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Ta có hệ \(\hept{\begin{cases}\left(4x^2+1\right)x+\left(y-3\right)\sqrt{5-2y}=0\left(1\right)\\4x^2+y^2+2\sqrt{3-4x}=7\left(2\right)\end{cases}}\)
ĐK \(\hept{\begin{cases}y\ge\frac{5}{2}\\x\le\frac{3}{4}\end{cases}}\)
Đặt \(\hept{\begin{cases}2x=a\\\sqrt{5-2y}=b\ge0\end{cases}\Rightarrow\hept{\begin{cases}4x^2=a^2\\5-2y=b^2\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}4x^2=a^2\\y-3=\frac{5-b^2}{2}-3=\frac{-1-b^2}{2}\end{cases}}\)
Thế vào (1) ta có \(\left(a^2+1\right)\frac{a}{2}+\frac{-1-b^2}{2}b=0\)
\(\Leftrightarrow\frac{a^3+a}{2}+\frac{-b^3-b}{2}=0\Leftrightarrow a^3-b^3+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+1\right)=0\)
\(\Leftrightarrow a=b\)vì \(a^2+ab+b^2+1>0\forall a,b\)
\(\Rightarrow2x=\sqrt{5-2y}\Rightarrow4x^2=5-2y\Rightarrow y=\frac{5-4x^2}{2}\)
Thế y vào (2) ta có \(4x^2+\left(\frac{5-4x^2}{2}\right)^2+2.\sqrt{3-4x}=7\)
\(\Leftrightarrow16x^2+\left(5-4x^2\right)^2+8\sqrt{3-4x}=28\)\(\Leftrightarrow16x^2+25-40x^2+16x^4+8\sqrt{3-4x}-28=0\)
\(\Leftrightarrow16x^4-24x^2+8\sqrt{3-4x}-3=0\)
\(\Leftrightarrow\left(16x^4-1\right)-\left(24x^2-6\right)+\left(8\sqrt{3-4x}-8\right)=0\)
\(\Leftrightarrow\left(4x^2-1\right)\left(4x^2+1\right)-6\left(4x^2-1\right)+\left(8\sqrt{3-4x}-8\right)=0\)
\(\Leftrightarrow\left(4x^2-1\right)\left(4x^2+1\right)-6\left(4x^2-1\right)+8.\frac{2-4x}{\sqrt{3-4x}+1}=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-1\right)\left(4x^2+1\right)-6\left(2x+1\right)\left(2x-1\right)-8.2.\frac{2x-1}{\sqrt{3-4x}+1}=0\)
\(\Leftrightarrow\left(2x-1\right)\left[\left(2x+1\right)\left(4x^2+1\right)-6\left(2x+1\right)-\frac{16.1}{\sqrt{3-4x}+1}\right]=0\)
\(\Leftrightarrow\left(2x-1\right)\left[\left(2x+1\right)\left(4x^2-5\right)-\frac{16}{\sqrt{3-4x}+1}\right]=0\)
\(\Leftrightarrow2x-1=0\)
Vì với \(y=\frac{5-4x^2}{2}\ge\frac{5}{2}\Rightarrow4x^2-5< 0\Rightarrow\left(2x+1\right)\left(4x^2-5\right)-\frac{16}{\sqrt{3-4x}+1}< 0\)
\(\Leftrightarrow x=\frac{1}{2}\Rightarrow y=\frac{5-4\left(\frac{1}{2}\right)^2}{2}=2\)
Vậy hệ có nghiệm \(\left(x;y\right)=\left(\frac{1}{2};2\right)\)
<=> xy+5x+3y+15=xy+8x+y+8 <=> 3x-2y=7 <=> 9x-6y=21 <=> x=3 <=> x=3
10xy+14x-15y-21=10xy+10x-12y-12 4x-3y=9 8x-6y=18 8.3-6y=18 y=1
x/y=2/3 nên x=2/3.y
\(\int^{x=\frac{2}{3}y}_{x+y=10}\Leftrightarrow\int^{^{x=\frac{2}{3}y}}_{^{\frac{5}{3}y=10}}\Leftrightarrow\int^{^{x=\frac{2}{3}y}}_{y=6}\Leftrightarrow\int^{x=4}_{y=6}\)
ĐKXĐ: x > y
Ta có hệ \(\hept{\begin{cases}\sqrt{x+y}+\sqrt{x-y}=4\\x^2+y^2=18\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x+y+2\sqrt{\left(x+y\right)\left(x-y\right)}+x-y=16\\x^2+y^2=18\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2\sqrt{x^2-y^2}=16-2x\\x^2+y^2=18\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x^2-y^2}=8-x\\x^2+y^2=18\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}8-x\ge0\\x^2-y^2=\left(8-x\right)^2\\x^2+y^2=18\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\le8\\x^2-y^2=64-16x+x^2\\x^2+y^2=18\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\le8\\-y^2=64-16x\\x^2+y^2=18\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\le8\\y^2=16x-64\\x^2+y^2-y^2=18-16x+64\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\le8\left(1\right)\\y^2=16x-64\left(2\right)\\x^2+16x-82=0\left(3\right)\end{cases}}\)
Giải (3) \(x^2+16x-82=0\)
\(\Leftrightarrow x^2+16x+64=146\)
\(\Leftrightarrow\left(x+8\right)^2=146\)
\(\Leftrightarrow x+8=\pm\sqrt{146}\)
\(\Leftrightarrow x=\pm\sqrt{146}-8\)(Thỏa mãn (1) )
Thay vào (2) tìm được y rồi so sánh ĐKXĐ => KL
@Fabulous Joker cảm ơn ông nhiều lắm
mai tôi phải nộp bài r
\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=4\\2\left(x+y\right)+4\left(x-y\right)=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=6\\2\left(x+y\right)+24=10\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=6\\x+y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-\dfrac{13}{2}\end{matrix}\right.\)