HOC24
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Tổng số hạt là 115 \(\Rightarrow\) p+e+n=115\(\Leftrightarrow\) 2p+n=115
Mà hạt mang điện nhiều hơn hạt không mang điện là 25
\(\Rightarrow\) 2p-n=25
Ta có hệ pt : \(\left\{{}\begin{matrix}2p+n=115\\2p-n=25\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4p=140\\2p-n=25\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}p=35\\n=70-25=45\end{matrix}\right.\) \(\Rightarrow\) Số khối : A=35+45=80
Số hiệu nguyên tử : Z = số p = số e = 35
Ta có :
\(2^{300}=2^{3^{100}}=8^{100}\)
\(3^{200}=3^{2^{100}}=9^{100}\)
Mà \(8^{100}< 9^{100}\)
\(\Rightarrow\) \(2^{300}< 3^{200}\)
\(\frac{1}{2a-1}.\sqrt{25a^4-100a^5+100a^6}\)
= \(\frac{1}{2a-1}.\sqrt{25a^4\left(1-4a+4a^2\right)}\)
= \(\frac{1}{2a-1}.5a^2\sqrt{\left(1-2a\right)^2}\)
= \(\frac{5a^2}{2a-1}.\left(1-2a\right)\)
= \(-5a^2\)
5/2+5x=2x-1/3
\(\Leftrightarrow\) \(\frac{5}{2}+\frac{1}{3}=-\left(5x-2x\right)\)
\(\Leftrightarrow\) \(-3x=\frac{17}{6}\)
\(\Leftrightarrow\) \(x=\frac{-17}{18}\)
Vậy \(S=\left\{-\frac{17}{18}\right\}\)
B = \(4\sqrt{20}+\sqrt{6-2\sqrt{5}}-15\sqrt{\frac{1}{5}}\)
= \(4\sqrt{4\times5}+\sqrt{5-2\sqrt{5}+1}-\sqrt{15}^2\times\sqrt{\frac{1}{5}}\)
= \(8\sqrt{5}+\sqrt{\left(\sqrt{5}-1\right)^2}-3\sqrt{5}\)
= \(5\sqrt{5}+\sqrt{5}-1=6\sqrt{5}-1\)
+ He has written an email for 15 minutes .
_ He has not written an email for 15 minutes .
? Has he written an email for 15 minutes ?
1) read
2) will do
3) to drink
Tìm x biết rằng : \(\frac{4}{x}+\sqrt{x-\frac{1}{x}}=x+\sqrt{2x-\frac{5}{x}}\)
\(a^2+4b^2+9=2ab+3a+6b\)
\(\Leftrightarrow2a^2+8b^2+18=4ab+6a+12b\)
\(\Leftrightarrow\left(a^2-4ab+4b^2\right)+\left(a^2-6a+9\right)+\left(4b^2-12b+9\right)=0\)
\(\Leftrightarrow\left(a-2b\right)^2+\left(a-3\right)^2+\left(2b-3\right)^2=0\) \(\Rightarrow\left\{{}\begin{matrix}\left(a-2b\right)^2=0\\\left(a-3\right)^2=0\\\left(2b-3\right)^2=0\end{matrix}\right.\)
(do \(\left(a-2b\right)^2\ge0;\left(a-3\right)^2=0;\left(2b-3\right)^2=0\) )
\(\Leftrightarrow\left\{{}\begin{matrix}a=3\\b=\frac{3}{2}\end{matrix}\right.\) Vậy (a;b)=(3;3/2)