giúp mình nhé

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

 \(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=4\\2x+4y=-10\end{matrix}\right.\)

 \(\Leftrightarrow\left\{{}\begin{matrix}-7y=14\\x+2y=-5\end{matrix}\right.\) 

\(\Leftrightarrow\left\{{}\begin{matrix}y=-2\\x+2.\left(-2\right)=-5\end{matrix}\right.\)

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

Vậy nghiệm của hệ phương trình là \(\left\{{}\begin{matrix}y=-2\\x=-1\end{matrix}\right.\)

Ta có: \(\left\{{}\begin{matrix}2x-3y=4\\x+2y=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=4\\-2x-4y=10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-7y=14\\2x-3y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2\\2x-3\cdot\left(-2\right)=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=-2\\2x+6=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2\\2x=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2\\x=-1\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-1\\y=-2\end{matrix}\right.\)

\(a,\Leftrightarrow\left\{{}\begin{matrix}5x+15y=-10\\5x-4y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}19y=-21\\5x-4y=11\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{21}{19}\\5x-4\left(-\dfrac{21}{19}\right)=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{25}{19}\\y=-\dfrac{21}{19}\end{matrix}\right.\)

\(c,\Leftrightarrow\left\{{}\begin{matrix}3x+5y=1\\10x-5y=-40\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+5y=1\\13x=-39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=2\end{matrix}\right.\\ d,\Leftrightarrow\left\{{}\begin{matrix}5x-10y=-30\\5x-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-3y=5\\-7y=-35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=5\end{matrix}\right.\\ e,\Leftrightarrow\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=4\\2\left(x+y\right)+4\left(x-y\right)=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=6\\2\left(x+y\right)+3\cdot6=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=6\\x+y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-\dfrac{13}{2}\end{matrix}\right.\)

1) \(-2x^2+x+1-2\sqrt[]{x^2+x+1}=0\)

\(\Leftrightarrow2\sqrt[]{x^2+x+1}=-2x^2+x+1\left(1\right)\)

Ta có :

\(2\sqrt[]{x^2+x+1}=2\sqrt[]{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\ge\sqrt[]{3}\)

Dấu "=" xảy ra khi và chỉ khi \(x+\dfrac{1}{2}=0\Leftrightarrow x=-\dfrac{1}{2}\)

\(\left(1\right)\Leftrightarrow-2x^2+x+1=\sqrt[]{3}\)

\(\Leftrightarrow2x^2-x+\sqrt[]{3}-1=0\)

\(\Delta=1-8\left(\sqrt[]{3}-1\right)=9-8\sqrt[]{3}\)

\(pt\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt[]{9-8\sqrt[]{3}}}{4}\left(loại\right)\\x=\dfrac{1-\sqrt[]{9-8\sqrt[]{3}}}{4}\left(loại\right)\end{matrix}\right.\) \(\left(vì.x=-\dfrac{1}{2}\right)\)

Vậy phương trình cho vô nghiệm

nhâ vế 1 vs 2

nhân vế 2 vs 3 là ra thôi bn

trừ 2 vế cho nhau nữa

\(\left\{{}\begin{matrix}3x+2y=4\\2x-3y=7\end{matrix}\right.< =>\left\{{}\begin{matrix}6x+4y=8\\6x-9y=21\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}13y=-13\\3x+2y=4\end{matrix}\right.< =>\left\{{}\begin{matrix}y=-1\\3x=4+2=6\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}y=-1\\x=2\end{matrix}\right.\)

4:

x+3y=4m+4 và 2x+y=3m+3

=>2x+6y=8m+8 và 2x+y=3m+3

=>5y=5m+5 và x+3y=4m+4

=>y=m+1 và x=4m+4-3m-3=m+1

x+y=4

=>m+1+m+1=4

=>2m+2=4

=>2m=2

=>m=1

3:

x+2y=3m+2 và 2x+y=3m+2

=>2x+4y=6m+4 và 2x+y=3m+2

=>3y=3m+2 và x+2y=3m+2

=>y=m+2/3 và x=3m+2-2m-4/3=m+2/3

Vì \(\dfrac{1}{-2}< >-\dfrac{2}{3}\) nên phương trình có nghiệm duy nhất