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\(\left\{{}\begin{matrix}7x^3+y^3+3xy\left(x-y\right)-12x^2+6x=1\left(1\right)\\\sqrt[3]{4x+y+1}+\sqrt{3x+2y}=4\left(2\right)\end{matrix}\right.\)
(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)
a: \(\Leftrightarrow\left\{{}\begin{matrix}8x-4y+12-3x+6y-9=48\\9x-12y+9+16x-8y-36=48\end{matrix}\right.\)
=>5x+2y=48-12+9=45 và 25x-20y=48+36-9=48+27=75
=>x=7; y=5
b: \(\Leftrightarrow\left\{{}\begin{matrix}6x+6y-2x+3y=8\\-5x+5y-3x-2y=5\end{matrix}\right.\)
=>4x+9y=8 và -8x+3y=5
=>x=-1/4; y=1
c: \(\Leftrightarrow\left\{{}\begin{matrix}-4x-2+1,5=3y-6-6x\\11,5-12+4x=2y-5+x\end{matrix}\right.\)
=>-4x-0,5=-6x+3y-6 và 4x-0,5=x+2y-5
=>2x-3y=-5,5 và 3x-2y=-4,5
=>x=-1/2; y=3/2
e: \(\Leftrightarrow\left\{{}\begin{matrix}x\cdot2\sqrt{3}-y\sqrt{5}=2\sqrt{3}\cdot\sqrt{2}-\sqrt{5}\cdot\sqrt{3}\\3x-y=3\sqrt{2}-\sqrt{3}\end{matrix}\right.\)
=>\(x=\sqrt{2};y=\sqrt{3}\)
\(\left\{{}\begin{matrix}4x+5y=3\\x-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4\left(5+3y\right)+5y=3\\x=5+3y\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}20+12y+5y=3\\x=5+3y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}20+17y=3\\x=5+3y\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}17y=-17\\x=5+3y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=2\end{matrix}\right.\)
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)
\(\left\{{}\begin{matrix}7x^3+y^3+3xy\left(x-y\right)-12x^2+6x=1\left(1\right)\\\sqrt[3]{4x+y+1}+\sqrt{3x+2y}=4\left(2\right)\end{matrix}\right.\)(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)
Em cảm ơn ạ ^^ !!