Giải các phương trình...
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Giải các phương trình sau:
a)
b)
c)
d)
e)
f)
g)
h)
i)
K
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H
5 tháng 7 2021
$A : HCl ; B : NaCl ; X : H_2O ; D : CO_2 ; F : Cl_2 ; G : H_2 ; E : NaOH ; H : AgNO_3$
$Na_2CO_3 + 2HCl \to 2NaCl + H_2O + CO_2$
$2NaCl + 2H_2O \xrightarrow{dpdd} 2NaOH + H_2 + Cl_2$
$2NaOH + + CO_2 \to Na_2CO_3 + H_2O$
$NaCl + AgNO_3 \to AgCl + NaNO_3$
22 tháng 3 2022
2NaCl + H2SO4 ---> Na2SO4 (B) + 2HCl (A)
4HCl + MnO2 ---> MnCl2 (D) + Cl2 (C) + 2H2O
Cl2 + 2NaBr ---> 2NaCl (G) + Br2 (F)
Br2 + 2NaI ---> 2NaBr (I) + I2 (H)
NaCl + AgNO3 ---> NaNO3 (K) + AgCl (J)
HCl + NaOH ---> NaCl (G) + H2O (E)
H
6 tháng 2 2021
\(A:SO_2\\ B : Fe_2O_3\\ D : SO_3\\ E : H_2O\\ F: H_2SO_4\\ G : CuSO_4\\ H : K_2SO_3\\ I : BaSO_3\\ K : KCl\\ L : BaSO_4 \\ M : HCl\)
\(4FeS_2 + 11O_2 \xrightarrow{t^o} 2Fe_2O_3 + 8SO_2\\ 2SO_2 + O_2 \xrightarrow{t^o,V_2O_5} 2SO_3\\ SO_3 + H_2O \to H_2SO_4 \\ 2H_2SO_4 + Cu \\ CuSO_4 + SO_2 + 2H_2O\\ SO_2 + 2KOH \to K_2SO_3 + H_2O\\ K_2SO_3 + BaCl_2 \to BaSO_3 + 2KCl\\ BaSO_3 + H_2SO_4 \to BaSO_4 + SO_2 + H_2O\\ SO_2 + Cl_2 + 2H_2O \to 2HCl + H_2SO_4\)
15 tháng 3 2022
a: 3x-15=0
nên 3x=15
hay x=5
b: 4x+20=0
nên 4x=-20
hay x=-5
c: -5x-20=0
nên -5x=20
hay x=-4
c/ 
d/ 
)(x-
) = 0 g/ (3x-1)(2x-3)(x+5) = 0 
h)
m)
= 8 – x k) 
= – 4x +7
q) 
20 tháng 6 2023
6:
k: =>x^2-9<x^2+2x+3
=>2x+3>-9
=>2x>-12
=>x>-6
1:
h: =>x(x-1)=0
=>x=0; x=1
i: =>x(x-3)=0
=>x=0; x=3
Trả lời:
a, \(\frac{2}{2x+1}-\frac{3}{2x-1}=\frac{4}{4x^2-1}\)\(\left(đkxđ:x\ne\pm\frac{1}{2}\right)\)
\(\Leftrightarrow\frac{2\left(2x-1\right)-3\left(2x+1\right)}{4x^2-1}=\frac{4}{4x^2-1}\)
\(\Rightarrow4x-2-6x-3=4\)
\(\Leftrightarrow-2x-5=4\)
\(\Leftrightarrow-2x=9\)
\(\Leftrightarrow x=\frac{-9}{2}\left(tm\right)\)
Vậy \(S=\left\{\frac{-9}{2}\right\}\)
b, \(\frac{2x}{x-1}+\frac{18}{x^2+2x-3}=\frac{2x-5}{x+3}\)\(\left(đkxđ:x\ne1;x\ne-3\right)\)
\(\Leftrightarrow\frac{2x\left(x+3\right)+18}{x^2+2x-3}=\frac{\left(2x-5\right)\left(x-1\right)}{x^2+2x-3}\)
\(\Rightarrow2x^2+6x+18=2x^2-7x+5\)
\(\Leftrightarrow2x^2+6x-2x^2+7x=5-18\)
\(\Leftrightarrow13x=-13\)
\(\Leftrightarrow x=-1\)\(\left(tm\right)\)
Vậy \(S=\left\{-1\right\}\)
c, \(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)\(\left(đkxđ:x\ne1\right)\)
\(\Leftrightarrow\frac{x^2+x+1+2x^2-5}{x^3-1}=\frac{4\left(x-1\right)}{x^3-x}\)
\(\Rightarrow3x^2+x-4=4x-4\)
\(\Leftrightarrow3x^2+x-4x=-4+4\)
\(\Leftrightarrow3x^2-3x=0\)
\(\Leftrightarrow3x\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\left(tm\right)\\x=1\left(ktm\right)\end{cases}}}\)
Vậy \(S=\left\{0\right\}\)
d, \(\frac{11}{x}=\frac{9}{x+1}+\frac{2}{x-4}\)\(\left(đkxđ:x\ne0;x\ne-1;x\ne4\right)\)
\(\Leftrightarrow\frac{11\left(x+1\right)\left(x-4\right)}{x\left(x+1\right)\left(x-4\right)}=\frac{9x\left(x-4\right)+2x\left(x+1\right)}{x\left(x+1\right)\left(x-4\right)}\)
\(\Rightarrow11\left(x^2-3x-4\right)=9x^2-36x+2x^2+2x\)
\(\Leftrightarrow11x^2-33x-44=11x^2-34x\)
\(\Leftrightarrow11x^2-33x-11x^2+34x=44\)
\(\Leftrightarrow x=44\)\(\left(tm\right)\)
Vậy \(S=\left\{44\right\}\)
e, \(\frac{14}{3x-12}-\frac{2+x}{x-4}=\frac{3}{8-2x}-\frac{5}{6}\)\(\left(đkxđ:x\ne4\right)\)
\(\Leftrightarrow\frac{14}{3\left(x-4\right)}-\frac{2+x}{x-4}=\frac{-3}{2\left(x-4\right)}-\frac{5}{6}\)
\(\Leftrightarrow\frac{28-6\left(2+x\right)}{6\left(x-4\right)}=\frac{-9-5\left(x-4\right)}{6\left(x-4\right)}\)
\(\Rightarrow28-12-6x=-9-5x+20\)
\(\Leftrightarrow16-6x=11-5x\)
\(\Leftrightarrow-6x+5x=11-16\)
\(\Leftrightarrow-x=-5\)
\(\Leftrightarrow x=5\)\(\left(tm\right)\)
Vậy \(S=\left\{5\right\}\)
a)\(\frac{2}{2x+1}\)\(-\frac{3}{2x-1}\)\(=\frac{4}{4x^2-1}\)
<=> \(\frac{2\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}\)\(-\frac{3\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}\)\(=\frac{4}{\left(2x-1\right)\left(2x+1\right)}\)
<=>\(\frac{4x-2-6x-3}{\left(2x-1\right)\left(2x+1\right)}\)\(=\frac{4}{\left(2x-1\right)\left(2x+1\right)}\)
=> -2x-5=4
<=>-2x=9
<=>\(x=\frac{-9}{2}\)