a.

\(\left\{{}\begin{matrix}x^3-y^3=16x-4y\\-4=5x^2-y^2\end{matrix}\right.\)

Nhân vế:

\(-4\left(x^3-y^3\right)=\left(16x-4y\right)\left(5x^2-y^2\right)\)

\(\Leftrightarrow21x^3-5x^2y-4xy^2=0\)

\(\Leftrightarrow x\left(7x-4y\right)\left(3x+y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4y}{7}\\y=-3x\end{matrix}\right.\)

Thế vào \(y^2=5x^2+4...\)

b. Đề bài không hợp lý ở \(4x^2\)

c.

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

Trừ vế:

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

\(\Leftrightarrow x^3-3x^2+3x-1=y^3+6y^2+12y+8\)

\(\Leftrightarrow\left(x-1\right)^3=\left(y+2\right)^3\)

\(\Leftrightarrow x-1=y+2\)

\(\Leftrightarrow y=x-3\)

Thế vào \(x^2=2y^2=x-4y\) ...

b.

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

\(\Rightarrow y^4-2y^2-4xy^3+4xy=-1\)

\(\Leftrightarrow\left(y^2-1\right)^2-4xy\left(y^2-1\right)=0\)

\(\Leftrightarrow\left(y^2-1\right)\left(y^2-1-4xy\right)=0\)

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

Thế vào \(2x^2+y^2-2xy=1\) ...

Với \(x=\dfrac{y^2-1}{4y}\) ta được:

\(2\left(\dfrac{y^2-1}{4y}\right)^2+y^2-2\left(\dfrac{y^2-1}{4y}\right)y=1\)

\(\Leftrightarrow5y^4-6y^2+1=0\)

a) hpt \(\Leftrightarrow\left\{{}\begin{matrix}x+y+xy=11\\\left(x+y\right)^2-2xy-\left(x+y\right)=8\end{matrix}\right.\)

Đặt S=x+y; P =xy, ta có hệ :

\(\left\{{}\begin{matrix}S+P=11\\S^2-S-2P=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}P=11-S\\S^2-S-2\left(11-S\right)=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}P=11-S\\S^2+S-30=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}P=11-S\\\left[{}\begin{matrix}S=5\\S=-6\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy=11-\left(x+y\right)\\\left[{}\begin{matrix}x+y=5\\x+y=-6\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\curlyvee\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\\\text{hệ vô nghiệm}\end{matrix}\right.\)

Vậy...

b)ĐK : \(\left\{{}\begin{matrix}x\le0\\x\ge2\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{2x^2-4x}-4=\frac{2x^2-2x+12}{x+8}-4\)

\(\Leftrightarrow\frac{2x^2-4x-16}{\sqrt{2x^2-4x}+4}=\frac{2x^2-6x-20}{x+8}\)

\(\Leftrightarrow\frac{\left(x-4\right)\left(x+2\right)}{\sqrt{2x^2-4x}+4}=\frac{\left(x-5\right)\left(x+2\right)}{x+8}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\\frac{x-4}{\sqrt{2x^2-4x}+4}=\frac{x+5}{x+8}\left(1\right)\end{matrix}\right.\)

Giải tiếp pt 1 và kết hợp vs đk t tìm được nghiệm

(mượn tạm chỗ comment) @Nguyễn Việt Lâm anh có làm việc bên online math không

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.\(\left\{{}\begin{matrix}4x+2y=14\\2x-2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x=18\\2x-2y=4\end{matrix}\right.\)

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

⇔\(\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)

vậy  hệ pt có ndn \(\left\{2;0\right\}\)

b.\(\left\{{}\begin{matrix}2x-4y=0\\3x+2y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-4y=0\\6x+4y=16\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}8x=16\\2x-4y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\4-4y=0\end{matrix}\right.\)

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

vậy hệ pt có ndn \(\left\{2;1\right\}\)

d.\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)

đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b\) ta có hệ pt:

\(\left\{{}\begin{matrix}a+b=\dfrac{1}{12}\\8a+15b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8a+8b=\dfrac{2}{3}\\8a+15b=1\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}7b=\dfrac{1}{3}\\8a+15b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{1}{21}\\8a+15\times\dfrac{1}{21}=1\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}b=\dfrac{1}{21}\\8a+\dfrac{5}{7}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{1}{21}\\8a=\dfrac{2}{7}\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}b=\dfrac{1}{21}\\a=\dfrac{1}{28}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{21}\\\dfrac{1}{x}=\dfrac{1}{28}\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}y=21\\x=28\end{matrix}\right.\)

vậy hệ pt có ndn\(\left\{28;21\right\}\)

ĐKXĐ: \(\left\{{}\begin{matrix}2x+y\ge1\\x+2y\ge2\\x+4y\ge0\end{matrix}\right.\)

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

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

Thế vào pt 2 => x;y

Đặt \(\left\{{}\begin{matrix}\sqrt{2x+y-1}=a\ge0\\\sqrt{x+2y-2}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^2-b^2=x-y+1\)

Phương trình thứ nhất trở thành:

\(a-b+a^2-b^2=0\)

\(\Leftrightarrow\left(a-b\right)\left(1+a+b\right)=0\Leftrightarrow a=b\)

\(\Leftrightarrow\sqrt{2x+y-1}=\sqrt{x+2y-2}\Rightarrow y=x+1\)

Thay xuống pt dưới:

\(4x^2-\left(x+1\right)^2+x+4-\sqrt{3x+1}-\sqrt{5x+4}=0\)

\(\Leftrightarrow3x^2-x+3-\sqrt{3x+1}-\sqrt{5x+4}=0\)

\(\Leftrightarrow3x^2-3x+x+1-\sqrt{3x+1}+x+2-\sqrt{5x+4}=0\)

\(\Leftrightarrow3x\left(x-1\right)+\frac{\left(x+1\right)^2-\left(3x+1\right)}{x+1+\sqrt{3x+1}}+\frac{\left(x+2\right)^2-\left(5x+4\right)}{x+2+\sqrt{5x+4}}=0\)

\(\Leftrightarrow3x\left(x-1\right)+\frac{x\left(x-1\right)}{x+1+\sqrt{3x+1}}+\frac{x\left(x-1\right)}{x+2+\sqrt{5x+4}}=0\)

\(\Leftrightarrow x\left(x-1\right)\left(3+\frac{1}{x+1+\sqrt{3x+1}}+\frac{1}{x+2+\sqrt{5x+4}}\right)=0\)

a: \(\left\{{}\begin{matrix}2x-2y+z=3\\2x+y-2z=-3\\3x-4y-z=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x-4y+2z=6\\8x+4y-8z=-3\\3x-4y-z=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}12x-6z=3\\11x-9z=1\\3x-4y-z=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\z=\dfrac{1}{2}\\4y=3x-z-4=\dfrac{3}{2}-\dfrac{1}{2}-4=1-4=-3\end{matrix}\right.\)

=>x=1/2;z=1/2;y=-3/4