Xét tam giác \(ABC\)vuông tại \(A\)đường cao \(AH\): 

\(\frac{1}{AH^2}=\frac{1}{AB^2}+\frac{1}{AC^2}=\frac{1}{6^2}+\frac{1}{8^2}\Rightarrow AH=4,8\left(cm\right)\).

\(BC^2=AB^2+AC^2\)(định lí Pythagore) 

\(=6^2+8^2=100\)

\(\Rightarrow BC=10\left(cm\right)\)

\(HC=\frac{AC^2}{BC}=\frac{8^2}{10}=6,4\left(cm\right)\)

\(HB=BC-HC=10-6,4=3,6\left(cm\right)\)

Xét tam giác \(AHB\)vuông tại \(H\)đường cao \(HQ\): 

\(AQ=\frac{AH^2}{AB}=\frac{4,8^2}{6}=3,84\left(cm\right)\)

Xét tam giác \(ACQ\)vuông tại \(A\): 

\(CQ^2=AC^2+AQ^2=8^2+3,84^2\Rightarrow CQ=\frac{8\sqrt{769}}{25}\left(cm\right)\)

\(a,A=\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-3}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)

\(A=\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}:\frac{x-9-x+4+\sqrt{x}+2}{x-5\sqrt{x}+6}\)

\(A=\frac{1}{\sqrt{x}+1}:\frac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

\(A=\frac{\sqrt{x}-2}{\sqrt{x}+1}=\frac{\sqrt{x}+1-3}{\sqrt{x}+1}=1-\frac{3}{\sqrt{x}+1}\)

\(< =>b,3⋮\sqrt{x}+1\)

lập bảng thì ra đc

\(x=4;0\)

??? ở đâu

\(x=\sqrt{2+\sqrt{\frac{5+\sqrt{5}}{2}}}+\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}\)

\(\Leftrightarrow x^2=4+2\sqrt{\frac{3-\sqrt{5}}{2}}\)

\(\Leftrightarrow x^2=4+\sqrt{6-2\sqrt{5}}\)

\(\Leftrightarrow x^2=4+\sqrt{5}-1\)

\(\Leftrightarrow x^2=3+\sqrt{5}\)

\(\Leftrightarrow x^2=\frac{6+2\sqrt{5}}{2}\)

\(\Leftrightarrow x=\frac{\sqrt{5}+1}{\sqrt{2}}\)

Ta có:

\(P=\left(x-\sqrt{3-\sqrt{5}}-\sqrt{2}+1\right)^{2019}\)

\(=\left(\frac{\sqrt{5}+1}{\sqrt{2}}-\sqrt{3-\sqrt{5}}-\sqrt{2}+1\right)^{2019}\)

\(=\left(\frac{\sqrt{5}+1-\sqrt{6-2\sqrt{5}}-2+\sqrt{2}}{\sqrt{2}}\right)^{2019}\)

\(=\left(\frac{\sqrt{5}+1-\left(\sqrt{5}-1\right)-2+\sqrt{2}}{\sqrt{2}}\right)^{2019}\)

\(=\left(1\right)^{2019}=1\)   

Ta có x³ < y³ (1)

Lại có (x + 2)³ = x³ + 6x² + 12x + 8 = (x³ + x² + x + 1) + (5x² + 11x + 7) 

=  \(y^3+5\left(x+\frac{11}{10}\right)^2+\frac{19}{20}\)

Nhận thấy \(5\left(x+\frac{11}{10}\right)^2+\frac{19}{20}>0\)

=> \(y^3+5\left(x+\frac{11}{10}\right)^3+\frac{19}{20}>y^3\)

=> \(\left(x+2\right)^3>y^3\)(2)

Từ (1) và (2) => y = (x + 1)³ 

Khi đó 1 + x + x² + x³ = x³ + 3x² + 3x + 1 

<=> 2x² + 2x = 0

<=> 2x(x + 1) = 0

<=> \(\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

Khi x = 0 => y = 1

Khi x = -1 => y = 0

Vậy các cặp (x ; y) thỏa mãn là (0;1) ; (-1;0) 

\(A=\frac{9\left(a+b+c\right)^2}{ab+bc+ca}+\frac{a^2+b^2+c^2}{ab+bc+ca}\)

\(A=\frac{9\left(a^2+b^2+c^2+2ab+2bc+2ca\right)}{ab+bc+ca}+\frac{a^2+b^2+c^2}{ab+bc+ca}\)

\(A=\frac{9a^2+9b^2+9c^2+18ab+18bc+18ca}{ab+bc+ca}+\frac{a^2+b^2+c^2}{ab+bc+ca}\)

\(A=\frac{9a^2+9b^2+9c^2+18ab+18bc+18ca+a^2+b^2+c^2}{ab+bc+ca}\)

dễ thấy \(9a^2+9b^2+9c^2\ge9ab+9bc+9ca\)(bđt tương đương)

\(a^2+b^2+c^2\ge ab+bc+ca\)

\(A\ge\frac{28ab+28bc+28ca}{ab+bc+ca}=28\)dấu "=" xảy ra khi và chỉ khi \(a=b=c=1\)

\(< =>MIN:A=28\)