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Câu 1:
\(\left\{{}\begin{matrix}m^2x+y=3m\\-4x-y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m^2x-4x=3m+6\\-4x-y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(m^2-4\right)=3m+6\\-4x-y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3m+6}{m^2-4}=\dfrac{3}{m-2}\\y=6-\dfrac{3}{m-2}=\dfrac{6m-15}{m-2}\end{matrix}\right.\)Câu 2:
\(\left\{{}\begin{matrix}5x-y=13\\x+3y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}15x-3y=39\\x+3y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}16x=32\\x+3y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-3\end{matrix}\right.\)
a) Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x+1}{x-2}=\dfrac{x+2}{x-9}=\dfrac{x+1-x-2}{x-2-x+9}=-\dfrac{1}{7}\)
Hay \(\dfrac{x+1}{x-2}=-\dfrac{1}{7}\Leftrightarrow-x+2=7x+7\Leftrightarrow-x=7x+5\Leftrightarrow-x-7x=5\Leftrightarrow-8x=5\Leftrightarrow x=-\dfrac{5}{8}\)b) phải sử dụng \(\left\{{}\begin{matrix}x\left(x+y\right)=10\\y\left(x+y\right)=6\end{matrix}\right.\)(sửa đề)
\(\Leftrightarrow\left(x+y\right)^2=16\Leftrightarrow\left[{}\begin{matrix}x+y=4\\x+y=-4\end{matrix}\right.\)
Nên \(\left[{}\begin{matrix}x=-\dfrac{5}{2}\\y=-\dfrac{3}{2}\end{matrix}\right.\)
1.
\(\dfrac{2}{3}.\left\{\dfrac{2}{3}-\dfrac{2}{3}.\left[\dfrac{2}{3}-\dfrac{2}{3}.\left(\dfrac{2}{3}-\dfrac{1}{2}\right)\right]\right\}\)
=\(\dfrac{2}{3}.\left\{\dfrac{2}{3}-\dfrac{2}{3}.\left[\dfrac{2}{3}-\dfrac{2}{3}.\dfrac{1}{6}\right]\right\}\)
=\(\dfrac{2}{3}.\left\{\dfrac{2}{3}-\dfrac{2}{3}.\left[\dfrac{2}{3}-\dfrac{1}{9}\right]\right\}\)
=\(\dfrac{2}{3}.\left\{\dfrac{2}{3}-\dfrac{2}{3}.\dfrac{5}{9}\right\}\)
=\(\dfrac{2}{3}.\left\{\dfrac{2}{3}-\dfrac{10}{27}\right\}\)
=\(\dfrac{2}{3}.\dfrac{8}{27}\)
=...
Lại Signum ư -.-
Ta có \(x\ne0\Rightarrow x>0\Rightarrow Sgnx^2=1\)
\(\left(-1\right)^{2n}=+1\Rightarrow Sgn\left(\left(-1\right)^{2n}\right)=Sgn1=1\)
\(\left(-1\right)^{2n+1}=-1\Rightarrow Sgn\left(\left(-1\right)^{2n+1}\right)=Sgn\left(-1\right)=-1\)
KL: ...