HOC24
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Chủ đề / Chương
Bài học
\(\left(1\right)P_2O_5+3H_2O\rightarrow2H_3PO_4\)
\(\left(2\right)2H_3PO_4+3Mg\rightarrow Mg_3\left(PO_4\right)_2+3H_2\)
\(\left(3\right)2H_2+O_2\rightarrow2H_2O\)
\(\sqrt{53+12\sqrt{10}}-\sqrt{47-6\sqrt{10}}=\sqrt{45+2.3\sqrt{5}.2\sqrt{2}+8}-\)\(\sqrt{45-2.3\sqrt{5}.\sqrt{2}+2}\)
\(=\sqrt{\left(3\sqrt{5}+2\sqrt{2}\right)^2}-\sqrt{\left(3\sqrt{5}-\sqrt{2}\right)^2}\) \(=\left|3\sqrt{5}+2\sqrt{2}\right|-\left|3\sqrt{5}-\sqrt{2}\right|\) \(=3\sqrt{5}+2\sqrt{2}-\left(3\sqrt{5}-\sqrt{2}\right)=3\sqrt{2}\)
\(\left\{{}\begin{matrix}5\left(x+2y\right)=4x-1\\2x+4=3\left(x-5y\right)-20\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5x+10y=4x-1\\2x+4=3x-15y-20\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+10y=-1\left(1\right)\\x-15y=24\left(2\right)\end{matrix}\right.\)
Lấy (1)-(2): \(25y=-25\Leftrightarrow y=-1\) thay vào (1) \(\Leftrightarrow x=9\)
a,ĐK:\(x\ge\frac{3}{2}\)
\(PT\Leftrightarrow\left(3x+2\right)\sqrt{2x-3}-\left(3x+2\right)-2x^2+8=0\)
\(\Leftrightarrow\left(3x+2\right)\left(\sqrt{2x-3}-1\right)-2\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(3x+2\right).\frac{2\left(x-2\right)}{\sqrt{2x-3}+1}-2\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow2\left(x-2\right)\left[\frac{3x+2}{\sqrt{2x-3}+1}-\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\\frac{3x+2}{\sqrt{2x-3}+1}=x+2\left(1\right)\end{matrix}\right.\)
Giải (1)\(\Leftrightarrow3x+2=\sqrt{2x-3}\left(x+2\right)+x+2\)
\(\Leftrightarrow2x=\sqrt{2x-3}\left(x+2\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{3}{2}\\4x^2=\left(2x-3\right)\left(x^2+4x+4\right)\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{3}{2}\\2x^3+x^2-4x-12=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{3}{2}\\\left(x-2\right)\left(2x^2+5x+6\right)=0\end{matrix}\right.\) \(\Leftrightarrow x=2\left(tm\right)\)
Vậy \(x=2\)
b, Đề là \(5\sqrt{x+1}\) hay \(5\sqrt{x+4}\) vậy?
\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}\)
\(=\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}\)
\(=\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}\)
\(=\frac{1}{x+4}-\frac{1}{x+7}=\frac{3}{x^2+11x+28}\)
Với x,y,z dương, áp dụng BĐT AM-GM:
\(\left\{{}\begin{matrix}x^3+x^3+y^3\ge3x^2y\\x^3+y^3+y^3\ge3xy^2\end{matrix}\right.\) \(\Rightarrow3\left(x^3+y^3\right)\ge3\left(x^2y+xy^2\right)\)
Tương tự:\(3\left(y^3+z^3\right)\ge3\left(y^2z+yz^2\right)\);\(3\left(x^3+z^3\right)\ge3\left(x^2z+xz^2\right)\)
Cộng vế theo vế:
\(\Leftrightarrow6\left(x^3+y^3+z^3\right)\ge3\left(x^2y+xy^2\right)+3\left(y^2z+yz^2\right)+3\left(x^2z+xz^2\right)\)
\(\Leftrightarrow8\left(x^3+y^3+z^3\right)\ge x^3+y^3+3xy\left(x+y\right)+y^3+z^3+3yz\left(y+z\right)+x^3+z^3+3xz\left(x+z\right)\) \(\Leftrightarrow8\left(x^3+y^3+z^3\right)\ge\left(x+y\right)^3+\left(y+z\right)^3+\left(x+z\right)^3\) (đpcm)
ĐK: \(x\ne1\)
PT\(\Leftrightarrow\frac{1}{x-1}-\frac{4}{x^2+x+1}=\frac{2x^2-5}{x^3-1}\)
\(\Leftrightarrow\frac{x^2+x+1-4\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{2x^2-5}{x^3-1}\)
\(\Leftrightarrow\frac{x^2-3x+5}{x^3-1}=\frac{2x^2-5}{x^3-1}\) với \(x\ne1\)
\(\Leftrightarrow x^2-3x+5=2x^2-5\)
\(\Leftrightarrow x^2+3x-10=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\) (tm)
Vậy...
ĐK: \(0\le x\le3\)
(Mà VT phương trình không âm nên điều kiện kéo theo \(x^2-x-2\ge0\))
\(PT\Leftrightarrow\sqrt{x}-\left(x-1\right)+\sqrt{3-x}-\left(x-2\right)+2x-3=x^2-x-2\)
\(\Leftrightarrow\frac{x-x^2+2x-1}{\sqrt{x}+x-1}+\frac{3-x-x^2+4x-4}{\sqrt{3-x}+x-2}-x^2+3x-1=0\)
\(\Leftrightarrow\left(-x^2+3x-1\right)\left(\frac{1}{\sqrt{x}+x-1}+\frac{1}{\sqrt{3-x}+x-2}+1\right)=0\)
Với \(0\le x\le3\) \(\Rightarrow\left\{{}\begin{matrix}x+\sqrt{x}-1\le\sqrt{3}+2\\x+\sqrt{3-x}-2\le\sqrt{3}+1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\frac{1}{x+\sqrt{x}-1}>0\\\frac{1}{x+\sqrt{3-x}-2}>0\end{matrix}\right.\)
\(\Rightarrow\frac{1}{x+\sqrt{x}-1}+\frac{1}{x+\sqrt{3-x}-2}+1=0\) vô no
\(\Leftrightarrow-x^2+3x-1=0\Leftrightarrow x=\frac{3\pm\sqrt{5}}{2}\)
Thử lại thấy \(x=\frac{3+\sqrt{5}}{2}\) thỏa mãn.