Đáp án:
\(\begin{array}{l}
6)\,a)\,y = \dfrac{2}{3}x\,\,\,b)\,y = 2x\,\,c)\,y = 3x\\
7)\,a)\,y = x;\,\,\,b)y = \left( {2 - \sqrt 2 } \right)x + 1 - \sqrt 2 \,\,\\
c)\,\,y = \left( {1 + \sqrt 3 } \right)x + \sqrt 3
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
6)\,y = ax + b\,\left( {a \ne 0} \right)\\
Dt\,di\,qua\,goc\,toa\,do\,nen\,b = 0 \Rightarrow y = ax\\
a)\,Thay\,x = 3;\,y = 2\\
\Rightarrow 2 = a.3 \Rightarrow a = \dfrac{2}{3} \Rightarrow y = \dfrac{2}{3}x\\
b)\,a = 2 \Rightarrow y = 2x\\
c)\, \Rightarrow a = 3\\
\Rightarrow y = 3x\\
7)\,a)\,Thay\,x = 0;y = 0\,vao\,ham\,so\,ta\,duoc:\\
0 = \left( {k + 1} \right).0 + k \Rightarrow k = 0\\
\Rightarrow y = x\\
b)Thay\,x = 0;y = 1 - \sqrt 2 \,vao\,ham\,so\,ta\,duoc:\\
1 - \sqrt 2 = \left( {k + 1} \right).0 + k \Rightarrow k = 1 - \sqrt 2 \\
\Rightarrow y = \left( {2 - \sqrt 2 } \right)x + 1 - \sqrt 2 \,\,\\
c)\, \Rightarrow \left\{ \begin{array}{l}
k + 1 = \sqrt 3 + 1\\
k \ne 3
\end{array} \right. \Leftrightarrow k = \sqrt 3 \\
\Rightarrow y = \left( {1 + \sqrt 3 } \right)x + \sqrt 3
\end{array}\)
