2: Để (d)//y=(m²+1)x-4 thì \(\left\{{}\begin{matrix}m^2=1\\m-5\ne-4\end{matrix}\right.\Leftrightarrow m=1\)

a)\(2Ca+O_2\underrightarrow{t^o}2CaO\)

   \(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

   \(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3\downarrow+H_2O\)

   \(CaCO_3\underrightarrow{t^o}CaO+CO_2\)

   \(CaO+2HCl\rightarrow CaCl_2+H_2O\)

a: ĐKXĐ: x<>0; x<>1(A)

(B): x<>0; x<>3

(C): x<>2; x<>-2

b: \(A=\dfrac{2\left(x-1\right)}{x\left(x-1\right)}=\dfrac{2}{x}\)

\(B=\dfrac{2\left(x-3\right)}{x\left(x-3\right)}=\dfrac{2}{x}\)

\(C=\dfrac{3\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{3}{x+2}\)

c: Khi x=0 thì A và B không xác định

Khi x=0 thì C=3/2

Khi x=3 thì B ko xác định, A=2/3; C=3/5

\(a,DKXD:\)

\(+x^2-x\ne0\Leftrightarrow x\ne0;1\)

\(+x^2-3x\ne0\Leftrightarrow x\ne0;3\)

+\(x^2-4\ne0\Leftrightarrow x\ne\pm4\)

\(b,\)

\(\dfrac{2x-2}{x^2-x}=\dfrac{2\left(x-1\right)}{x\left(x-1\right)}=\dfrac{2}{x}\)

\(\dfrac{2x-6}{x^2-3x}=\dfrac{2\left(x-3\right)}{x\left(x-3\right)}=\dfrac{2}{x}\)

\(\dfrac{3x-6}{x^2-4}=\dfrac{3\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{3}{x+2}\)

\(c,\)

Vì phân thức 1,2 cùng kết quả nên mk lm 1 cái thôi nhé

+ Thay \(x=0\) vào \(\dfrac{2}{x}\Leftrightarrow\dfrac{2}{0}=0\)

   Thay \(x=3\) vào \(\dfrac{2}{x}\Leftrightarrow\dfrac{2}{3}\) 

+ Thay \(x=0\) vào \(\dfrac{3}{x+2}\Leftrightarrow\dfrac{3}{0+2}=\dfrac{3}{2}\)

   Thay \(x=3\) vào \(\dfrac{3}{x+2}\Leftrightarrow\dfrac{3}{3+2}=\dfrac{3}{5}\)

Câu 2

\((1) MnO_2 + 4HCl \to MnCl_2 + Cl_2 + 2H_2O\\ (2) Cl_2 + H_2 \xrightarrow{as} 2HCl\\ (3) 3Cl_2 + 2Fe \xrightarrow{t^o} 2FeCl_3\\ (4) 2FeCl_3 + Fe \to 3FeCl_2\\ (5) 2NaOH + Cl_2 \to NaCl + NaClO + H_2O\)

\((1) 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ (2) 2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3\\ (3) C + O_2 \xrightarrow{t^o} CO_2\\ (4) 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ (5) 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\\ (6) 2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\\ (7) Fe + H_2SO_4 \to FeSO_4 + H_2\\ (8) Cu + 2H_2SO_4 \to CuSO_4 + SO_2 + 2H_2O\\ (9) 2Fe + 6H_2SO_4 \to Fe_2(SO_4)_3 + 3SO_2 + 6H_2O\\ (10) 2Al + 6H_2SO_4 \to Al_2(SO_4)_3 + 3SO_2 + 6H_2O\)

\(4\sin^22x-4\cos2x-1=0\)

\(\Leftrightarrow4\left(1-\cos^22x\right)-4\cos2x-1=0\)

\(\Leftrightarrow4-4\cos^22x-4\cos2x-1=0\)

\(\Leftrightarrow-4\cos^22x-4\cos2x+3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\dfrac{1}{2}\\cos2x=\dfrac{-3}{2}\left(L\right)\end{matrix}\right.\Leftrightarrow\cos2x=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{3}+k2\pi\\2x=-\dfrac{\pi}{3}-k2\pi\end{matrix}\right.\)

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

Chọn A

4.(1-cos²2x) - 4cos2x -1 =0

-4cos²2x - 4cos2x + 3 = 0

bấm máy tính giải phương trình bậc 2

cos x = \(\dfrac{1}{2}\) => cos x = cos \(\dfrac{\pi}{3}\)  => x = \(\dfrac{\pi}{3}\) + k2\(\pi\)

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

cos x = \(\dfrac{-3}{2}\) ( vô nghiệm)

\(\Leftrightarrow2\sqrt{x-4}=5\left(x\ge4\right)\\ \Leftrightarrow\sqrt{x-4}=\dfrac{5}{2}\\ \Leftrightarrow x-4=\dfrac{25}{4}\\ \Leftrightarrow x=\dfrac{41}{4}\left(tm\right)\)