\(n_{H2SO4}=n.M=0,5\left(mol\right)\)

\(BTNT\left(H\right):n_{H2O}=n_{H2SO4}=0,5\left(mol\right)\)

\(BTNT\left(O\right):n_O=n_{H2O}=0,5\left(mol\right)\)

\(\Rightarrow m_O=n.M=8g\)

Mà \(m_{oxit}=m_{KL}+m_O=27,3\)

\(\Rightarrow m=m_{Kl}=19,3g\)
 

Chắc là \(\dfrac{1}{y+2}\) chứ lẻ 2 vậy chuyển sang = 3 ??

- Đặt \(\left\{{}\begin{matrix}\dfrac{x}{x-1}=a\\\dfrac{1}{y+2}=b\end{matrix}\right.\)

HPTTT : \(\left\{{}\begin{matrix}3a-2b=4\\2a+b=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3a-2b=4\\4a+2b=10\end{matrix}\right.\)

\(\Rightarrow7a=14\)

\(\Rightarrow a=2\)

\(\Rightarrow b=1\)

- Thay lại ta được : \(\left\{{}\begin{matrix}\dfrac{x}{x-1}=2\\\dfrac{1}{y+2}=1\end{matrix}\right.\)

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

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

Vậy ...

SO3 : S+6

K2SO4 : K+1; S+6

P2O5 : P+5

H3PO4 : P+5

FeO : Fe+2

Fe3O4 : Fe+8/3

H2S : S-2

CH4 : C-4

HNO3 : N+5

NO : N+2

NO2 : N+4

N2O : N+1

FeS : Fe+2; S-2

Fe(NO3)3 : Fe+3 , N+5

Al2(SO4)3 : Al+3; S+6

CaCO3 : Ca+2; C+4

Ta có : \(m_{oxit}=m_{KL}+m_O=58,4\)

\(\Rightarrow m_O=14,4g\)

\(\Rightarrow n_O=\dfrac{m}{M}=0,9\left(mol\right)\)

\(BTNT\left(O\right):n_{H2O}=n_O=0,9\left(mol\right)\)

\(BTNT\left(H\right):n_{H2SO4}=n_{H2O}=0,9\left(mol\right)\)

\(\Rightarrow V_{H2SO4}=0,45l\)

2. Ta có : \(\sin x-\sqrt{3}\cos x=1\)

\(\Leftrightarrow\dfrac{1}{2}\sin x-\dfrac{\sqrt{3}}{2}\cos x=\dfrac{1}{2}\)

\(\Leftrightarrow\cos60.\sin x-\sin60.\cos x=\dfrac{1}{2}\)

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

\(\Leftrightarrow-x+\dfrac{\pi}{3}=-\dfrac{\pi}{6}+k2\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{2}-k2\pi\)

Vậy ...

4. Tương tự câu 2 ta được : \(\sin\left(\dfrac{\pi}{3}+4x\right)=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow4x+\dfrac{\pi}{3}=\dfrac{\pi}{4}+k2\pi\)

\(\Leftrightarrow x=-\dfrac{\pi}{48}+\dfrac{k}{2}\pi\)

Vậy ..

Bài 1 :

Ta có : \(\dfrac{3x+5}{2}-1\le\dfrac{x+2}{3}+x\)

\(\Leftrightarrow\dfrac{3x+5}{2}-1-\dfrac{x+2}{3}-x\le0\)

\(\Leftrightarrow\dfrac{3\left(3x+5\right)-6-2\left(x+2\right)-6x}{6}\le0\)

\(\Leftrightarrow9x+15-6-2x-4-6x\le0\)

\(\Leftrightarrow x\le-5\)

Mà \(\left\{{}\begin{matrix}x\in Z\\x>-10\end{matrix}\right.\)

Vậy \(x\in\left\{-5;-6;-7;-8;-9\right\}\)

Bài 27 :

\(a,M=6x^2+9xy-y^2-\left(5x^2-2xy\right)=6x^2+9xy-y^2-5x^2+2xy\)

\(=x^2+11xy-y^2\)

\(b,M=x^2-7xy+8y^2+\left(3xy-4y^2\right)=x^2-7xy+8y^2+3xy-4y^2\)

\(=x^2-4xy+4y^2=\left(x-2y\right)^2\)

\(c,M=25x^2y-13xy^2+y^3-11x^2y+2y^2\)

\(=14x^2y-13xy^2+y^3+2y^2\)

\(d,M=-\left(12x^4-15x^2y+2xy^2+7\right)=-12x^4+15x^2y-2xy^2-7\)