\(a,\Leftrightarrow A\left(0;2\right)\in\left(d\right)\Leftrightarrow3m-1=2\Leftrightarrow m=1\\ b,\Leftrightarrow m-2=-2\Leftrightarrow m=0\\ c,\Leftrightarrow\left\{{}\begin{matrix}m-2=3\\3m-1\ne-2\end{matrix}\right.\Leftrightarrow m=5\\ d,\text{PT hoành độ giao điểm: }\left(m-2\right)x+3m-1=3x-2\\ \Leftrightarrow x\left(m-2-3\right)+3m-1+2=0\\ \Leftrightarrow x\left(m-5\right)=-3m-1\Leftrightarrow x=\dfrac{-3m-1}{m-5}\)

Vì 2 đt cắt bên trái trục tung nên hoành độ âm

\(\Leftrightarrow x< 0\Leftrightarrow\dfrac{-3m-1}{m-5}< 0\Leftrightarrow\dfrac{3m+1}{m-5}>0\Leftrightarrow\left[{}\begin{matrix}m>5\\m< -\dfrac{1}{3}\end{matrix}\right.\)

\(e,\text{Gọi điểm cố định mà }\left(d\right)\text{ luôn đi qua là }M\left(x_0;y_0\right)\\ \Leftrightarrow\left(m-2\right)x_0+3m-1=y_0\\ \Leftrightarrow mx_0-2x_0+3m-1-y_0=0\\ \Leftrightarrow m\left(x_0+3\right)-\left(2x_0+y_0+1\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x_0+3=0\\2x_0+y_0+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_0=-3\\y_0=5\end{matrix}\right.\Leftrightarrow M\left(-3;5\right)\\ \text{Vậy }\left(d\right)\text{ luôn đi qua }M\left(-3;5\right)\)

\(a,=4x^4y+3x^3-5\\ b,=x^2-x^3-2+2x+x^3+27=x^2+2x+25\)

\(=\dfrac{\left(1-3x\right)\left(2x-1\right)+2x\left(3x-2\right)-\left(3x-2\right)}{2x\left(2x-1\right)}\\ =\dfrac{-6x^2+5x-1+6x^2-4x-3x+2}{2x\left(2x-1\right)}=\dfrac{-2x+1}{2x\left(2x-1\right)}=\dfrac{-1}{2x}\\ \text{Thay }x=\dfrac{1}{682}\\ =\dfrac{-1}{2\cdot\dfrac{1}{682}}=-\dfrac{1}{\dfrac{1}{341}}=-341\)

Gọi 2 cgv là \(a,b(a,b>0)\)

Áp dụng tc dtsbn:

\(\dfrac{a}{2}=\dfrac{b}{5}=\dfrac{b-a}{5-2}=\dfrac{12}{3}=4\\ \Rightarrow\left\{{}\begin{matrix}a=8\\b=20\end{matrix}\right.\)

Vậy diện tích là \(\dfrac{1}{2}ab=\dfrac{1}{2}\cdot8\cdot20=80\left(đvdt\right)\)

\(A=3\left(1+3+3^2+3^3+...+3^9\right)⋮3\)

Làm trong 8 ngày cần \(6\cdot12:8=9\left(\text{công nhân}\right)\)

Vậy cần thêm 9-6=3 công nhân

\(a,\Leftrightarrow2x^2+x+a=\left(x+3\right)\cdot g\left(x\right)\\ \text{Thay }x=-3\Leftrightarrow18-3+a=0\Leftrightarrow a=-15\\ b,\Leftrightarrow x^3+ax^2-4=\left(x^2+4x+4\right)\cdot f\left(x\right)=\left(x+2\right)^2\cdot f\left(x\right)\\ \text{Thay }x=-2\Leftrightarrow-8+4a-4=0\\ \Leftrightarrow4a-12=0\Leftrightarrow a=3\)

Nguyễn Minh Sơn lớp 6 đòi hỏi j?