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a: =>x-4=0 hoặc x+5=0
=>x=4 hoặc x=-5
b: =>39/7:x=13
hay x=3/7
c: \(\Leftrightarrow\left(4.5-2x\right)=\dfrac{11}{4}:\dfrac{4}{9}=\dfrac{99}{16}\)
\(\Leftrightarrow2x=-\dfrac{27}{16}\)
hay x=-27/32
d: \(\Leftrightarrow x\cdot\dfrac{19}{15}=684\)
hay x=540
a. \(\left[{}\begin{matrix}x-4=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)
b.\(\Leftrightarrow\dfrac{39}{7}:x=13\)
\(\Leftrightarrow x=13.\dfrac{39}{7}\)
\(\Leftrightarrow x=\dfrac{507}{7}\)
c.\(\Leftrightarrow4,5-2x=\dfrac{99}{16}\)
\(\Leftrightarrow-2x=\dfrac{27}{16}\)
\(\Leftrightarrow x=-\dfrac{27}{32}\)
11)\(\dfrac{3x+1}{x-5}+\dfrac{2x}{x-5}=\dfrac{3x+2x+1}{x-5}=\dfrac{5x+1}{x-5}\)
12)\(\dfrac{4-x^2}{x-3}+\dfrac{2}{x^2-9}=\dfrac{4-x^2}{x-3}+\dfrac{2}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(4-x^2\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{2}{\left(x-3\right)\left(x+3\right)}=\dfrac{2+\left(2-x\right)\left(2+x\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
13)
\(\dfrac{3}{4x-2}+\dfrac{2x}{4x^2-1}=\dfrac{3}{2\left(2x-1\right)}+\dfrac{2x}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{3\left(2x+1\right)}{2\left(2x-1\right)\left(2x+1\right)}+\dfrac{2.2x}{2\left(2x-1\right)\left(2x+1\right)}=\dfrac{6x+3+4x}{2\left(2x-1\right)\left(2x+1\right)}=\dfrac{10x+3}{2\left(2x-1\right)\left(2x+1\right)}\)
14)
\(\dfrac{2x+1}{2x-4}+\dfrac{5}{x^2-4}=\dfrac{2x+1}{2\left(x-2\right)}+\dfrac{5}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(2x+1\right)\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}+\dfrac{5.2}{2\left(x-2\right)\left(x+2\right)}=\dfrac{2x^2+5x+12}{2\left(x-2\right)\left(x+2\right)}\)
\(a)7x-21=0.\\ \Leftrightarrow x=3.\)
\(b)3x-5=2x+7.\\ \Leftrightarrow x=12.\)
\(c)6x+18=0.\\ \Leftrightarrow x=-3.\\ d)5x+2=4x-1.\\ \Leftrightarrow x=-3.\)
Bài 1:
a: 7x-21=0
=>7x=21
hay x=3
b: 3x-5=2x+7
=>3x-2x=7+5
=>x=12
c: 6x+18=0
=>6x=-18
hay x=-3
d: 5x+2=4x-1
=>5x-4x=-1-2
=>x=-3
\(Mg + 2HCl \to MgCl_2 + H_2\\ CuO + 2HCl \to CuCl_2 + H_2O\\ n_{Mg}= n_{H_2}= \dfrac{2,24}{22,4} = 0,1(mol)\\ \Rightarrow n_{CuO} = \dfrac{10,8-0,1.24}{80}=0,105(mol)\\ n_{HCl} = 2n_{Mg} + 2n_{CuO} = 0,1.2 + 0,105.2 = 0,41(mol)\\ m_{HCl} = 0,41.36,5 = 14,965(gam)\)