Câu I:
1. a) Ta có: \(\left(\dfrac{1}{2}\right)^{1000}=\left[\left(\dfrac{1}{2}\right)^5\right]^{200}=\left(\dfrac{1}{32}\right)^{200}\)
mà \(\dfrac{1}{16}>\dfrac{1}{32}\) nên \(\left(\dfrac{1}{16}\right)^{200}>\left(\dfrac{1}{32}\right)^{200}\)
Vậy \(\left(\dfrac{1}{16}\right)^{200}>\left(\dfrac{1}{2}\right)^{1000}\)
b) Ta có: \(32^{27}=\left(2^5\right)^{27}=2^{135}\); \(16^{39}=\left(2^4\right)^{39}=2^{156}\)
Vì \(2^{135}< 2^{156}\Rightarrow32^{27}< 16^{39}\) mà \(16^{39}< 18^{39}\)
\(\Rightarrow32^{27}< 18^{39}\text{}\Rightarrow\text{}\text{}\left(-32\right)^{27}>\left(-18\right)^{39}\)
Vậy \(\left(-32\right)^{27}>\left(-18\right)^{39}\)
2. Với mọi \(x,y,z,t\in N\)*, ta có:
\(\left\{{}\begin{matrix}\dfrac{x}{x+y+z}>\dfrac{x}{x+y+z+t}\\\dfrac{y}{x+y+t}>\dfrac{y}{x+y+z+t}\\\dfrac{z}{y+z+t}>\dfrac{z}{x+y+z+t}\\\dfrac{t}{x+z+t}>\dfrac{t}{x+y+z+t}\end{matrix}\right.\)
\(\Rightarrow M>\dfrac{x+y+z+t}{x+y+z+t}\Rightarrow M>1\)
Lại có: \(\left\{{}\begin{matrix}\dfrac{x}{x+y+z}< \dfrac{x+t}{x+y+z+t}\\\dfrac{y}{x+y+t}< \dfrac{y+z}{x+y+z+t}\\\dfrac{z}{y+z+y}< \dfrac{z+x}{x+y+z+t}\\\dfrac{t}{x+z+t}< \dfrac{t+y}{x+y+z+t}\end{matrix}\right.\)
\(\Rightarrow M< \dfrac{2\left(x+y+z+t\right)}{x+y+z+t}\Rightarrow M< 2\)
Vì \(1< M< 2\) nên \(M\) không phải là số tự nhiên
3. a) Vì \(\left(3x-5\right)^{2006}\ge0\) với mọi \(x\). Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{5}{3}\)
\(\left(y^2-1\right)^{2018}\ge0\) với mọi \(y\). Dấu "=" xảy ra \(\Leftrightarrow y=1\)
\(\left(x-z\right)^{2100}\ge0\) với mọi \(x,z\). Dấu "=" xảy ra \(\Leftrightarrow x=z=\dfrac{5}{3}\)
\(\Rightarrow\left(3x-5\right)^{2006}+\left(y^2-1\right)^{2018}+\left(x-z\right)^{2100}\ge0\) với mọi \(x,y,z\). Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{5}{3};y=1;z=\dfrac{5}{3}\)
Vậy \(x=\dfrac{5}{3};y=1;z=\dfrac{5}{3}\)
b) Vì \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}=\dfrac{x^2+y^2+z^2}{4+9+16}=\dfrac{116}{29}=4\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{x^2}{4}=4\\\dfrac{y^2}{9}=4\\\dfrac{z^2}{16}=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\y=6\\z=8\end{matrix}\right.\)
Vậy \(x=1;y=6;z=8\)
Câu II:
1. a) Biểu thức \(A\) xác định
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-4\ne0\\2x^2-x^3\ne0\\x^2-3x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm2\\x\ne0\\x\ne3\end{matrix}\right.\)
Với \(x\ne0;x\ne\pm2;x\ne3\), ta có:
\(A=\left(\dfrac{2+x}{2-x}-\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\right):\left(\dfrac{x^2-3x}{2x^2-x^3}\right)\)
\(A=\dfrac{\left(2+x\right)^2+4x^2-\left(2-x\right)^2}{\left(2-x\right)\left(2+x\right)}:\dfrac{x\left(x-3\right)}{x^2\left(2-x\right)}\)
\(A=\dfrac{4+4x+x^2+4x^2-4+4x-x^2}{4-x^2}:\dfrac{x-3}{x\left(2-x\right)}\)
\(A=\dfrac{4x^2+8x}{4-x^2}\cdot\dfrac{x\left(2-x\right)}{x-3}\)
\(A=\dfrac{4x^2\left(x+2\right)\left(2-x\right)}{\left(2-x\right)\left(2+x\right)\left(x-3\right)}\)
\(A=\dfrac{4x^2}{x-3}\)
Vậy với \(x\ne0;x\ne\pm2;x\ne3\) thì \(A=\dfrac{4x^2}{x-3}\)
b) Với \(x\ne0;x\ne\pm2;x\ne3\) thì \(A>0\Rightarrow\dfrac{4x^2}{x-3}>0\) mà \(4x^2>0\) với mọi \(x\)
