- Đặt \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\Rightarrow27a+24b=10,2\left(1\right)\)

Khí thu được sau p/ứ là khí H₂: \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

2                                         3       (mol)

a                                        3/2 a   (mol)

\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

1                                        1   (mol)

b                                         b   (mol)

Từ hai PTHH trên ta có: \(\dfrac{3}{2}a+b=0,5\left(2\right)\)

\(\left(1\right),\left(2\right)\) ta có hệ: \(\left\{{}\begin{matrix}27a+24b=10,2\\\dfrac{3}{2}a+b=0,5\end{matrix}\right.\)

Giải ra ta có \(\left\{{}\begin{matrix}a=0,2\left(mol\right)\\b=0,2\left(mol\right)\end{matrix}\right.\)

a) \(\%Al=\dfrac{m_{Al}}{m_{hh}}.100\%=\dfrac{0,2.27}{10,2}.100\%\approx52,94\%\)

\(\%Mg=100\%-\%Al=100\%-52,94=47,06\%\)

b)

 \(3H_2+Fe_2O_3\rightarrow^{t^0}2Fe+3H_2O\)

3               1              2  (mol)

0,5            1/6         1/3  (mol)

\(m_{Fe}=\dfrac{1}{3}.56=\dfrac{56}{3}\left(g\right)\)

\(m_{Fe_2O_3\left(pứ\right)}=\dfrac{1}{6}.160=\dfrac{80}{3}\left(g\right)\)

\(m_{Fe_2O_3\left(dư\right)}=60-m_{Fe}=60-\dfrac{56}{3}=\dfrac{124}{3}\left(g\right)\)

\(a=\dfrac{124}{3}+\dfrac{80}{3}=68\left(g\right)\)

à bỏ khúc (2) ==>....m=-9 (tmđk) á

Chỉnh lại nhé:

\(\dfrac{1}{x_1^2}+\dfrac{1}{x_2^2}=\dfrac{25}{16}\)

\(\Leftrightarrow\dfrac{x_1^2+x_2^2}{x_1^2x_2^2}=\dfrac{25}{16}\)

\(\Leftrightarrow\dfrac{\left(x_1+x_2\right)^2-2x_1x_2}{x_1^2x_2^2}=\dfrac{25}{16}\)

\(\Rightarrow\dfrac{\dfrac{\left(m-1\right)^2}{4}+2.\dfrac{m+1}{2}}{\dfrac{\left(m+1\right)^2}{4}}=\dfrac{25}{16}\)

\(\Leftrightarrow4.\dfrac{\dfrac{\left(m-1\right)^2+4m+4}{4}}{\left(m+1\right)^2}=\dfrac{25}{16}\)

\(\Leftrightarrow\dfrac{m^2-2m+1+4m+4}{\left(m+1\right)^2}=\dfrac{25}{16}\)

\(\Leftrightarrow\dfrac{m^2+2m+5}{\left(m+1\right)^2}=\dfrac{25}{16}\)

\(\Leftrightarrow1+\dfrac{4}{\left(m+1\right)^2}=\dfrac{25}{16}\)

\(\Leftrightarrow\dfrac{1}{\left(m+1\right)^2}=\dfrac{9}{64}\)

Giải ra ta được \(\left[{}\begin{matrix}m=\dfrac{5}{3}\left(loại\right)\\m=-\dfrac{11}{3}\left(nhận\right)\end{matrix}\right.\)

Giao thừa còn làm bài gì đấy :)

\(2x^2+\left(m-1\right)x-m-1=0\left(1\right)\)

- Gọi x₁, x₂ lần lượt là 2 nghiệm của phương trình (1).

Ta có \(x_1>0,x_2>0\) và \(\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{5}{4}\left(2\right)\).

Để phương trình có nghiệm thì: \(\Delta\ge0\)

\(\Rightarrow\left(m-1\right)^2+4.2\left(m+1\right)\ge0\)

\(\Leftrightarrow m^2-2m+1+8m+8\ge0\)

\(\Leftrightarrow m^2+6m+9\ge0\)

\(\Leftrightarrow\left(m+3\right)^2\ge0\).

Ta thấy bất đẳng thức cuối cùng là luôn đúng nên phương trình (1) luôn có nghiệm với \(\forall m\).

