NaOH + CuSO₄ --> Cu(OH)₂ + Na₂SO₄

           0,105<---0,0525------------------>0,0525   

=> \(\left\{{}\begin{matrix}C_{M\left(NaOH_{dư}\right)}=\dfrac{0,625-0,105}{0,5}=1,04M\\C_{M\left(Na_2SO_4\right)}=\dfrac{0,0525}{0,5}=0,105M\end{matrix}\right.\)

a, n_H2 = \(\dfrac{\dfrac{1176}{1000}}{22,4}=0,0525\left(mol\right)\)

PTHH: Cu + 2H₂SO₄(đặc, nguội) ---> CuSO₄ + SO₂ + 2H₂O

       0,0525   0,105                              0,0525  0,0525

=> \(\left\{{}\begin{matrix}m_{Cu}=0,0525.64=3,36\left(g\right)\\m_{Fe}=27-3,36=23,64\left(g\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{3,36}{27}=12,44\%\\\%m_{Fe}=100\%-12,44\%=87,56\%\end{matrix}\right.\)

b, \(C_{MddH_2SO_4}=\dfrac{0,105}{\dfrac{800}{1000}}0,13125M\)

c, n_NaOH = 1,25.\(\dfrac{500}{1000}\) = 0,625 (mol)

PTHH: CuSO₄ + 2NaOH ---> Cu(OH)₂ + Na₂SO₄

LTL: 0,0525 < \(\dfrac{0,625}{2}\) => NaOH dư

Theo pthh: \(\left\{{}\begin{matrix}n_{NaOH\left(pư\right)}=2n_{CuSO_4}=2.0,0525=0,105\left(mol\right)\\n_{Cu\left(OH\right)_2}=n_{Na_2SO_4}=n_{CuSO_4}=0,0525\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}C_{MNaOH\left(dư\right)}=\dfrac{0,625-0,105}{0,5}=1,04M\\C_{MNa_2SO_4}=\dfrac{0,0525}{0,5}=0,105M\end{matrix}\right.\)

Bài 11:

a: Ta có: \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}\)

\(=\sqrt{x}\cdot\left(\sqrt{x}-1\right)\)

\(=x-\sqrt{x}\)

b: Để P=2 thì \(x-\sqrt{x}-2=0\)

hay x=4

Bài 10:

a: Ta có: \(A=\left(1+\dfrac{\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)\)

\(=\dfrac{x+\sqrt{x}+1}{x+1}:\dfrac{x-2\sqrt{x}+1}{\left(x+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x+\sqrt{x}+1}{x+1}\cdot\dfrac{x+1}{\sqrt{x}-1}\)

\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}\)

b: Để A<0 thì \(\sqrt{x}-1< 0\)

hay x<1

Kết hợp ĐKXĐ, ta được: \(0\le x< 1\)

Để A=-1 thì \(x+\sqrt{x}+1=-\sqrt{x}+1\)

\(\Leftrightarrow x=0\)

c: Thay x=4 vào A, ta được:

\(A=\dfrac{4+2+1}{2-1}=7\)

Gọi vận tốc ca nô là x ( x > 0 ) 

Theo bài ra ta có pt \(\dfrac{72}{x+3}+\dfrac{54}{x-3}=6\Rightarrow x=21\left(tm\right)\)