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

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

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

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 \(\left(2;1\right),\left(3;0\right)\)

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

Đặt \(x-2=a\)

Khi đó hệ đã cho trở thành \(\left\{{}\begin{matrix}a^2+y^2=1\\a^3+y^3=1\end{matrix}\right.\)

Đến đây đưa về hệ đối xứng loại 1 rồi đó, đặt tổng và tích làm là ra

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

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

\(\Leftrightarrow2x-y=2\)

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

Thế xuống pt dưới:

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

Đặt \(x^2-2x=t\)

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

\(\Leftrightarrow...\)

Tham Khảo:

https://olm.vn/hoi-dap/detail/264041645597.html

Sai thì hong bít j đâu ;-;

\(7x^3+y^3+3xy\left(x-y\right)-12x^2+6x=1\)

\(\Leftrightarrow\left(8x^3-12x^2+6x-1\right)-\left(x^3-3x^2y+3xy^2-y^3\right)=0\)

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

\(\Leftrightarrow2x-1=x-y\)

\(\Leftrightarrow y=1-x\)

Thế xuống dưới:

\(\sqrt[3]{3x+2}+\sqrt{x+2}=4\)

\(\Leftrightarrow\sqrt[3]{3x+2}-2+\sqrt{x+2}-2=0\)

\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{\sqrt[3]{\left(3x+2\right)^2}+2\sqrt[3]{3x+2}+4}+\dfrac{1}{\sqrt{x+2}+2}\right)=0\)

\(\left\{{}\begin{matrix}7x^3+y^3+3xy\left(x-y\right)-12x^2+6x=1\left(1\right)\\\sqrt[3]{4x+y+1}+\sqrt{3x+2y}=4\left(2\right)\end{matrix}\right.\)

(1)⇔ y³ - 3y²x + 3x²y + 7x³ = 1 - 6x + 12x² <=> y³ - 3y²x + 3x²y - x³ = 1 - 6x + 12x² - 8x³ <=> (y - x)³ = (1 - 2x)³ <=> y - x = 1 - 2x <=> y = 1 - x

Thế vào (2)\(\Leftrightarrow\sqrt[3]{4x+1-x+1}+\sqrt{3x+2\left(1-x\right)}=4\Leftrightarrow\sqrt[3]{3x+2}+\sqrt{x+2}=4\)

Đặt a=\(\sqrt[3]{3x+2}\Leftrightarrow a^3=3x+2\)

b=\(\sqrt{x+2}\left(b\ge0\right)\Leftrightarrow b^2=x+2\Leftrightarrow3b^2=3x+6\)

Vậy 3b²-a³=4

Vậy ta sẽ có hệ phương trình \(\left\{{}\begin{matrix}3b^2-a^3=4\\a+b=4\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}3b^2-a^3=4\left(3\right)\\b=4-a\end{matrix}\right.\)

(3)\(\Leftrightarrow3\left(4-a\right)^2-a^3=4\Leftrightarrow a^3-3a^2+24a-44=0\Leftrightarrow\left(a-2\right)\left(a^2-a+22\right)=0\)(*)

Ta có a²-a+22>0

Vậy (*)\(\Leftrightarrow a-2=0\Leftrightarrow a=2\Leftrightarrow b=2\)

Vậy \(\left\{{}\begin{matrix}\sqrt[3]{3x+2}=2\\\sqrt{x+2}=2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}3x+2=8\\x+2=4\end{matrix}\right.\)\(\Leftrightarrow x=2\Leftrightarrow y=-1\)

Vậy (x;y)=(2;-1)

các bn ơi giúp mình với

a) ĐK:x\(\ge\dfrac{3}{4}\)

\(3\left(x^2-1\right)+4x=4x\sqrt{4x-3}\Leftrightarrow3x^2-3+4x=4x\sqrt{4x-3}\Leftrightarrow4x-3-4x\sqrt{4x-3}+4x^2-x^2=0\Leftrightarrow\left(\sqrt{4x-3}-2x\right)^2-x^2=0\Leftrightarrow\left(\sqrt{4x-3}-2x-x\right)\left(\sqrt{4x-3}-2x+x\right)^2=0\Leftrightarrow\left(\sqrt{4x-3}-3x\right)\left(\sqrt{4x-3}-x\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}\sqrt{4x-3}-3x=0\\\sqrt{4x-3}-x=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}\sqrt{4x-3}=3x\left(x\ge0\right)\\\sqrt{4x-3}=x\left(x\ge0\right)\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}4x-3=9x^2\\4x-3=x^2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}9x^2-4x+3=0\\x^2-4x+3=0\end{matrix}\right.\)(*)

Vì 9x²-4x+3>0 nên 9x²-4x+3=0(loại)

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

Vậy S={1;3}

b)

(1)⇔ y³ - 3y²x + 3x²y + 7x³ = 1 - 6x + 12x² <=> y³ - 3y²x + 3x²y - x³ = 1 - 6x + 12x² - 8x³ <=> (y - x)³ = (1 - 2x)³ <=> y - x = 1 - 2x <=> y = 1 - x

Thế vào (2)\(\Leftrightarrow\sqrt[3]{4x+1-x+1}+\sqrt{3x+2\left(1-x\right)}=4\Leftrightarrow\sqrt[3]{3x+2}+\sqrt{x+2}=4\)

Đặt a=\(\sqrt[3]{3x+2}\Leftrightarrow a^3=3x+2\)

b=\(\sqrt{x+2}\left(b\ge0\right)\Leftrightarrow b^2=x+2\Leftrightarrow3b^2=3x+6\)

Vậy 3b²-a³=4

Vậy ta sẽ có hệ phương trình \(\left\{{}\begin{matrix}3b^2-a^3=4\\a+b=4\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}3b^2-a^3=4\left(3\right)\\b=4-a\end{matrix}\right.\)

(3)\(\Leftrightarrow3\left(4-a\right)^2-a^3=4\Leftrightarrow a^3-3a^2+24a-44=0\Leftrightarrow\left(a-2\right)\left(a^2-a+22\right)=0\)(*)

Ta có a²-a+22>0

Vậy (*)\(\Leftrightarrow a-2=0\Leftrightarrow a=2\Leftrightarrow b=2\)

Vậy \(\left\{{}\begin{matrix}\sqrt[3]{3x+2}=2\\\sqrt{x+2}=2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}3x+2=8\\x+2=4\end{matrix}\right.\)\(\Leftrightarrow x=2\Leftrightarrow y=-1\)

Vậy (x;y)=(2;-1)

Câu hỏi của Nguyễn Thu Trà - Toán lớp 9 | Học trực tuyến