Mọi người giúp em giải cầu đánh dấu với
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Câu 16:
PTHH: \(Cl_2+2NaOH\rightarrow NaCl+NaClO+H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{Cl_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\\n_{NaOH}=\dfrac{600\cdot20\%}{40}=3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) 2 chất p/ứ hết
Mặt khác: \(m_{Cl_2}=1,5\cdot71=106,5\left(g\right)\)
\(\Rightarrow m_{nướcjaven}=m_{Cl_2}+m_{ddNaOH}=706,5\left(g\right)\)
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Bài 1:
a: \(=-10x^3+20x^4-5x\)
b: \(=\dfrac{1}{3}a^2b+7a^5-1\)
c: \(=a^3+8+25-a^3=33\)
d: \(=x^2-16+8-x^3=-x^3+x^2-8\)
e: \(=a^3+1+8-a^3=9\)
f: \(=\dfrac{7-2x+4x-8}{2x+3}=\dfrac{2x-1}{2x+3}\)
g: \(=\dfrac{3}{2\left(x+3\right)}-\dfrac{2}{x\left(x+3\right)}\)
\(=\dfrac{3x-4}{2x\left(x+3\right)}\)
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(2/3×x-1/3)=2/3+1/3
(2/3×x-1/3)=3/3
2/3×x=3/3+1/3
2/3×x=4/3
x=4/3:3/2
x=4/3×2/3
x=8/9
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Dạ 2 đề là 1 ạ tại em muốn ghi lại cho mọi người hiểu ạ
\(2^{\sqrt{3x+2y-1}}+3^{\sqrt{2x-y-2}}=2\)
Ta có: \(\left\{{}\begin{matrix}\sqrt{3x+2y-1}\ge0\\\sqrt{2x-y-2}\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2^{\sqrt{3x+2y-1}}\ge1\\3^{\sqrt{2x-y-2}}\ge1\end{matrix}\right.\)
\(\Rightarrow2^{\sqrt{3x+2y-1}}+3^{\sqrt{2x-y-2}}\ge2\)
Dấu = xảy ra khi
\(\left\{{}\begin{matrix}3x+2y-1=0\\2x-y-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{7}\\y=-\dfrac{4}{7}\end{matrix}\right.\)