Câu I :
1 ) a ) Ta có : \(\left\{{}\begin{matrix}\left(\dfrac{1}{16}\right)^{200}=\left(\dfrac{1}{2}\right)^{800}=\dfrac{1}{2^{800}}\\\left(\dfrac{1}{2}\right)^{1000}=\dfrac{1}{2^{1000}}\end{matrix}\right.\)
Do : \(2^{1000}>2^{800}\Rightarrow\dfrac{1}{2^{800}}>\dfrac{1}{2^{1000}}\Rightarrow\left(\dfrac{1}{16}\right)^{200}>\left(\dfrac{1}{2}\right)^{1000}\)
Vậy \(\left(\dfrac{1}{16}\right)^{200}>\left(\dfrac{1}{2}\right)^{1000}\)
b ) Ta có : \(\left(-32\right)^{27}=\left(-2\right)^{135}>\left(-2\right)^{156}=\left(-16\right)^{39}>\left(-18\right)^{39}\)
Vậy \(\left(-32\right)^{27}>\left(-18\right)^{39}\)
2) Vì \(x,y,z,t\in\) N* nên ta có các BĐT :
\(\dfrac{x}{x+y+z+t}< \dfrac{x}{x+y+z}< \dfrac{x}{x+y}\)
\(\dfrac{y}{x+y+z+t}< \dfrac{y}{x+y+t}< \dfrac{y}{x+y}\)
\(\dfrac{z}{x+y+z+t}< \dfrac{z}{y+z+t}< \dfrac{z}{z+t}\)
\(\dfrac{t}{x+y+z+t}< \dfrac{t}{x+z+t}< \dfrac{t}{z+t}\)
Cộng từng vế BĐT ta thu được :
\(\dfrac{x+y+z+t}{x+y+z+t}< \dfrac{x}{x+y+z}+\dfrac{y}{x+y+t}+\dfrac{z}{y+z+t}+\dfrac{t}{z+x+t}< \dfrac{x+y}{x+y}+\dfrac{z+t}{z+t}\)
\(\Leftrightarrow1< M< 2\)
Vậy M không phải là số tự nhiên .
3a ) Ta có : \(\left\{{}\begin{matrix}\left(3x-5\right)^{2006}\ge0\\\left(y^2-1\right)^{2008}\ge0\\\left(x-z\right)^{2100}\ge0\end{matrix}\right.\Rightarrow\left(3x-5\right)^{2006}+\left(y^2-1\right)^{2008}+\left(x-z\right)^{2100}\ge0\)
Vậy dấu bằng sẽ xảy ra .
\(\Rightarrow\left\{{}\begin{matrix}3x-5=0\\y^2-1=0\\x-z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}\\\left[{}\begin{matrix}y=1\left(N\right)\\y=-1\left(L\right)\end{matrix}\right.\\z=x=\dfrac{5}{3}\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)=\left(\dfrac{5}{3};1;\dfrac{5}{3}\right)\) .
b ) Theo tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x^2+y^2+z^2}{2^2+3^2+4^2}=\dfrac{116}{29}=4\)
\(\left\{{}\begin{matrix}\dfrac{x^2}{4}=4\\\dfrac{y^2}{9}=4\\\dfrac{z^2}{16}=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\y=6\\z=8\end{matrix}\right.\left(TM\right)\)
Vậy \(\left(x;y;z\right)=\left(4;6;8\right)\)
Câu II :
1a ) ĐKXĐ : \(x\ne-2;x\ne0;x\ne2\)
\(A=\left(\dfrac{2+x}{2-x}-\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\right):\left(\dfrac{x^2-3x}{2x^2-x^3}\right)\)
\(=\dfrac{\left(2+x\right)\left(2+x\right)+4x^2-\left(2-x\right)\left(2-x\right)}{\left(2-x\right)\left(2+x\right)}:\dfrac{x\left(x-3\right)}{x^2\left(2-x\right)}\)
\(=\dfrac{x^2+4x+4+4x^2-x^2+4x-4}{\left(2-x\right)\left(2+x\right)}:\dfrac{x-3}{x\left(2-x\right)}\)
\(=\dfrac{4x}{2-x}.\dfrac{x\left(2-x\right)}{x-3}\)
\(=\dfrac{4x^2}{x-3}\)
b ) Để \(A>0\Leftrightarrow\dfrac{4x^2}{x-3}>0\Leftrightarrow x-3>0\Leftrightarrow x>3\) ( Do : \(4x^2>0\) )
c ) Ta có : \(\left|x-7\right|=4\Leftrightarrow\left[{}\begin{matrix}x=3\left(L\right)\\x=11\left(N\right)\end{matrix}\right.\) .
