\(n_{CaCO_3}=\dfrac{50}{100}=0,5\left(mol\right)\\ a,PTHH:CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ b,n_{CO_2}=n_{CaCl_2}=n_{CaCO_3}=0,5\left(mol\right)\\ V_{CO_2\left(25^oC,1bar\right)}=0,5.24,79=12,395\left(l\right)\\ c,n_{HCl}=2.0,5=1\left(mol\right)\\ V_{ddHCl}=\dfrac{1}{0,2}=5\left(l\right)\\ d,m_{CaCl_2}=111.0,5=55,5\left(g\right)\)

chịu thôi. 100x A +90000=100000

Bài 2:

a) ĐKXĐ: \(\left\{{}\begin{matrix}x+3\ne0\\x-3\ne0\\9-x^2\ne0\end{matrix}\right.\Leftrightarrow x\ne\pm3\)

b) \(A=\dfrac{3}{x+3}+\dfrac{1}{x-3}-\dfrac{18}{9-x^2}\)

\(A=\dfrac{3}{x+3}+\dfrac{1}{x-3}+\dfrac{18}{x^2-9}\)

\(A=\dfrac{3\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{x+3}{\left(x+3\right)\left(x-3\right)}+\dfrac{18}{\left(x+3\right)\left(x-3\right)}\)

\(A=\dfrac{3x-9+x+3+18}{\left(x+3\right)\left(x-3\right)}\)

\(A=\dfrac{4x+12}{\left(x+3\right)\left(x-3\right)}\)

\(A=\dfrac{4\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\)

\(A=\dfrac{4}{x-3}\) 

c) Thay `x=-1` vào A ta có:

\(A=\dfrac{4}{-1-3}=\dfrac{4}{-4}=-1\)

d) `A=-4` khi: \(\dfrac{4}{x-3}=-4\)

\(\Leftrightarrow x-3=-1\)

\(\Leftrightarrow x=2\left(tm\right)\)

Bài 1:

a: ĐKXĐ: x<>3

\(\dfrac{9}{x-3}+\dfrac{3x}{3-x}\)

\(=\dfrac{9}{x-3}-\dfrac{3x}{x-3}=\dfrac{9-3x}{x-3}\)

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

b: \(\dfrac{5}{x+5}+\dfrac{-4}{x+4}\)

\(=\dfrac{5\left(x+4\right)-4\left(x+5\right)}{\left(x+5\right)\left(x+4\right)}\)

\(=\dfrac{5x+20-4x-20}{\left(x+5\right)\left(x+4\right)}=\dfrac{x}{\left(x+5\right)\left(x+4\right)}\)

c: \(\dfrac{x+5}{2x-3}-\dfrac{2x-7}{3-2x}-\dfrac{x+4}{3-2x}\)

\(=\dfrac{x+5}{2x-3}+\dfrac{2x-7}{2x-3}+\dfrac{x+4}{2x-3}\)

\(=\dfrac{x+5+2x-7+x+4}{2x-3}\)

\(=\dfrac{4x+2}{2x-3}\)

d: \(\dfrac{x^2-y^2}{10x^3y}:\dfrac{x-y}{5xy}\)

\(=\dfrac{\left(x-y\right)\left(x+y\right)}{10x^3y}\cdot\dfrac{5xy}{x-y}\)

\(=\dfrac{x+y}{1}\cdot\dfrac{5xy}{10x^3y}\)

\(=\dfrac{x+y}{2x^2}\)

e: \(\dfrac{2x^2-20x+50}{3x+3}\cdot\dfrac{x^2-1}{4\left(x-5\right)^3}\)

\(=\dfrac{2\left(x^2-10x+25\right)}{3\left(x+1\right)}\cdot\dfrac{\left(x+1\right)\left(x-1\right)}{4\left(x-5\right)^3}\)

\(=\dfrac{2\left(x-5\right)^2}{4\left(x-5\right)^3}\cdot\dfrac{x-1}{3}\)

\(=\dfrac{x-1}{3\cdot2\left(x-5\right)}=\dfrac{x-1}{6x-30}\)

f: \(\dfrac{x-2}{x+1}:\dfrac{x^2-5x+6}{x^2-2x-3}\)

\(=\dfrac{x-2}{x+1}:\dfrac{\left(x-2\right)\left(x-3\right)}{\left(x-3\right)\left(x+1\right)}\)

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

g: \(\dfrac{x}{x-2y}+\dfrac{x}{x+2y}+\dfrac{4xy}{4y^2-x^2}\)

\(=\dfrac{x}{x-2y}+\dfrac{x}{x+2y}-\dfrac{4xy}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\dfrac{x\left(x+2y\right)+x\left(x-2y\right)-4xy}{\left(x-2y\right)\left(x+2y\right)}\)

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

h: \(\dfrac{1}{x-y}+\dfrac{3xy}{y^3-x^3}+\dfrac{x-y}{x^2+xy+y^2}\)

\(=\dfrac{1}{x-y}-\dfrac{3xy}{\left(x-y\right)\cdot\left(x^2+xy+y^2\right)}+\dfrac{x-y}{x^2+xy+y^2}\)

\(=\dfrac{x^2+xy+y^2-3xy+\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)

\(=\dfrac{2\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{2\left(x-y\right)}{x^2+xy+y^2}\)

i: \(\left(\dfrac{2}{x+2}+\dfrac{2}{x-1}\right)\cdot\dfrac{x^2-4}{4x^2-1}\)

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

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

j: \(1+\dfrac{x^3-x}{x^2+1}\cdot\left(\dfrac{1}{1-x}-\dfrac{1}{1-x^2}\right)\)

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

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

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

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

tradition

chất tham gia: Hydrogen, Oxygen

Pt: \(\dfrac{3}{x^2+x+1}+\dfrac{4}{x^2+x+2}-\dfrac{6}{x^2+x+4}=1\) (*)

ĐK: \(\left\{{}\begin{matrix}x^2+x+1\ne0\\x^2+x+2\ne0\\x^2+x+4\ne0\end{matrix}\right.\)(luôn đúng)

Đặt: \(x^2+x+2=t\ge\dfrac{7}{4}\)

(*) trở thành:

\(\dfrac{3}{t-1}+\dfrac{4}{t}-\dfrac{6}{t+2}=1\)  

\(\Leftrightarrow\dfrac{3t\left(t+2\right)}{t\left(t-1\right)\left(t+2\right)}+\dfrac{4\left(t-1\right)\left(t+2\right)}{t\left(t-1\right)\left(t+2\right)}-\dfrac{6t\left(t-1\right)}{t\left(t-1\right)\left(t+2\right)}=1\)

\(\Leftrightarrow3t\left(t+2\right)+4\left(t-1\right)\left(t+2\right)-6t\left(t-1\right)=t\left(t-1\right)\left(t+2\right)\)

\(\Leftrightarrow3t^2+6t+4\left(t^2+t-2\right)-6t^2+6t=t\left(t^2+t-2\right)\)

\(\Leftrightarrow-3t^2+12t+4t^2+4t-8=t^3+t^2-2t\)

\(\Leftrightarrow t^2+16t-8=t^3+t^2-2t\)

\(\Leftrightarrow t^3-18t+8=0\)

\(\Leftrightarrow\left(t-4\right)\left(t^2+4t-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=4\left(tm\right)\\t=\sqrt{6}-2\left(ktm\right)\\t=-\sqrt{6}-2\left(ktm\right)\end{matrix}\right.\)

\(\Rightarrow x^2+x+2=4\)

\(\Leftrightarrow x^2+x-2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)

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

Vậy: ...