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a) \(\left\{{}\begin{matrix}2x+3y=5\\4x-5y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=10\\4x-5y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=5\\11y=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+3\cdot\dfrac{9}{11}=5\\y=\dfrac{9}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{27}{11}=5\\y=\dfrac{9}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=\dfrac{28}{11}\\y=\dfrac{9}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{14}{11}\\y=\dfrac{9}{11}\end{matrix}\right.\)
Vậy: \(x=\dfrac{14}{11};y=\dfrac{9}{11}\)
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PT (1) <=> x = 3y + 3. Thay x = 3y + 3 vào PT (2) ta có: \(\left(3y+3\right)^2+y^2-2\left(3y+3\right)-2y-9=0\Leftrightarrow10y^2+10y-6=0\Leftrightarrow y=\frac{-5+\sqrt{85}}{10}\)hoặc \(y=\frac{-5-\sqrt{85}}{10}\)
- Nếu \(y=\frac{-5+\sqrt{85}}{10}\) \(\Rightarrow x=3y+3=\frac{15+3\sqrt{85}}{10}\)
- Nếu \(y=\frac{-5-\sqrt{85}}{10}\Rightarrow x=3y+3=\frac{15-3\sqrt{85}}{10}\)
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ĐK: \(x\ge\frac{1}{2}\)
\(\hept{\begin{cases}x\left(2x-2y-1\right)=3\left(y+2\right)\left(1\right)\\3y+6\sqrt{2x-1}=y^2-x+23\left(2\right)\end{cases}}\)
pt (1) <=> \(2x^2-2xy-x-3y-6=0\)
<=> \(2x^2-x\left(2y+1\right)-\left(3y+6\right)=0\)
có \(\Delta=\left(2y+1\right)^2+4\left(3y+6\right)=4y^2+28y+49=\left(2y+7\right)^2\)
=> (1) có hai nghiệm: \(\orbr{\begin{cases}x_1=\frac{\left(2y+1\right)-\left(2y+7\right)}{4}=-\frac{3}{2}\left(loai\right)\\x_2=\frac{\left(2y+1\right)+\left(2y+7\right)}{4}=y+2\end{cases}}\)
+) Với \(x=y+2\) thế vào (2) ta có:
\(3y+6\sqrt{2\left(y+2\right)-1}=y^2-\left(y+2\right)+23\)
<=> \(6\sqrt{2y+3}=y^2-4y+21\)
ĐK: \(y\ge-\frac{3}{2}\)
\(6\sqrt{2y+3}=y^2-4y+21\)
<=> \(6\sqrt{2y+3}-2y-12=y^2-6y+9\)
<=> \(\frac{2\left(9\left(2y+3\right)-\left(y+6\right)^2\right)}{3\sqrt{2y+3}+y+6}-\left(y-3\right)^2=0\)
<=> \(\frac{-2\left(y-3\right)^2}{3\sqrt{2y+3}+y+6}-\left(y-3\right)^2=0\)
<=> \(\left(y-3\right)^2\left(\frac{-2}{3\sqrt{2y+3}+y+6}-1\right)=0\)
<=> y - 3 = 0
<=> y = 3 thỏa mãn
khi đó x = y + 2 = 3 + 2 = 5 thỏa mãn
Kết luận:...
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\(\hept{\begin{cases}x^2=y^3-3y^2+2y\\y^2=x^3-3x^2+2x\end{cases}\Leftrightarrow\hept{\begin{cases}x^2=y^3-3y^2+2y\\x^2-y^2=y^3-x^3-3y^2+3x^2+2y-2x\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}x^2=y^3-3y^2+2y\\2\left(y-x\right)\left(y+x\right)=\left(y-x\right)\left(y^2+xy+x^2\right)+2\left(y-x\right)\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x^2=y^3-3y^2+2y\\\left(y-x\right)\left[xy+\left(x-1\right)^2+\left(y-1\right)^2\right]=0\end{cases}}\)
Theo Cauchy-schwarz có: \(\frac{\left(x-1\right)^2}{1}+\frac{\left(1-y\right)^2}{1}\ge\frac{\left(x-y\right)^2}{2}\)Dấu "=" xảy ra <=> x+y=2 (1)
\(\Rightarrow xy+\left(x-1\right)^2+\left(y-1\right)^2\ge xy+\frac{x^2-2xy+y^2}{2}=x^2+y^2\ge0\) Dấu bằng xảy ra <=> x=y=0 (2)
Từ (1) và (2) => \(xy+\left(x-1\right)^2+\left(y-1\right)^2>0\)
\(\Rightarrow x=y\)
=> Hệ phương trình \(\Leftrightarrow\hept{\begin{cases}x^2=y^3-3y^2+2y\\y^2=y^3-3y^2+2y\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x^2=y^3-3y^2+2y\\0=y^3-4y^2+2y\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x^2=y^3-3y^2+2y\\0=y^3-4y^2+2y\end{cases}}\)
Tự làm nốt nhé
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Câu 1:
Thay \(x=\sqrt{2};y=2\sqrt{2}\) vào đồ thị hàm số \(y=ax^2\) ta có:
\(\left(\sqrt{2}\right)^2.a=2\sqrt{2}\Leftrightarrow2a=2\sqrt{2}\Leftrightarrow a=\sqrt{2}\)
Vậy \(a=\sqrt{2}\) thì đồ thị hàm số \(y=ax^2\) đi qua điểm \(\left(\sqrt{2};2\sqrt{2}\right)\)
b) \(\left\{{}\begin{matrix}2x+3y=-1\\x-2y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2.\left(3+2y\right)+3y=-1\\x=3+2y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7y=-7\\x=3+2y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=3+2.\left(-1\right)=1\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm duy nhất là \(\left(1;-1\right)\)
Đặt 1/(x+2y)=a; y=b
=>a+b=-2 và 2a-3b=1
=>a=-1; b=-1
=>y=-1; x+2y=-1
=>y=-1; x=-1-2y=-1-2*(-1)=-1+2=1