1)

HPT \(\Leftrightarrow\left\{{}\begin{matrix}15x-6y=-27\\8x+6y=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2y=5x+9\\23x=-23\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(-1;2\right)\)

2)

HPT \(\Leftrightarrow\left\{{}\begin{matrix}2x+y=4\\2x+4y=10\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-3y=-6\\x=5-2y\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=1\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(1;2\right)\)

3)

HPT \(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=14\\3x+6y=12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\2y=4-x\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(2;1\right)\)

4) 

HPT \(\Leftrightarrow\left\{{}\begin{matrix}5x+6y=17\\54x-6y=42\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}59x=59\\y=9x-7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(1;2\right)\)

Ấy, nhìn không kỹ nên sai sót kỹ thuật rồi, bước đặt nhân tử chung bị nhầm.

Làm lại cho chính xác hơn:

Hệ đã cho tương đương \(\left\{{}\begin{matrix}\left(x^2-2\right)^2-4+\left(y-3\right)^2=0\left(1\right)\\y=\dfrac{22-x^2}{x^2+2}\end{matrix}\right.\)

Đặt \(x^2-2=t\Rightarrow x^2=t+2\Rightarrow y=\dfrac{20-t}{t+4}\Rightarrow y-3=\dfrac{4\left(2-t\right)}{t+4}\left(2\right)\)

Thay (2) vào (1):

\(t^2-4+\dfrac{16\left(2-t\right)^2}{\left(t+4\right)^2}=0\Leftrightarrow\left(t-2\right)\left(t+2\right)+\dfrac{16\left(t-2\right)^2}{\left(t+4\right)^2}=0\)

\(\Leftrightarrow\left(t-2\right)\left(t+2+\dfrac{16\left(t-2\right)}{\left(t+4\right)^2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t-2=0\\t+2+\dfrac{16\left(t-2\right)}{\left(t+4\right)^2}=0\end{matrix}\right.\)

TH1: \(t-2=0\Rightarrow t=2\Rightarrow x^2=4\) \(\Rightarrow\left[{}\begin{matrix}x=-2;y=3\\x=2;y=3\end{matrix}\right.\)

TH2: \(t+2+\dfrac{16\left(t-2\right)}{\left(t+4\right)^2}=0\Leftrightarrow\left(t+2\right)\left(t^2+8t+16\right)+16t-32=0\)

\(\Leftrightarrow t^3+8t^2+16t+2t^2+16t+32+16t-32=0\)

\(\Leftrightarrow t^3+10t^2+48t=0\)

\(\Leftrightarrow t\left(t^2+10t+48\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}t=0\\t^2+10t+48=0\left(vn\right)\end{matrix}\right.\) \(\Rightarrow x^2=2\) \(\Rightarrow\left[{}\begin{matrix}x=-\sqrt{2};y=5\\x=\sqrt{2};y=5\end{matrix}\right.\)

Vậy hệ đã cho có 4 cặp nghiệm:

\(\left(x;y\right)=\left(-2;3\right);\left(2;3\right);\left(-\sqrt{2};5\right);\left(\sqrt{2};5\right)\)

\(\left\{{}\begin{matrix}\left(x^2-2\right)^2-4+\left(y-3\right)^2=0\\\left(x^2+2\right).y=22-x^2\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}\left(x^2-2\right)^2-4+\left(y-3\right)^2=0\\y=\dfrac{22-x^2}{x^2+2}\end{matrix}\right.\)

Đặt \(x^2-2=t\ge-2\)

\(\Rightarrow x^2=t+2\Rightarrow y=\dfrac{20-t}{t+4}\Rightarrow y-3=\dfrac{8-4t}{t+4}=\dfrac{4\left(2-t\right)}{t+4}\)

Thay vào pt trên ta được:

\(t^2-4+\dfrac{16\left(2-t\right)^2}{\left(t+4\right)^2}=0\Leftrightarrow\left(t-2\right)\left(t+2\right)+\dfrac{16\left(t-2\right)^2}{\left(t+4\right)^2}=0\)

\(\Leftrightarrow\left(t-2\right)\left(t+2+\dfrac{16}{\left(t+4\right)^2}\right)=0\)

\(\Leftrightarrow t-2=0\) (do \(t+2+\dfrac{16}{\left(t+4\right)^2}>0\) \(\forall t\ge-2\) )

\(\Rightarrow t=2\Rightarrow x^2-2=2\Rightarrow x^2=4\)

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

Vậy hệ đã cho có 2 cặp nghiệm:

\(\left(x;y\right)=\left(-2;3\right);\left(2;3\right)\)

các bn ơi giúp mình với

a.

\(2x^3-x^2y+x^2+y^2-2xy-y=0\)

\(\Leftrightarrow x^2\left(2x-y+1\right)-y\left(2x-y+1\right)=0\)

\(\Leftrightarrow\left(x^2-y\right)\left(2x-y+1\right)=0\)

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

Thế vào pt đầu:

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

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

\(\Leftrightarrow...\)

b.

\(x^2-2xy+x=-y\)

Thế vào \(y^2\) ở pt dưới:

\(x^2\left(x^2-4y+3\right)+\left(x^2-2xy+x\right)^2=0\)

\(\Leftrightarrow x^2\left(x^2-4y+3\right)+x^2\left(x-2y+1\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\Rightarrow y=0\\x^2-4y+3+\left(x-2y+1\right)^2=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2x^2-4xy+2x+4y^2-8y+4=0\)

\(\Leftrightarrow2\left(x^2-2xy+x\right)+4y^2-8y+4=0\)

\(\Leftrightarrow-2y+4y^2-8y+4=0\)

\(\Leftrightarrow...\)

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

Pt đầu chắc là sai đề (chắc chắn), bạn kiểm tra lại

Với pt sau:

Nhận thấy một ẩn bằng 0 thì 2 ẩn còn lại cũng bằng 0, do đó \(\left(x;y;z\right)=\left(0;0;0\right)\) là 1 nghiệm

Với \(x;y;z\ne0\)

Từ pt đầu ta suy ra \(y>0\) , từ đó suy ra \(z>0\) từ pt 2 và hiển nhiên \(x>0\) từ pt 3

Do đó:

\(\left\{{}\begin{matrix}y=\dfrac{2x^2}{x^2+1}\le\dfrac{2x^2}{2x}=x\\z=\dfrac{3y^3}{y^4+y^2+1}\le\dfrac{3y^3}{3\sqrt[3]{y^4.y^2.1}}=y\\x=\dfrac{4z^4}{z^6+z^4+z^2+1}\le\dfrac{4z^4}{4\sqrt[4]{z^6z^4z^2}}=z\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y\le x\\z\le y\\x\le z\end{matrix}\right.\) \(\Rightarrow x=y=z\)

Dấu "=" xảy ra khi và chỉ khi \(x=y=z=1\)

Vậy nghiệm của hệ là \(\left(x;y;z\right)=\left(0;0;0\right);\left(1;1;1\right)\)