Để phương trình có 2 nghiệm dương thì:

\(\left\{{}\begin{matrix}x_1+x_2>0\\x_1x_2>0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}-\dfrac{m-1}{2}>0\\-\dfrac{m+1}{2}>0\end{matrix}\right.\)\(\Rightarrow m< -1\)

\(\left(2\right)\Rightarrow\dfrac{x_1+x_2}{x_1x_2}=\dfrac{5}{4}\)

\(\Rightarrow\dfrac{-\dfrac{m-1}{2}}{-\dfrac{m+1}{2}}=\dfrac{5}{4}\)

\(\Rightarrow\dfrac{m-1}{m+1}=\dfrac{5}{4}\)

Giải ra ta có \(m=-9\left(tmđk\right)\)

\(A=\dfrac{x^2}{x-1}=\dfrac{x^2-1+1}{x-1}=\dfrac{\left(x-1\right)\left(x+1\right)+1}{x-1}=x+1+\dfrac{1}{x-1}\)

\(=\left(x-1\right)+\dfrac{1}{x-1}+2\)

Dễ dàng nhận thấy \(x-1>0\).

Áp dụng BĐT Cauchy cho 2 số dương ta có:

\(A=\left(x-1\right)+\dfrac{1}{x-1}+2\ge2\sqrt{\left(x-1\right).\dfrac{1}{x-1}}+2=4\)

Dấu "=" xảy ra khi \(\sqrt{x-1}=\dfrac{1}{\sqrt{x-1}}\Leftrightarrow x=2\)

Vậy \(MinA=4\), đạt tại x=2.

\(P=\dfrac{4}{a^2+b^2}+\dfrac{3}{ab}\)

Áp dụng BĐT Bunhiacopxki ta có:

\(\left(\dfrac{4}{a^2+b^2}+\dfrac{3}{ab}\right)\left[4\left(a^2+b^2\right)+12ab\right]\ge\left[\sqrt{\dfrac{4}{a^2+b^2}.4\left(a^2+b^2\right)}+\sqrt{\dfrac{3}{ab}.12ab}\right]^2=100\)

\(\Rightarrow P\ge\dfrac{100}{4\left(a^2+b^2\right)+12ab}=\dfrac{100}{4\left(a+b\right)^2+4ab}=\dfrac{25}{\left(a+b\right)^2+ab}\)

\(\Rightarrow P\ge\dfrac{25}{4^2+ab}=\dfrac{25}{16+ab}\) (vì \(a+b\le4\)).

Mặt khác ta có: \(ab\le\dfrac{\left(a+b\right)^2}{4}\le\dfrac{4^2}{4}=4\)

\(\Rightarrow P\ge\dfrac{25}{16+4}=\dfrac{5}{4}\)

Dấu "=" xảy ra khi \(a=b=2\).

Vậy \(MinP=\dfrac{5}{4}\), đạt tại \(a=b=2\)

Xem cách mk làm là đc rồi bạn.

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)

\(\Rightarrow\dfrac{1}{a}=\left(\dfrac{1}{2}-\dfrac{1}{b}\right)+\left(\dfrac{1}{2}-\dfrac{1}{c}\right)\)

\(\Rightarrow\dfrac{1}{a}=\dfrac{b-2}{2b}+\dfrac{c-2}{2c}\)

Dễ dàng chứng minh \(\dfrac{b-2}{2b},\dfrac{c-2}{2c}\) là các số dương.

Áp dụng BĐT Cauchy cho 2 số dương ta có:

\(\dfrac{b-2}{2b}+\dfrac{c-2}{2c}\ge2\sqrt{\dfrac{\left(b-2\right)\left(c-2\right)}{4bc}}=\sqrt{\dfrac{\left(b-2\right)\left(c-2\right)}{bc}}\)

\(\Rightarrow\dfrac{1}{a}\ge\sqrt{\dfrac{\left(b-2\right)\left(c-2\right)}{bc}}\left(1\right)\)

CMTT ta có: \(\left\{{}\begin{matrix}\dfrac{1}{b}\ge\sqrt{\dfrac{\left(c-2\right)\left(a-2\right)}{ca}}\left(2\right)\\\dfrac{1}{c}\ge\sqrt{\dfrac{\left(a-2\right)\left(b-2\right)}{ab}}\left(3\right)\end{matrix}\right.\)