\(\Rightarrow x-3>0\Leftrightarrow x>3\) kết hợp với điều kiện xác định
Vậy \(x>3\) thì \(A>0\)
c) Ta có: \(\left|x-7\right|=4\Leftrightarrow\left[{}\begin{matrix}x-7=4\\x-7=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\left(tmđk\right)\\x=3\left(ktmđk\right)\end{matrix}\right.\)
Thay \(x=11\) vào \(A\), ta có: \(A=\dfrac{4\cdot11^2}{11-3}=\dfrac{484}{8}=60,5\)
Vậy với \(\left|x-7\right|=4\) thì \(A=60,5\)
2. a) \(3x^2-7x+2=3\left(x^2-\dfrac{7}{3}x+\dfrac{2}{3}\right)=3\left(x^2-\dfrac{1}{3}x-2x+\dfrac{2}{3}\right)\)
\(=3\left[x\left(x-\dfrac{1}{3}\right)-2\left(x-\dfrac{1}{3}\right)\right]=3\left(x-\dfrac{1}{3}\right)\left(x-2\right)\)
b) \(a\left(x^2+1\right)-x\left(a^2+1\right)=ax^2+a-a^2x-x=ax^2-a^2x+a-x\)
\(=ax\left(x-a\right)-\left(x-a\right)=\left(x-a\right)\left(ax-1\right)\)
Câu III:
1. a) \(\sqrt{5-x}+\sqrt{y-2005}+\sqrt{z+2007}=\dfrac{1}{2}\left(x+y+z\right)\) \(\left(ĐKXĐ:x\ge5;y\ge2005;z\ge-2007\right)\)
\(\Leftrightarrow2\sqrt{x-5}+2\sqrt{y-2005}+2\sqrt{z+2007}=x+y+z\)
\(\Leftrightarrow\left(x-5+2\sqrt{x-5}+1\right)+\left(y-2005+2\sqrt{y-2005}+1\right)+\left(z+2007+2\sqrt{z+2007}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-5}+1\right)^2+\left(\sqrt{y-2005}+1\right)^2+\left(\sqrt{z+2007}+1\right)^2=0\)
Vì \(\sqrt{x-5}\ge0\) với mọi \(x\) \(\Rightarrow\sqrt{x-5}+1\ge1>0\) với mọi \(x\)
\(\sqrt{y-2005}\ge0\) với mọi \(y\) \(\Rightarrow\sqrt{y-2005}+1\ge1>0\) với mọi \(y\)
\(\sqrt{z+2007}\ge0\) với mọi \(z\) \(\Rightarrow\sqrt{z+2007}+1\ge1>0\) với mọi \(z\)
\(\Rightarrow\left(\sqrt{x-5}+1\right)^2+\left(\sqrt{y-2005}+1\right)^2+\left(\sqrt{z+2007}+1\right)^2>0\) với mọi \(x,y,z\)
Vậy phương trình đã cho vô nghiệm
b) \(\sqrt{3x^2-12x+21}+\sqrt{5x^2-20x+24}=-2x^2+8x-3\)
Ta có: \(VT=\sqrt{3\left(x-2\right)^2+9}+\sqrt{5\left(x-2\right)^2+4}\ge5\) với mọi \(x\). Dấu "=" xảy ra \(\Leftrightarrow x=2\)
\(VP=-\left(\sqrt{2}x-2\sqrt{2}\right)^2+5\le5\) với mọi \(x\). Dấu "=" xảy ra \(\Leftrightarrow x=2\)
Vì \(VT\ge5,VP\le5\) nên dấu "=" xảy ra \(\Leftrightarrow VT=VP=5\Leftrightarrow x=2\)
Vậy phương trình đã cho có 1 nghiệm \(x=2\)
2. Ta có: \(2^x+2^y+2^z=672\)
\(\Leftrightarrow2^x\left(1+2^{y-x}+2^{z-x}\right)=2^5\cdot21\) (do \(x< y< z\))
Vì \(x< y< z\Rightarrow\left\{{}\begin{matrix}y-x>0\\z-x>0\end{matrix}\right.\Rightarrow1+2^{y-x}+2^{z-x}\) là số lẻ
\(\Rightarrow\left\{{}\begin{matrix}2^x=2^5\\1+2^{y-x}+2^{z-x}=21\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\2^{y-x}+2^{z-x}=20\left(2\right)\end{matrix}\right.\)
Thay \(x=5\) vào \(\left(2\right)\), ta có:
\(2^{y-5}+2^{z-5}=20\)
\(\Leftrightarrow2^{y-5}\left(1+2^{z-y}\right)=2^2\cdot5\)
Vì \(y< z\Rightarrow z-y>0\Rightarrow1+2^{z-y}\) là số lẻ, mà \(2^{y-5}\) chỉ chứa ước nguyên tố là \(2\)
\(\Rightarrow\left\{{}\begin{matrix}2^{y-5}=2^2\\1+2^{z-y}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=7\\2^{z-y}=4\left(3\right)\end{matrix}\right.\)
Thay \(y=7\) vào \(\left(3\right)\), ta có:
\(2^{z-7}=2^2\Leftrightarrow z-7=2\Leftrightarrow z=9\)
Vậy \(\left(x;y;z\right)=\left(5;7;9\right)\)