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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.
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:
\(\left\{{}\begin{matrix}\Delta>0\\S>0\\P>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(2m+3\right)^2-4\left(m^2-1\right)>0\\2m+3>0\\m^2-1>0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}-\frac{13}{12}< m< -1\\m>1\end{matrix}\right.\)
\(f\left(x\right)=x^2-\left(2m+3\right)x+m^2-1\)
\(f\left(x\right)=x^2-2\left(m+\frac{3}{2}\right)x+\left(m+\frac{3}{2}\right)^2-3m-\frac{5}{4}\)
\(f\left(x\right)=\left(x-m-\frac{3}{2}\right)^2-3m-\frac{5}{4}\ge-3m-\frac{5}{4}\)
\(\Rightarrow-3m-\frac{5}{4}=\frac{2017}{4}\Rightarrow-3m=\frac{1011}{2}\Rightarrow m=-\frac{337}{2}\)
Có : \(x-2y-\sqrt{xy}+\sqrt{x}-2\sqrt{y}=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)+\sqrt{x}-2\sqrt{y}=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}+1\right)=0\)
\(\Leftrightarrow\sqrt{x}=2\sqrt{y}\) (Do \(\sqrt{x}+\sqrt{y}+1>0,\forall x;y>0\))
\(\Leftrightarrow x=4y\)
Khi đó \(P=\dfrac{7y}{\left(2\sqrt{y}+3\sqrt{y}\right).\left(\sqrt{x}+2\sqrt{y}\right)}\)
\(=\dfrac{7y}{5\sqrt{y}.4\sqrt{y}}=\dfrac{7}{20}\)
\(A=\dfrac{\sqrt{20}-6}{\sqrt{14-6\sqrt{5}}}-\dfrac{\sqrt{20}-\sqrt{28}}{\sqrt{12-2\sqrt{35}}}=\dfrac{-2\left(3-\sqrt{5}\right)}{\sqrt{\left(3-\sqrt{5}\right)^2}}+\dfrac{2\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}}\)
\(=\dfrac{-2\left(3-\sqrt{5}\right)}{3-\sqrt{5}}+\dfrac{2\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}=-2+2=0\)
\(B=\sqrt{\dfrac{\left(9-4\sqrt{3}\right)\left(6-\sqrt{3}\right)}{\left(6-\sqrt{3}\right)\left(6+\sqrt{3}\right)}}-\sqrt{\dfrac{\left(3+4\sqrt{3}\right)\left(5\sqrt{3}+6\right)}{\left(5\sqrt{3}-6\right)\left(5\sqrt{3}+6\right)}}\)
\(=\sqrt{\dfrac{66-33\sqrt{3}}{33}}-\sqrt{\dfrac{78+39\sqrt{3}}{39}}=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}\right)=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{3}-1-\sqrt{3}-1\right)=-\sqrt{2}\)
a) Ta có: \(A=\dfrac{\sqrt{10}-3\sqrt{2}}{\sqrt{7-3\sqrt{5}}}-\dfrac{\sqrt{10}-\sqrt{14}}{\sqrt{6-\sqrt{35}}}\)
\(=\dfrac{2\sqrt{5}-6}{3-\sqrt{5}}-\dfrac{2\sqrt{5}-2\sqrt{7}}{\sqrt{7}-\sqrt{5}}\)
\(=\dfrac{\left(2\sqrt{5}-6\right)\left(3+\sqrt{5}\right)}{4}-\dfrac{\left(2\sqrt{5}-2\sqrt{7}\right)\left(\sqrt{7}+\sqrt{5}\right)}{2}\)
\(=\dfrac{\left(\sqrt{5}-3\right)\left(3+\sqrt{5}\right)-\left(2\sqrt{5}-2\sqrt{7}\right)\left(\sqrt{7}+\sqrt{5}\right)}{2}\)
\(=\dfrac{5-9-2\left(5-7\right)}{2}\)
\(=\dfrac{-4-2\cdot\left(-2\right)}{2}\)
\(=0\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
Lời giải:
$\frac{1}{x}+\frac{1}{y}+\frac{2}{x+y}=\frac{x+y}{xy}+\frac{2}{x+y}$
$=x+y+\frac{2}{x+y}$
$=\frac{x+y}{2}+\frac{x+y}{2}+\frac{2}{x+y}$
$\geq \frac{x+y}{2}+2\sqrt{\frac{x+y}{2}.\frac{2}{x+y}}$ (áp dụng BDT Cô-si)
$\geq \frac{2\sqrt{xy}}{2}+2=\frac{2}{2}+2=3$
Vậy ta có đpcm
Dấu "=" xảy ra khi $x=y=1$
\(\Leftrightarrow x^2+\left(y+4\right)x-6y^2+12y-30=0\)
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.
rút gọn sai rồi bạn cái khúc cuối là -26 chứ k phải -30