\(\left(1\right),\left(2\right),\left(3\right)\Rightarrow\dfrac{1}{abc}\ge\dfrac{\left(a-2\right)\left(b-2\right)\left(c-2\right)}{abc}\)

\(\Rightarrow\left(a-2\right)\left(b-2\right)\left(c-2\right)\le1\left(đpcm\right)\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=b=c\\\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\end{matrix}\right.\Leftrightarrow a=b=c=3\)

\(J=\dfrac{1}{\sqrt{1-x}+\sqrt{1+x}}\)

\(\Rightarrow J^2=\dfrac{1}{\left(1-x\right)+2\sqrt{\left(1-x\right)\left(1+x\right)}+\left(1+x\right)}=\dfrac{1}{2+2\sqrt{\left(1-x\right)\left(1+x\right)}}\)

Ta có: \(-1\le x\le1\Rightarrow\left(1-x\right)\left(1+x\right)\ge0\)

\(\Rightarrow2+2\sqrt{\left(1-x\right)\left(1+x\right)}\ge2\)

\(\Rightarrow0< \dfrac{1}{2+2\sqrt{\left(1-x\right)\left(1+x\right)}}\le\dfrac{1}{2}\) hay \(J^2\le\dfrac{1}{2}\)

\(\Rightarrow0< J\le\dfrac{\sqrt{2}}{2}\)

\(MaxJ=\dfrac{\sqrt{2}}{2}\Leftrightarrow\left(1+x\right)\left(1-x\right)=0\Leftrightarrow x=\pm1\)

Vậy \(a=\dfrac{\sqrt{2}}{2}\)

\(x^2-11x+m-2=0\left(1\right)\)

Để phương trình (1) có 2 nghiệm phân biệt thì:

\(\Delta>0\Rightarrow\left(-11\right)^2-4.1.\left(m-2\right)>0\)

\(\Leftrightarrow121-4m+8>0\)

\(\Leftrightarrow m< \dfrac{129}{4}\)

Theo hệ thức Vi-et ta có:

\(\left\{{}\begin{matrix}x_1+x_2=11\left(1'\right)\\x_1x_2=m-2\end{matrix}\right.\).

Ta có: \(\sqrt{x^2_1-10x_1+m-1}=5-\sqrt{x_2+1}\left(2\right)\)

Đk: \(\left\{{}\begin{matrix}x_1^2-10x_1+m-1\ge0\\-1\le x_2\le24\end{matrix}\right.\)

\(\left(2\right)\Rightarrow x^2_1-10x_1+m-1=25-10\sqrt{x_2+1}+x_2+1\)

\(\Leftrightarrow x_1^2-10x_1+\left(m-2\right)-25+10\sqrt{11-x_1+1}-x_2=0\)

\(\Rightarrow x_1^2-\left(x_1+x_2\right)-9x_1+x_1x_2-25+10\sqrt{12-x_1}=0\)

\(\Rightarrow x_1\left(x_1+x_2\right)-11-9x_1-25+10\sqrt{12-x_1}=0\)

\(\Rightarrow11x_1-9x_1-36+10\sqrt{12-x_1}=0\)

\(\Leftrightarrow2x_1+10\sqrt{12-x_1}-36=0\)

\(\Leftrightarrow x_1+5\sqrt{12-x_1}-18=0\)

\(\Leftrightarrow18-x_1=5\sqrt{12-x_1}\left(x_1\le12\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}18-x_1\ge0\\\left(18-x_1\right)^2=25\left(12-x_1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}18-x_1\ge0\\324-36x_1+x_1^2=300-25x_1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_1\le18\\x_1^2-11x_1+24=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_1\le18\\\left[{}\begin{matrix}x=3\\x=8\end{matrix}\right.\left(nhận\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x_1=3\\x_1=8\end{matrix}\right.\left(nhận\right)\)

Thay \(x_1=3\) vào (1') ta được:

\(3+x_2=11\Rightarrow x_2=8\left(nhận\right)\)

\(\Rightarrow m=x_1x_2+2=3.8+2=26\left(thỏa\Delta>0\right)\)

Thay \(x_1=8\) vào (1') ta được:'

\(8+x_2=11\Rightarrow x_2=3\left(nhận\right)\)

\(\Rightarrow m=x_1x_2+2=8.3+2=26\left(thỏa\Delta>0\right)\)

Vậy giá trị m cần tìm là 26.