a) \(\left\{{}\begin{matrix}2x+3y=5\\4x-5y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=10\\4x-5y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=5\\11y=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+3\cdot\dfrac{9}{11}=5\\y=\dfrac{9}{11}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{27}{11}=5\\y=\dfrac{9}{11}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=\dfrac{28}{11}\\y=\dfrac{9}{11}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{14}{11}\\y=\dfrac{9}{11}\end{matrix}\right.\)

Vậy: \(x=\dfrac{14}{11};y=\dfrac{9}{11}\)

giúp mình nhé

\(\left\{{}\begin{matrix}\dfrac{xy}{4x+3y}=\dfrac{4}{7}\\\dfrac{xy}{2x+y}=\dfrac{4}{5}\end{matrix}\right.\)\(\left(đk:4x\ne-3y,-2x\ne y,xy\ne0\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4x+3y}{xy}=\dfrac{7}{4}\\\dfrac{2x+y}{xy}=\dfrac{5}{4}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4x+3y}{xy}=\dfrac{7}{4}\\\dfrac{4x+2y}{xy}=\dfrac{5}{2}\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x^3-y^3=35\\6x^2+9y^2=12x-27y\end{matrix}\right.\)

\(\Rightarrow x^3-y^3-6x^2-9y^2=35-12x+27y\)

\(\Leftrightarrow x^3-6x^2+12x-8=y^3+9y^2+27y+27\)

\(\Leftrightarrow\left(x-2\right)^3=\left(y+3\right)^3\)

\(\Leftrightarrow x-2=y+3\)

\(\Leftrightarrow y=x-5\)

Thay vào pt dưới: \(2x^2+3\left(x-5\right)^2=4x-9\left(x-5\right)\)

\(\Leftrightarrow...\)

a/ ĐKXĐ:...

\(\Leftrightarrow x^2+2x+1+2x+3-2\sqrt{2x+3}+1=0\)

\(\Leftrightarrow\left(x+1\right)^2+\left(\sqrt{2x+3}-1\right)^2=0\)

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

b/ \(\Leftrightarrow\left\{{}\begin{matrix}x^2+3xy=4\\4y^2+xy=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5x^2+15xy=20\\16y^2+4xy=20\end{matrix}\right.\)

\(\Rightarrow5x^2+11xy-16y^2=0\)

\(\Leftrightarrow\left(x-y\right)\left(5x+16y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-\frac{16}{5}y\end{matrix}\right.\)

Bạn tự thế vào một trong hai pt giải tiếp

Woa nghiệm đẹp:) Nhưng em giải đúng hay ko là một chuyện:v

ĐK: \(x\ge-\frac{3}{2}\)

PT \(\Leftrightarrow x^2+4x+3+\left(2-2\sqrt{2x+3}\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+3\right)+\frac{4-4\left(2x+3\right)}{2+\sqrt{2x+3}}=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+3\right)-\frac{8\left(x+1\right)}{2+\sqrt{2x+3}}=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+3-\frac{8}{2+\sqrt{2x+3}}\right)=0\)

Giải cái ngoặc nhỏ suy ra x = -1

Giải cái ngoặc to:

\(\Leftrightarrow x+3=\frac{8}{2+\sqrt{2x+3}}\)

Nghiệm xấu quá :( => em bí.

Đánh máy ẩu và sai lầm chết người -_-" Ai đó xóa giúp em bài kia với ạ. Em cảm ơn. Nói gì thì nói chứ cách em phức tạp quá:( Mà chưa chắc đúng.

ĐK: \(x\ge-\frac{3}{2}\)

PT \(\Leftrightarrow\left(x^2+4x+3\right)+\left(2-2\sqrt{2x+3}\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+3\right)-\frac{8\left(x+1\right)}{2+2\sqrt{2x+3}}=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+3-\frac{8}{2+2\sqrt{2x+3}}\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+3-\frac{4}{1+\sqrt{2x+3}}\right)=0\)

\(\Leftrightarrow x=-1\)

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

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

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

=>\(\left\{{}\begin{matrix}x=2\\y=5-2x=5-2\cdot2=1\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}-x+2y=2\\2x-y=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-2x+4y=4\\2x-y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3y=3\\x-2y=-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=1\\x=-2+2y=-2+2\cdot1=0\end{matrix}\right.\)

c: \(\left\{{}\begin{matrix}2x-y=13\\y-5=-7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-y=13\\y=-7+5=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=y+13=-2+13=11\\y=-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{11}{2}\\y=-2\end{matrix}\right.\)

d: \(\left\{{}\begin{matrix}3x+y=8\\2x-3y=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}9x+3y=24\\2x-3y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11x=25\\3x+y=8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{25}{11}\\y=8-3x=8-3\cdot\dfrac{25}{11}=8-\dfrac{75}{11}=\dfrac{13}{11}\end{matrix}\right.\)

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

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}2x+3y=1\\3x-3y=9\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}5x=10\\x-y=3\end{matrix}\right.\)

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

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