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Giúp mình bài tập này với ạ, mình cần gấp ạ( chỉ cần bài 4,5 phương pháp t...

HT.Phong (9A5) · Accepted Answer

Bài 5:

a) Để hpt có nghiệm duy nhất thì $\dfrac{m}{2}\ne\dfrac{2}{m}\Leftrightarrow m\ne\pm2$

$\left\{{}\begin{matrix}mx+2y=m+1\2x+my=2m-1\end{matrix}\right.\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{m-mx+1}{2}\2x+m\cdot\dfrac{m-mx+1}{2}=2m-1\end{matrix}\right.\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{m-mx+1}{2}\2x+\dfrac{m^2-m^2x+m}{2}=2m-1\end{matrix}\right.\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{m-mx+1}{2}\4x+m^2-m^2x+m=4m-2\end{matrix}\right.\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{m-mx+1}{2}\\left(m^2-4\right)x=m^2-3m+2\end{matrix}\right.\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{m-m\cdot\dfrac{m-1}{m+2}+1}{2}=\dfrac{\dfrac{m\left(m+2\right)-m\left(m-1\right)+m+2}{m+2}}{2}=\dfrac{2m+1}{m+2}\x=\dfrac{m^2-3m+2}{m^2-4}=\dfrac{m-1}{m+2}\end{matrix}\right.$

Để x,y nguyên thì $\dfrac{m-1}{m+2};\dfrac{2m+1}{m+2}$ phải nguyên

+) Ta có: $\dfrac{m-1}{m+2}=\dfrac{m+2-3}{m+2}=1-\dfrac{3}{m+2}$

=> m + 2 ∈ Ư(3) = {1; -1; 3; -3}

=> m ∈ {-1; -3; 1; -5} (1)

+) Ta có: $\dfrac{2m+1}{m+2}=\dfrac{2m+4-3}{m+2}=2-\dfrac{3}{m+2}$

=> m + 2 ∈ Ư(3) = {1; -1; 3; -3}

=> m ∈ {-1; -3; 1; -5} (2)

Từ (1) và (2) => m ∈ {1; -1; 3; -3}

Câu trả lời:

Bài 5:

a) Để hpt có nghiệm duy nhất thì $\dfrac{m}{2}\ne\dfrac{2}{m}\Leftrightarrow m\ne\pm2$

$\left\{{}\begin{matrix}mx+2y=m+1\2x+my=2m-1\end{matrix}\right.\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{m-mx+1}{2}\2x+m\cdot\dfrac{m-mx+1}{2}=2m-1\end{matrix}\right.\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{m-mx+1}{2}\2x+\dfrac{m^2-m^2x+m}{2}=2m-1\end{matrix}\right.\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{m-mx+1}{2}\4x+m^2-m^2x+m=4m-2\end{matrix}\right.\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{m-mx+1}{2}\\left(m^2-4\right)x=m^2-3m+2\end{matrix}\right.\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{m-m\cdot\dfrac{m-1}{m+2}+1}{2}=\dfrac{\dfrac{m\left(m+2\right)-m\left(m-1\right)+m+2}{m+2}}{2}=\dfrac{2m+1}{m+2}\x=\dfrac{m^2-3m+2}{m^2-4}=\dfrac{m-1}{m+2}\end{matrix}\right.$

Để x,y nguyên thì $\dfrac{m-1}{m+2};\dfrac{2m+1}{m+2}$ phải nguyên

+) Ta có: $\dfrac{m-1}{m+2}=\dfrac{m+2-3}{m+2}=1-\dfrac{3}{m+2}$

=> m + 2 ∈ Ư(3) = {1; -1; 3; -3}

=> m ∈ {-1; -3; 1; -5} (1)

+) Ta có: $\dfrac{2m+1}{m+2}=\dfrac{2m+4-3}{m+2}=2-\dfrac{3}{m+2}$

=> m + 2 ∈ Ư(3) = {1; -1; 3; -3}

=> m ∈ {-1; -3; 1; -5} (2)

Từ (1) và (2) => m ∈ {1; -1; 3; -3}

Nguyễn Huy Tú · Answer

Bài 4

a, $\left\{{}\begin{matrix}-2\sqrt{3}x+3\sqrt{5}y=-21\4x-2\sqrt{3}y=2\sqrt{3}\left(2+\sqrt{5}\right)\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{21-3\sqrt{5}y}{-2\sqrt{3}}\\dfrac{4\left(21-3\sqrt{5}y\right)}{-2\sqrt{3}}-2\sqrt{3}y=2\sqrt{3}\left(2+\sqrt{5}\right)\left(1\right)\end{matrix}\right.$

$\left(1\right)\Rightarrow84-21\sqrt{5}y+12y=-12\left(2+\sqrt{5}\right)$

$\Leftrightarrow84+y\left(-21\sqrt{5}+12\right)=-24-12\sqrt{5}\Leftrightarrow y=\dfrac{-108-12\sqrt{5}}{-21\sqrt{5}+12}$

$\Rightarrow x=\dfrac{\dfrac{\left(21-3\sqrt{5}\right).\left(-108-12\sqrt{5}\right)}{-21\sqrt{5}+12}}{-2\sqrt{3}}$

b, $\left\{{}\begin{matrix}\left(x-1\right)^2+\left(y-2\right)^2=\left(x+1\right)^2+1+\left(y+1\right)^2\\left(x-y-3\right)^2=\left(x-y-1\right)^2\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2-\left(x+1\right)^2=1+\left(y+1\right)^2-\left(y-2\right)^2\\left(x-y-3-x+y+1\right)\left(x-y-3+x-y-1\right)=0\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}-4x-1=-\left(2y-1\right)\-2\left(2x-2y-4\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-4x+2y=2\x-y-2=0\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}-x+y=1\x=y+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-y-2+y=1\x=y+2\end{matrix}\right.$( vô lí )

Vậy hpt vô nghiệm