\(x^2-y=y^2-x\Leftrightarrow x^2-y^2+x-y=0\Leftrightarrow\left[{}\begin{matrix}x=y\left(l\right)\\x+y+1=0\left(nh\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}A=\left(x+y\right)\left(x^2-xy+y^2\right)-3xy\left(x^2+y^2\right)+6x^2y^2\left(x+y\right)\\x+y=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}A=-\left(x^2-xy+y^2\right)-3xy\left(x^2+y^2\right)-6x^2y^2\\x+y=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}A=-\left(1-3xy\right)-3xy\left(1-2xy\right)-6x^2y^2\\x^2+y^2=1-2xy\end{matrix}\right.\)

\(\Leftrightarrow A=-1+3xy-3xy+6x^2y^2-6x^2y^2=-1\)

x = 9

x = 2

câu 3 mình ko hỉu

\(B=\frac{\sqrt{x}}{\sqrt{x}-1}:\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)

\(B=\frac{\sqrt{x}}{\sqrt{x}-1}:\left(\frac{\sqrt{x}-1+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)

\(B=\frac{\sqrt{x}}{\sqrt{x}-1}:\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(B=\frac{\sqrt{x}}{\sqrt{x}-1}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+1}=\sqrt{x}\)

Kudo thik

Hả?

Gọi H là trung điểm BC

\(\left\{{}\begin{matrix}\left(SBC\right)\perp\left(ABC\right)\\\left(SBC\right)\cap\left(ABC\right)=BC\\SH\perp BC\end{matrix}\right.\Rightarrow SH\perp\left(ABC\right)\)

\(\left\{{}\begin{matrix}\widehat{\left[\left(SAB\right);\left(ABC\right)\right]=60^O}\\\left(SAB\right)\cap\left(ABC\right)=AB\\HN\perp AB\end{matrix}\right.\Rightarrow\widehat{SHN}=60^O\)

\(SH=\sqrt{SB^2-BH^2}=\sqrt{\left(\frac{\sqrt{3}a}{2}\right)^2-\left(\frac{a}{2}\right)^2}=\frac{a\sqrt{2}}{2}\)

\(HN=cot60.SH=\frac{\sqrt{3}}{3}.a\frac{\sqrt{2}}{2}=\frac{a\sqrt{6}}{6}\)

Xét tam giácABC ta có:

H là trung điểm BC

HN//AC

Suy ra HN là đường trung bình của tam giác ABC

\(\Rightarrow AC=2HN=\frac{a\sqrt{6}}{3}\)

\(AB=\sqrt{BC^2-AC^2}=\sqrt{a^2-\frac{2}{3}a^2}=\frac{a\sqrt{3}}{3}\)

\(V=\frac{1}{3}SH.S_{ABC}=\frac{1}{3}SH.\frac{1}{2}AB.AC=\frac{1}{6}.a\frac{\sqrt{2}}{2}.a\frac{\sqrt{6}}{6}.a\frac{\sqrt{6}}{3}=\frac{a^3\sqrt{2}}{36}\)

\(\frac{\sqrt{\left(\sqrt{5}\right)^2-2.\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2}}{\sqrt{10}-\sqrt{6}}\)

\(\Leftrightarrow\frac{\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}{\sqrt{10}-\sqrt{6}}=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)}\)

\(\Leftrightarrow\frac{1}{\sqrt{2}}\)

\(2018^{2\left(x^2-y+1\right)}=\frac{2x+y}{x^2+2x+1}\)

\(\Leftrightarrow2\left(x^2-y+1\right)=log_{2018}\left(\frac{2x+y}{x^2+2x+1}\right)\)

\(\Leftrightarrow2\left(x^2+2x+1-2x-y\right)=log_{2018}\left(2x+y\right)-log_{2018}\left(x^2+2x+1\right)\)

\(\Leftrightarrow2\left(x^2+2x+1\right)+log_{2018}\left(x^2+2x+1\right)=log_{2018}\left(2x+y\right)+2\left(2x+y\right)\)

Đặt \(f\left(u\right)=log_{2018}u+2u\)

\(\begin{matrix}x^2+2x+1>0\\2x+y>0\end{matrix}\Rightarrow u>0\)

\(f'\left(u\right)=\frac{1}{u.ln2018}+2>0\)

Suy ra hàm số đồng biến

\(\Leftrightarrow f\left(x^2+2x+1\right)=f\left(2x+y\right)\)\(\Leftrightarrow x^2+2x+1=2x+y\) (tính chất hàm đồng biến)

\(\Leftrightarrow y=x^2+1\)

\(P=2y-3x=2x^2-3x+2\)

\(P=2\left(x-\frac{3}{4}\right)^2+\frac{7}{8}\)

\(P_{min}=\frac{7}{8}\) khi \(x=\frac{3}{4}\)

11

cho minh minh cho