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ta có B=-(\(\left(x^6-16x^3\right)=-\left(x^6-16x^3+64\right)+64=64-\left(x^3-8\right)^2\le64\)
dấu = xảy ra ,=> x=2
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\(\Leftrightarrow x^2+\left(y+4\right)x-6y^2+12y-30=0\)
Coi đây là pt bậc 2 ẩn x, y là tham số.
Giải pt, ta đc: \(x=\frac{-y-4\pm\sqrt{25y^2-40y+136}}{2}\)
Từ đó thay ngc vào pt để tìm x,y.
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\(x\ge-\frac{2}{3}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{3x+4}=a>0\\\sqrt{3x+2}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^2-b^2=2\)
Pt trở thành:
\(\left(a-b\right)\left(1+ab\right)=a^2-b^2\)
\(\Leftrightarrow\left(a-b\right)\left(1+ab\right)-\left(a-b\right)\left(a+b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(1+ab-a-b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left[a\left(b-1\right)-\left(b-1\right)\right]=0\)
\(\Leftrightarrow\left(a-b\right)\left(a-1\right)\left(b-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=1\\b=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x+4}=\sqrt{3x+2}\\\sqrt{3x+4}=1\\\sqrt{3x+2}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}4=2\left(vn\right)\\x=-1< -\frac{2}{3}\left(l\right)\\x=-\frac{1}{3}\end{matrix}\right.\)
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\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-2xy=13\\xy-\left(x+y\right)=1\end{matrix}\right.\)
Đặt S=x+y; P=xy\(\left(S^2\ge4P\right)\)
\(\Rightarrow\left\{{}\begin{matrix}S^2-2P=13\\P-S=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}S^2-2S-15=0\\P=S+1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}S=5\\P=6\end{matrix}\right.\\\left\{{}\begin{matrix}S=-3\\P=-2\end{matrix}\right.\end{matrix}\right.\)
x,y là nghiệm của pt:\(\left[{}\begin{matrix}X^2-5X+6=0\\X^2+3X-2=0\end{matrix}\right.\)
Đến đây tự giải ra nha.
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2 ) Ta có :
\(f\left(x\right)=x^2-\left(2m+3\right)x+m^2-1\ge\frac{2017}{4}\)
\(\Leftrightarrow x^2-\left(2m+3\right)x+m^2-\frac{2021}{4}\ge0\)
Hiển nhiên dấu bằng sẽ xảy ra
\(\Delta=\left(2m+3\right)^2-4\left(m^2-\frac{2021}{4}\right)=0\)
\(\Leftrightarrow4m^2+12m+9-4m^2+2021=0\)
\(\Leftrightarrow12m+2030=0\)
\(\Leftrightarrow m=-\frac{1015}{6}\)
Để pt có 2 nghiệm dương phân biệt:
\(\left\{{}\begin{matrix}\Delta>0\\S>0\\P>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(2m+3\right)^2-4\left(m^2-1\right)>0\\2m+3>0\\m^2-1>0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}-\frac{13}{12}< m< -1\\m>1\end{matrix}\right.\)
\(f\left(x\right)=x^2-\left(2m+3\right)x+m^2-1\)
\(f\left(x\right)=x^2-2\left(m+\frac{3}{2}\right)x+\left(m+\frac{3}{2}\right)^2-3m-\frac{5}{4}\)
\(f\left(x\right)=\left(x-m-\frac{3}{2}\right)^2-3m-\frac{5}{4}\ge-3m-\frac{5}{4}\)
\(\Rightarrow-3m-\frac{5}{4}=\frac{2017}{4}\Rightarrow-3m=\frac{1011}{2}\Rightarrow m=-\frac{337}{2}\)
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\(\left\{{}\begin{matrix}x^3+6x^2y=7\\8y^3+12xy^2=20\end{matrix}\right.\) \(\Leftrightarrow\left(x+2y\right)^3=27\Rightarrow x+2y=3\Rightarrow x=3-2y\)
Thay vào pt sau:
\(2y^3+3\left(3-2y\right)y^2-5=0\)
\(\Leftrightarrow4y^3-9y^2+5=0\)
\(\Leftrightarrow\left(y-1\right)\left(4y^2-5y-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}y=1\Rightarrow x=1\\y=\frac{5+\sqrt{105}}{8}\Rightarrow x=\frac{7-\sqrt{105}}{4}\\y=\frac{5-\sqrt{105}}{8}\Rightarrow x=\frac{7+\sqrt{105}}{4}\end{matrix}\right.\)
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AB=13cm
\(\widehat{C}=45^0\)
\(BC=13\sqrt{2}\left(cm\right)\)
a, Với \(x\ge0;x\ne25\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-5}-\frac{10\sqrt{x}}{x-25}-\frac{5}{\sqrt{x}+5}\)
\(=\frac{x+5\sqrt{x}-10\sqrt{x}-5\sqrt{x}+25}{x-25}=\frac{x-10\sqrt{x}+25}{x-25}=\frac{\sqrt{x}-5}{\sqrt{x}+5}\)
b, Thay x = 121 => \(\sqrt{x}=11\)ta được : \(\frac{11-5}{11+5}=\frac{6}{16}=\frac{3}{8}\)
c, Ta có : \(A=\left|\frac{\sqrt{x}-5}{\sqrt{x}+5}\right|=\frac{1}{4}\)
TH1 : \(\frac{\sqrt{x}-5}{\sqrt{x}+5}=\frac{1}{4}\Rightarrow4\sqrt{x}-20=\sqrt{x}+5\Leftrightarrow3\sqrt{x}=25\Leftrightarrow\sqrt{x}=\frac{25}{3}\Leftrightarrow x=\frac{625}{9}\)
TH2 : \(\frac{\sqrt{x}-5}{\sqrt{x}+5}=-\frac{1}{4}\Rightarrow4\sqrt{x}-20=-\sqrt{x}-5\Leftrightarrow5\sqrt{x}=15\Leftrightarrow x=9\)