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:

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x^3-x^2y-7\left(x-y\right)=x^2+y^2+2xy+4\\3x^2+y^2-8\left(x-y\right)+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2-7\right)\left(x-y\right)-x^2-2xy=y^2+4\\3x^2-8\left(x-y\right)=-y^2-4\end{matrix}\right.\)

Cộng vế:

\(\left(x^2-7\right)\left(x-y\right)-8\left(x-y\right)+2x^2-2xy=0\)

\(\Leftrightarrow\left(x^2-15\right)\left(x-y\right)+2x\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+2x-15\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x^2+2x-15=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

a.

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

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

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

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

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

Lần lượt thế vào \(y^2=5x^2+4\)...

b. Đề bài bất hợp lý, \(4x^2+y^4\) cần là \(4x^4+y^4\)

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\(\left\{28;21\right\}\)

a)\(\left\{{}\begin{matrix}8x+2y=4\\8x+3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\4x+1=2\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}y=1\\x=\frac{1}{4}\end{matrix}\right.\)b)

\(\left\{{}\begin{matrix}12x-8y=44\\12x-15y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y=35\\4x-5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\4x-5.5=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=7\end{matrix}\right.\)c)\(\left\{{}\begin{matrix}9x=-18\\4x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\4.\left(-2\right)+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=7\end{matrix}\right.\)

các câu khác tương tự

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

thay (3) vào (2)\(\Rightarrow3x-4\left(x-3\right)=2\)

\(\Leftrightarrow3x-4x+12=2\)

\(\Leftrightarrow-x=-10\Leftrightarrow x=10\)

thay x=10 vào (3)\(\Rightarrow y=10-3=7\)

Nghiệm của hệ \(\left\{10;7\right\}\)

b)\(\left\{{}\begin{matrix}7x-3y=5\left(1\right)\\4x+y=2\left(2\right)\Rightarrow y=2-4x\left(3\right)\end{matrix}\right.\)

thay (3) vào (1)\(\Rightarrow7x-3\left(2-4x\right)=5\)

\(\Leftrightarrow7x-6+12x=5\)

\(\Leftrightarrow19x=11\Leftrightarrow x=\dfrac{11}{19}\)

thay \(x=\dfrac{11}{19}vào\left(3\right)\)\(\Rightarrow y=2-4\dfrac{11}{19}=-\dfrac{6}{19}\)

nghiệm của hệ \(\left\{\dfrac{11}{19};\dfrac{-6}{19}\right\}\)

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

thay (3) vào (2)\(\Rightarrow5\left(-2-3y\right)-4y=1\)

\(\Leftrightarrow-10-15y-4y=1\)

\(\Leftrightarrow-19y=11\Leftrightarrow y=\dfrac{-11}{19}\)

thay \(y=\dfrac{-11}{19}vào\left(3\right)\Rightarrow x=-2-3\left(\dfrac{-11}{19}\right)=\dfrac{-5}{19}\)nghiệm của hệ \(\left\{\dfrac{-5}{9};\dfrac{-11}{19}\right\}\)

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

thay (3) vào (2)\(\Rightarrow5\left(-2-3y\right)-4y=1\)

\(\Leftrightarrow-10-15y-4y=1\)

\(\Leftrightarrow-19y=11\Leftrightarrow y=\dfrac{-11}{19}\)

thay \(y=\dfrac{-11}{19}vào\left(3\right)\Rightarrow x=-2-3\left(\dfrac{-11}{19}\right)=\dfrac{-5}{19}\)

nghiệm của hệ\(\left\{\dfrac{-5}{19};\dfrac{-11}{19}\right\}\)

CHÚC BẠN HỌC TỐT !

-có người nhờ t làm

\(\left\{{}\begin{matrix}x-y=3\\3x-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-3y=9\left(1\right)\\3x-4y=2\left(2\right)\end{matrix}\right.\) lấy (1)-(2) tìm được x;sau đó dễ dàng có y
\(\left\{{}\begin{matrix}7x-3y=5\\4x+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}28x-12y=20\left(1\right)\\28x+7y=14\left(2\right)\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x+3y=-2\\5x-4y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+15y=-10\left(1\right)\\5x-4y=11\left(2\right)\end{matrix}\right.\)

Gt: Nhân sao cho cả 2 pt xuất hiện chung 1 thừa số,trừ đi chỉ còn 1 x or y