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21 tháng 4 2023

\(\left\{{}\begin{matrix}x^3-y^3=35\\2x^2+3y^2=4x-9y\left(1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y^3-x^3=-35\\3y^2+9y+2x^2-4x=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y^3-x^3=-35\\9y^2+27y+6x^2-12x=0\end{matrix}\right.\)

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

\(\Rightarrow\left(y^3+9y^2+27y+27\right)-\left(x^3-6x^2+12x-8\right)=0\)

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

\(\Rightarrow\left(y-x+5\right)\left[\left(y+3\right)^2+\left(y+3\right)\left(x-2\right)+\left(x-2\right)^2\right]=0\)

*Với \(x=y+5\). Thay vào (1) ta được:

\(2\left(y+5\right)^2+3y^2=4\left(y+5\right)-9y\)

\(\Leftrightarrow2y^2+20y+50+3y^2=4y+20-9y\)

\(\Leftrightarrow5y^2+25y+30=0\Leftrightarrow y^2+5y+6=0\)

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

*\(y=-2\Rightarrow x=3\) ; \(y=-3\Rightarrow x=2\).

*Với \(\left(y+3\right)^2+\left(y+3\right)\left(x-2\right)+\left(x-2\right)^2=0\). Ta có:

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

\(=\left[\left(y+3\right)+\dfrac{\left(x-2\right)}{2}\right]^2+\dfrac{3}{4}\left(x-2\right)^2\ge0\)

Dấu "=" xảy ra khi \(x=2;y=-3\)

Vậy \(x=2;y=-3\)

Thử lại ta có nghiệm (x;y) của hệ đã cho là \(\left(3;-2\right),\left(2;-3\right)\)

 

25 tháng 3 2023

\(\left\{{}\begin{matrix}x^2+y^2+3=4x\\x^3+12x+y^3=6x^2+9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2-4x+4\right)+y^2=1\\\left(x^3-6x^2+12x-8\right)+y^3=1\end{matrix}\right.\)

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

Đặt \(a=x-2;b=y\). Hệ phương trình trở thành:

\(\left\{{}\begin{matrix}a^2+b^2=1\\a^3+b^3=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2ab=\left(a+b\right)^2-1\\\left(a+b\right)\left(a^2+b^2-ab\right)=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2ab=\left(a+b\right)^2-1\\\left(a+b\right)\left(1-\dfrac{\left(a+b\right)^2-1}{2}\right)=1\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left(a+b\right)\left[3-\left(a+b\right)^2\right]=2\)

\(\Leftrightarrow3\left(a+b\right)-\left(a+b\right)^3=2\)

\(\Leftrightarrow\left(a+b\right)^3-3\left(a+b\right)+2=0\)

\(\Leftrightarrow\left(a+b\right)^3-\left(a+b\right)^2+\left(a+b\right)^2-\left(a+b\right)-2\left(a+b-1\right)=0\)

\(\Leftrightarrow\left(a+b\right)^2\left(a+b-1\right)+\left(a+b\right)\left(a+b-1\right)-2\left(a+b-1\right)=0\)

\(\Leftrightarrow\left(a+b-1\right)\left[\left(a+b\right)^2+\left(a+b\right)-2\right]=0\)

\(\Leftrightarrow\left(a+b-1\right)^2\left(a+b+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b=1\\a+b=-2\end{matrix}\right.\)

Với \(\left\{{}\begin{matrix}a+b=1\\a^2+b^2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=1\\\left(a+b\right)^2-2ab=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=1\\ab=0\end{matrix}\right.\)

\(\Rightarrow\left(a;b\right)=\left(0;1\right),\left(1;0\right)\)

\(\Rightarrow\left(x-2;y\right)=\left(0;1\right),\left(1;0\right)\)

\(\Rightarrow\left(x;y\right)=\left(2;1\right),\left(3;0\right)\)

Với \(\left\{{}\begin{matrix}a+b=-2\\a^2+b^2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=-2\\\left(a+b\right)^2-2ab=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=-2=S\\ab=\dfrac{3}{2}=P\end{matrix}\right.\left(2\right)\)

Ta có: \(S^2-4P=\left(-2\right)^2-4.\dfrac{3}{2}=-2< 0\)

\(\Rightarrow\)Không tồn tại số a,b nào thỏa hệ phương trình (2).

Vậy nghiệm (x;y) của hpt đã cho là \(\left(2;1\right),\left(3;0\right)\)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

NV
8 tháng 3 2021

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

\(\Leftrightarrow\left(2x-y\right)^3=8\)

\(\Leftrightarrow2x-y=2\)

\(\Rightarrow y=2x-2\)

Thế xuống pt dưới:

\(\left(x^2-2x-2\right)\left(-3x^2+6x-9\right)=14\)

Đặt \(x^2-2x=t\)

\(\Rightarrow\left(t-2\right)\left(-3t-9\right)=14\)

\(\Leftrightarrow...\)

NV
26 tháng 3 2021

Từ pt dưới:

\(x^2+9y^2=6xy\Leftrightarrow x^2-6xy+9y^2=0\)

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

Thế lên pt trên: \(2.\left(3y\right)^2+y^2=19\)

\(\Leftrightarrow19y^2=19\Leftrightarrow y^2=1\Rightarrow\left[{}\begin{matrix}y=1\Rightarrow x=3\\y=-1\Rightarrow x=-3\end{matrix}\right.\)

7 tháng 11 2021

\(1,\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\2y+10+y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{16}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}3x=1-2y\\1-2y+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\3y+6+2y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

7 tháng 11 2021

\(1,\Leftrightarrow\left\{{}\begin{matrix}x=2y+4\\-4y-8+5y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\cdot5+4=14\\y=5\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}5x-30+6x=3\\y=10-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=4-2y\\6y-12+y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{7}\\y=\dfrac{19}{7}\end{matrix}\right.\)

a) Ta có: \(\left\{{}\begin{matrix}3x-2\left|y\right|=9\\2x+3\left|y\right|=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x-4\left|y\right|=18\\6x+9\left|y\right|=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-13\left|y\right|=15\\3x-2\left|y\right|=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left|y\right|=\dfrac{-15}{13}\\3x-2\left|y\right|=9\end{matrix}\right.\Leftrightarrow\)Phương trình vô nghiệmVậy: \(S=\varnothing\)

28 tháng 2 2021

$\begin{cases}3x-2|y|=9\\2x+3|y|=1\\\end{cases}$

`<=>` $\begin{cases}6x-4|y|=18\\6x+9|y|=3\\\end{cases}$

`<=>` $\begin{cases}13|y|=-15(loại)\\|3x|-2|y|=9\\\end{cases}$

Vậy HPT vô nghiệm