Thay \(x=11\) vào A ta được : \(A=\dfrac{4.11^2}{11-3}=\dfrac{484}{8}=\dfrac{121}{2}\)
2 )
a ) \(3x^2-7x+2\)
\(=3x^2-6x-x+2\)
\(=3x\left(x-2\right)-\left(x-2\right)\)
\(=\left(x-2\right)\left(3x-1\right)\)
b ) \(a\left(x^2+1\right)-x\left(a^2+1\right)\)
\(=ax^2+a-xa^2-x\)
\(=\left(ax^2-xa^2\right)+\left(a-x\right)\)
\(=ax\left(x-a\right)-\left(x-a\right)\)
\(=\left(x-a\right)\left(ax-1\right)\)
Câu III :
1a ) ĐKXĐ : \(\left\{{}\begin{matrix}x\ge5\\y\ge2005\\z\ge-2007\end{matrix}\right.\)
\(\sqrt{x-5}+\sqrt{y-2005}+\sqrt{z+2007}=\dfrac{1}{2}\left(x+y+z\right)\)
\(\Leftrightarrow x+y+z=2\sqrt{x-5}+2\sqrt{y-2005}+2\sqrt{z+2007}\)
\(\Leftrightarrow\left(x-5-2\sqrt{x-5}+1\right)+\left(y-2005-2\sqrt{y-2005}+1\right)+\left(z+2007-2\sqrt{z+2007}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-5}-1\right)^2+\left(\sqrt{y-2005}-1\right)^2+\left(\sqrt{z+2007}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-5}-1=0\\\sqrt{y-2005}-1=0\\\sqrt{z+2007}-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=2006\\z=-2006\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)=\left(6;2006;-2006\right)\)
b ) \(\sqrt{3x^2-12x+21}+\sqrt{5x^2-20x+24}=-2x^2+8x-3\)
\(\Leftrightarrow\sqrt{3\left(x^2-4x+7\right)}+\sqrt{5\left(x^2-4x+7\right)-11}=-2\left(x^2-4x+7\right)+11\)
Đặt \(x^2-4x+7=a\) . Phương trình trở thành :
\(\sqrt{3a}+\sqrt{5a-11}=-2a+11\) ( ĐK : \(a\le\dfrac{11}{2}\) )
\(\Leftrightarrow8a-11+2\sqrt{15a^2-33a}=4a^2-44a+121\)
\(\Leftrightarrow4a^2-52a+132=2\sqrt{15a^2-33a}\)
\(\Leftrightarrow2a^2-26a+66a=\sqrt{15a^2-33a}\)
\(\Leftrightarrow4a^4-104a^3+940a^2-3432a+4356=15a^2-33a\)
\(\Leftrightarrow4a^4-104a^3+925a^2-3399a+4356=0\)
\(\Leftrightarrow\left(2a-11\right)^2\left(a-12\right)\left(a-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2a-11=0\\a-12=0\\a-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=\dfrac{11}{2}\left(L\right)\\a=12\left(L\right)\\a=3\left(N\right)\end{matrix}\right.\)
Với \(a=3\)
\(\Leftrightarrow x^2-4x+4=0\)
\(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x=2\)
Vậy \(S=\left\{2\right\}\)
2 ) \(2^x+2^y+2^z=672\)
\(\Leftrightarrow2^x\left(1+2^{y-x}+2^{z-x}\right)=672\)
Do : \(z>y>x\) \(\Rightarrow1+2^{y-x}+2^{z-x}>0\)
Nên x phải là số dương
\(\Leftrightarrow1+2^{y-x}+2^{z-x}=\dfrac{672}{2^x}\)
\(\Leftrightarrow1+2^{y-x}+2^{z-x}=\dfrac{2^5.21}{2^x}\)
\(\Rightarrow x\le5\)
Do VT của phương trình là số lẻ nên \(\dfrac{2^5.21}{2^x}\) phải là số lẻ .
Thử x từ 1 đến 5 ta nhận thấy \(x=5\) là thỏa mãn .
\(\Rightarrow2^{y-5}+2^{z-5}=20\)
\(\Leftrightarrow2^{y-5}+2^{z-5}=2^2+2^4\)
\(\Rightarrow\left\{{}\begin{matrix}y-5=2\\z-5=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=7\\z=9\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)=\left(5;7;9\right)\)