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Do \(x+y=3\Rightarrow x=3-y\le1\Rightarrow y\ge2\)
Ta có :
\(P=\dfrac{y^3-x^3+\left(x+y\right)^2y}{6y^2+x^2}=\dfrac{2y^3+2xy^2+x^2y-x^3}{6y^2+x^2}\)
\(P=\dfrac{2y^2\left(x+y\right)-x^2\left(x-y\right)}{6y^2+x^2}=\dfrac{6y^2-x^2\left(x-y\right)}{6y^2+x^2}\) (1)
do \(y\ge2\Rightarrow-y\le-2\) mà \(x\le1\) nên ta có
\(x-y\le-1\)\(\Rightarrow-\left(x-y\right)\ge1\Rightarrow-x^2\left(x-y\right)\ge x^2\)
\(\Rightarrow\left(1\right)\ge\dfrac{6y^2+x^2}{6y^2+x^2}=1\)
do \(P\ge1\) nên \(MinP=1\) khi \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
â, đánh giá về trái ta có
\(\sqrt{x^2-4x+5}=\sqrt{\left(x-2\right)^2+1}>=1\)
\(\sqrt{9y^2-6y+1}>=0\)
do đó dấu bằng xảy ra khi x=2 va y=1/3
phần b làm tương tự
b, VT <=2-1=1
a/ Bạn nhầm đề
b/ \(\Leftrightarrow\left(\frac{3}{5}\right)^x+\left(\frac{4}{5}\right)^x=1\)
Với \(x=0;1\) không thỏa mãn
Nếu \(x=2\) ta thấy thỏa mãn
Nếu \(x>2\) do \(\left\{{}\begin{matrix}\frac{3}{5}< 1\\\frac{4}{5}< 1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(\frac{3}{5}\right)^x< \left(\frac{3}{5}\right)^2\\\left(\frac{4}{5}\right)^x< \left(\frac{4}{5}\right)^2\end{matrix}\right.\)
\(\Rightarrow\left(\frac{3}{5}\right)^x+\left(\frac{4}{5}\right)^x< \left(\frac{3}{5}\right)^2+\left(\frac{4}{5}\right)^2=1=VP\)
\(\Rightarrow\) Phương trình vô nghiệm
Vậy pt có nghiệm duy nhất \(x=2\)
c/ \(4x^2+4x+1+y^2-6y+9=74\)
\(\Leftrightarrow\left(2x+1\right)^2+\left(y-3\right)^2=74=7^2+5^2\)
\(\Rightarrow\left\{{}\begin{matrix}\left(2x+1\right)^2=7^2\\\left(y-3\right)^2=5^2\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}\left(2x+1\right)^2=5^2\\\left(y-3\right)^2=7^2\end{matrix}\right.\)
7. \(S=9y^2-12\left(x+4\right)y+\left(5x^2+24x+2016\right)\)
\(=9y^2-12\left(x+4\right)y+4\left(x+4\right)^2+\left(x^2+8x+16\right)+1936\)
\(=\left[3y-2\left(x+4\right)\right]^2+\left(x-4\right)^2+1936\ge1936\)
Vậy \(S_{min}=1936\) \(\Leftrightarrow\) \(\hept{\begin{cases}3y-2\left(x+4\right)=0\\x-4=0\end{cases}}\) \(\Leftrightarrow\) \(\hept{\begin{cases}x=4\\y=\frac{16}{3}\end{cases}}\)
8. \(x^2-5x+14-4\sqrt{x+1}=0\) (ĐK: x > = -1).
\(\Leftrightarrow\) \(\left(x+1\right)-4\sqrt{x+1}+4+\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow\) \(\left(\sqrt{x+1}-2\right)^2+\left(x-3\right)^2=0\)
Với mọi x thực ta luôn có: \(\left(\sqrt{x+1}-2\right)^2\ge0\) và \(\left(x-3\right)^2\ge0\)
Suy ra \(\left(\sqrt{x+1}-2\right)^2+\left(x-3\right)^2\ge0\)
Đẳng thức xảy ra \(\Leftrightarrow\) \(\hept{\begin{cases}\left(\sqrt{x+1}-2\right)^2=0\\\left(x-3\right)^2=0\end{cases}}\) \(\Leftrightarrow\) x = 3 (Nhận)
7. \(S=9y^2-12\left(x+4\right)y+\left(5x^2+24x+2016\right)\)
\(=9y^2-12\left(x+4\right)y+4\left(x+4\right)^2+\left(x^2+8x+16\right)+1936\)
\(=\left[3y-2\left(x+4\right)\right]^2+\left(x-4\right)^2+1936\ge1936\)
Vậy \(S_{min}=1936\) \(\Leftrightarrow\) \(\hept{\begin{cases}3y-2\left(x+4\right)=0\\x-4=0\end{cases}}\) \(\Leftrightarrow\) \(\hept{\begin{cases}x=4\\y=\frac{16}{3}\end{cases}}\)
\(x-3\sqrt{x}+2=x-2\sqrt{x}-\sqrt{x}+2=\sqrt{x}\left(\sqrt{x}-2\right)-\left(\sqrt{x}-2\right)=\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\)
\(2x-\sqrt{x}-3=2x+2\sqrt{x}-3\sqrt{x}-3=2\sqrt{x}\left(\sqrt{x}+1\right)-3\left(\sqrt{x}+1\right)=\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)\)
\(-6\sqrt{x}+5x-11=5x+5\sqrt{x}-11\sqrt{x}-11=5\sqrt{x}\left(\sqrt{x}+1\right)-11\left(\sqrt{x}+1\right)=\left(\sqrt{x}+1\right)\left(5\sqrt{x}-11\right)\)
\(6y^2-5y\sqrt{x}-x=\left(y^2-x\right)+\left(5y^2-5y\sqrt{x}\right)=\left(y-\sqrt{x}\right)\left(y+\sqrt{x}\right)+5y\left(y-\sqrt{x}\right)=\left(y-\sqrt{x}\right)\left(6y+\sqrt{x}\right)\)
\(x-2\sqrt{x-1}-a^2=x-1-2\sqrt{x-1}+1-a^2=\left(\sqrt{x-1}-1\right)^2-a^2=\left(\sqrt{x-1}-1-a\right)\left(\sqrt{x-1}-1+a\right)\)
ảnh nhỏ quá