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\(Đặt:n_{Al}=a\left(mol\right);n_{Mg}=b\left(mol\right)\left(a,b>0\right)\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ Ta.có.hpt:\left\{{}\begin{matrix}27a+24b=5,1\\22,4.1,5a+22,4b=5,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\\ \Rightarrow\%m_{Al}=\dfrac{0,1.27}{5,1}.100\%\approx52,941\%\)
a)2.x-138=2^3.3^2
2x-138=72
2x=72+138
2x=210
x=210:2=105
b)231-(x-6)=1339:13
231-(x-6)=103
x-6=231-103
x-6=128
x=128+6
x=134
\(x^{15}-\left(7+1\right)x^{14}+\left(7+1\right)x^{13}....+\left(7+1\right)x-5\)
\(=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}....+\left(x+1\right)x-5\)
\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}....-x^3-x^2+x^2+x-5\)
\(=x-5=7-5=2\)
a)\(\left(138^2-38^2\right)+\left(85^2-15^2\right)\)
\(=\left(138-38\right)\left(138+38\right)+\left(85-15\right)\left(85+15\right)\)
\(=100.176+70.100=100\left(176+70\right)\)
\(=100.246=24600\)
b) \(25,6\left(87+13\right)-5,6\left(4,9+5,1\right)\)
\(=25,6.100-0,56.100\)
\(=100\left(25,6-0,56\right)=100.25,04=2504\)
A=24600
B=2504
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