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K={1/x|x=k(k+1);k∈N và 1<=k<=5}
H={x∈N|1<=x<=16; x=3k+1; k∈N}
L={\(\frac{x}{x^2-1}\) ; 2<=x<=6}
I={\(\frac{1}{5^{x}}\) |x∈N và 0<=x<=3}
\(1,\left|2x-3\right|=x-5\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-5\ge0\\\left[{}\begin{matrix}2x-3=x-5\\2x-3=-x+5\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\text{≥}5\\\left[{}\begin{matrix}x=-2\\x=\frac{8}{3}\end{matrix}\right.\end{matrix}\right.\) (ko thỏa mãn)
=> pt vô nghiệm
\(2,\left|3x+2\right|=x+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1\text{≥}0\\\left[{}\begin{matrix}3x+2=x+1\\3x+2=-x-1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\text{≥}-1\\\left[{}\begin{matrix}x=-\frac{1}{2}\\x=-\frac{3}{4}\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\\x=-\frac{3}{4}\end{matrix}\right.\)
\(3,\left|2x+1\right|=7-x\)
\(\Leftrightarrow\left\{{}\begin{matrix}7-x\text{≥}0\\\left[{}\begin{matrix}2x+1=7-x\\2x+1=x-7\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\text{≥}7\\\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\end{matrix}\right.\) (loại)
=> pt vô nghiệm
\(4,\left|2x-5\right|=x+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1\text{≥}0\\\left[{}\begin{matrix}2x-5=x+1\\2x-5=-x-1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\text{≥}-1\\\left[{}\begin{matrix}x=6\\x=\frac{4}{3}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{4}{3}\end{matrix}\right.\)
\(5,\left|6x-2\right|=3x-4\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-4\text{≥}0\\\left[{}\begin{matrix}6x-2=3x-4\\6x-2=-3x+4\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\text{≥}\frac{4}{3}\\\left[{}\begin{matrix}x=-\frac{2}{3}\\x=\frac{2}{3}\end{matrix}\right.\end{matrix}\right.\) => pt vô nghiệm
\(6,\left|3x-2\right|=x-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2\text{≥}0\\\left[{}\begin{matrix}3x-2=x-2\\3x-2=-x+2\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\text{≥}2\\\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\end{matrix}\right.\) => pt vô nghiệm
\(7,\left|2x+3\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=1\\2x+3=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)
\(8,\left|2-x\right|=2x-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1\ge0\\\left[{}\begin{matrix}2-x=2x-1\\2-x=-2x+1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{1}{2}\\\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow x=1\)
\(9,\left|2x-1\right|=x-3\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3\ge0\\\left[{}\begin{matrix}2x-1=x-3\\2x-1=-x+3\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge3\\\left[{}\begin{matrix}x=-2\\x=\frac{4}{3}\end{matrix}\right.\end{matrix}\right.\) => pt vô nghiệm
\(10,2\left|x-1\right|=x+2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2\ge0\\\left[{}\begin{matrix}2x-2=x+2\\2x-2=-x-2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)
1/ \(3\left(2^2+1\right)\left(2^4+1\right)...\left(2^{128}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{128}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{128}+1\right)\)
..................................................................
\(=\left(2^{128}-1\right)\left(2^{128}+1\right)=2^{256}-1\)
2/ Ta có: \(a+b+c=0\Leftrightarrow a+b=-c\)
\(\Leftrightarrow a^2+2ab+b^2=c^2\Leftrightarrow a^2+b^2-c^2=-2ab\)
\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2-2b^2c^2-2c^2a^2=4a^2b^2\)
\(\Leftrightarrow a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2\)
Ta lại có: \(a^2+b^2+c^2=10\)
\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=100\)
\(\Leftrightarrow2\left(a^4+b^4+c^4\right)=100\Leftrightarrow a^4+b^4+c^4=50\)
\(\Leftrightarrow\frac{1}{a^4+b^4+c^4}=\frac{1}{50}\)
1: A(-1;1); B(0;-1); C(1;-3)
\(\overrightarrow{AB}=\left(0+1;-1-1\right)=\left(1;-2\right)\)
\(\overrightarrow{AC}=\left(1+1;-3-1\right)=\left(2;-4\right)\)
Vì \(\frac12=\frac{-2}{-4}\left(=\frac12\right)\)
nên A,B,C thẳng hàng
2: A(2;0); B(5;1); C(-1;-1)
\(\overrightarrow{AB}=\left(5-2;1-0\right)=\left(3;1\right);\overrightarrow{AC}=\left(-1-2;-1-0\right)=\left(-3;-1\right)\)
Vì \(\frac{3}{-3}=\frac{1}{-1}\left(=-1\right)\)
nên A,B,C thẳng hàng
3: A(4;3); B(2;0); C(0;-3)
\(\overrightarrow{AB}=\left(2-4;0-3\right)=\left(-2;-3\right);\overrightarrow{AC}=\left(0-4;-3-3\right)=\left(-4;-6\right)\)
Vì \(\frac{-2}{-4}=\frac{-3}{-6}\left(=\frac12\right)\)
nên A,B,C thẳng hàng
4: A(-1;2); B(2;3); C(4;-1)
\(\overrightarrow{AB}=\left(2+1;3-2\right)=\left(3;1\right);\overrightarrow{AC}=\left(4+1;-1-2\right)=\left(5;-3\right)\)
Vì 3/5<>1/-3
nên A,B,C không thẳng hàng