Ôn tập hệ hai phương trình bậc nhất hai ẩn

a,

\(\left\{{}\begin{matrix}mx+y=2\\4x+my=m+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2-mx\\4x+m\left(2-mx\right)-m-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=2-mx\\4x+2m-m^2x-m-2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=2-mx\\\left(4-m^2\right)x+m-2=0\left(\cdot\right)\end{matrix}\right.\)

+ Hệ pt có 1 nghiệm duy nhất khi pt (.) có 1 nghiệm duy nhất\(\Rightarrow4-m^2\ne0\Leftrightarrow m^2\ne4\Leftrightarrow m\ne\pm2\)

+ Hệ pt có vô số nghiệm khi pt (.) có vô số nghiệm

\(\Rightarrow\left\{{}\begin{matrix}4-m^2=0\\m-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m=-2\\m=2\end{matrix}\right.\\m=2\end{matrix}\right.\)

\(\Rightarrow m=2\)

+ Hệ pt vô nghiệm khi pt (.) vô nghiệm

\(\Rightarrow\left\{{}\begin{matrix}4-m^2=0\\m-2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m=2\\m=-2\end{matrix}\right.\\m\ne2\end{matrix}\right.\)

\(\Rightarrow m=-2\)

b, \(\left\{{}\begin{matrix}y=2-mx\\\left(4-m^2\right)x=2-m\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2-mx\\x=\dfrac{2-m}{4-m^2}=\dfrac{1}{m+2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=2-\dfrac{m}{m+2}=\dfrac{m+4}{m+2}\\x=\dfrac{1}{m+2}\end{matrix}\right.\)

Xét y-2x có:

y-2x = \(\dfrac{m+4}{m+2}-\dfrac{2}{m+2}=\dfrac{m+4-2}{m+2}=\dfrac{m+2}{m+2}=1\)

Vậy hệ thức y-2x không phụ thuộc vào m

\(2ab+6bc+2ac=7abc\\ \Rightarrow\dfrac{2}{c}+\dfrac{6}{a}+\dfrac{2}{b}=7\\ \)

Đặt x=1/a ; y=1/b ; z=1/c

\(\Rightarrow6x+2y+2z=7\)

\(H=\dfrac{4}{\dfrac{1}{b}+\dfrac{2}{a}}+\dfrac{9}{\dfrac{1}{c}+\dfrac{4}{a}}+\dfrac{4}{\dfrac{1}{c}+\dfrac{1}{b}}\\ =\dfrac{4}{2x+y}+\dfrac{9}{z+4x}+\dfrac{4}{y+z}\)

BĐT Cô si ;

\(\left(\dfrac{4}{2x+y}+\left(2x+y\right)\right)+\left(\dfrac{9}{4x+z}+\left(4x+z\right)\right)+\left(\dfrac{4}{y+z}+\left(y+z\right)\right)\\ \ge2\sqrt{4}+2\sqrt{9}+2\sqrt{4}=14\\ \Rightarrow C+7\ge14\\ \Rightarrow C\ge7\)

Min C=7 khi a=2;b=c=1

x là v lúc đi

\(\dfrac{24}{x+4}+1=\dfrac{24}{x-4}\)

x=14.4222051 km/h !

ta có: \(a^3-a^2b+ab^2-6b^3=0\)

\(\Leftrightarrow\left(a^3-2a^2b\right)+\left(a^2b-2ab^2\right)+\left(3ab^2-6b^3\right)=0\)

\(\Leftrightarrow\left(a-2b\right)\left(a^2+ab+3b^2\right)=0\)

vì \(a>b>0\Rightarrow\left(a^2+ab+3b^2\right)>0\)

\(\Rightarrow a-2b=0\)

\(\Leftrightarrow a=2b\)

Thế vào \(\dfrac{a^4-4b^4}{b^4-4a^4}=\dfrac{-4}{21}\)

Theo đề, ta có hệ phương trình:

\(\left\{{}\begin{matrix}2a-3\left(a-b-2\right)=a+3b-6\\4a+a-b-2=a+3b-6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2a-3a+3b+6=a+3b-6\\5a-b-2-a-3b+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-a+3b+6-a-3b+6=0\\4a-4b=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2a+12=0\\a-b=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=6\\b=7\end{matrix}\right.\)

a) \(4ab+a^2-3a-12b=a\left(4b+a\right)-3\left(4b+a\right)=\left(a-3\right)\left(4b+a\right)\)

b) \(x^3+3x^2+3x+1-27y^3=\left(x+1\right)^3-27y^3=\left(x+1-3y\right)\left[\left(x+1\right)^2+3y\left(x+1\right)+9y^2\right]=\left(x+1-3y\right)\left(x^2+2x+1+3xy+3y+9y^2\right)\)