a) \(x^2-64=0\)

\(\Leftrightarrow x^2-8^2=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x+8=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)

b) \(4x^2-4x+1=0\)

\(\Leftrightarrow\left(2x\right)^2-2.2x.1+1=0\)

\(\Leftrightarrow\left(2x-1\right)^2=0\)

\(\Leftrightarrow2x-1=0\) \(\Leftrightarrow x=\dfrac{1}{2}\)

c) \(9-6x+x^2=0\)

\(\Leftrightarrow3^2-2.3.x+x^2=0\)

\(\Leftrightarrow\left(3-x\right)^2=0\)

\(\Leftrightarrow3-x=0\Leftrightarrow x=3\)

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2. I'm going to the cinema on Monday

1. We visit our gandparents every Sunday

2. Broken bones in adults don't heal as fast as they do in children

\(\left(1\right)\)    \(2Fe+3Cl_2\underrightarrow{t^0}2FeCl_3\)

\(\left(2\right)\)  \(FeCl_3+3NaOH\rightarrow3NaCl+Fe\left(OH\right)_3\)

\(\left(3\right)\)   \(2Fe\left(OH\right)_3\underrightarrow{t^0}Fe_2O_3+3H_2O\)

\(\left(4\right)\)   \(Fe_2O_3+3CO\rightarrow2Fe+3CO_2\)

Áp dụng định lí Py - ta - go cho △ ABH ta có:

AB² = AH² + BH²

⇒ AH² = AB² - BH² = 10² - 5² = 75

⇒ AH = \(5\sqrt{3}\)

Tam giác ABC vuông tại A có đường kính AH nên theo tính chất đường cao trong tam giác vuông ta có:

AB² = BH . HC ⇒ BC = \(\dfrac{AB^2}{BH}=\dfrac{10^2}{5}=20\left(cm\right)\)

⇒ HC = 20 - 5 = 15 ( cm )

Áp dụng định lí Py - ta - go ta có:

BC² = AB² + AC² ⇒  AC² = 20² - 10² = 300

⇒  AC = \(10\sqrt{3}\)

b) Xét △ ABH vuông tại H ta có:

tan B = \(\dfrac{AH}{BH}=\dfrac{5\sqrt{3}}{5}=\sqrt{3}\)

Xét △ ACH vuông tại H ta có:

tan C = \(\dfrac{AH}{HC}=\dfrac{5\sqrt{3}}{15}=\dfrac{\sqrt{3}}{3}\)

⇒ tan B = 3 tan C

\(\sqrt{\dfrac{3\sqrt{3}-4}{2\sqrt{3}+1}}+\sqrt{\dfrac{\sqrt{3}+4}{5-2\sqrt{3}}}\)

\(=\sqrt{\dfrac{\left(3\sqrt{3}-4\right)\left(2\sqrt{3}-1\right)}{11}}-\sqrt{\dfrac{\left(\sqrt{3}+4\right)\left(5+2\sqrt{3}\right)}{13}}\)

\(=\sqrt{\dfrac{22-11\sqrt{3}}{11}}-\sqrt{\dfrac{26+13\sqrt{3}}{13}}\)

\(=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}\right)\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}\right)\)

\(=\dfrac{1}{\sqrt{2}}\left[\sqrt{3}-1-\left(\sqrt{3}+1\right)\right]\)

\(=-\sqrt{2}\)

Đặt \(a=\sqrt{x-1}\ge0\)

       \(b=\sqrt{6-x}\ge0\)

\(\Rightarrow a^2+b^2=x-1+6-x=5\)

Phương trình tương đương: 

\(a+b+ab=1\)

\(\Rightarrow a+b=1-ab\)

Lại có:   \(a^2+b^2=\left(a+b\right)^2-2ab=5\)

\(\Leftrightarrow\left(1-ab\right)^2-2ab=5\)

\(\Leftrightarrow\left(ab\right)^2-2ab+1-2ab=5\)

\(\Leftrightarrow\left(ab\right)^2-4ab-4=0\)

\(\Leftrightarrow\left(ab-2\right)^2\)  \(\Rightarrow ab=2\Rightarrow b=\dfrac{2}{a}\)

\(\Leftrightarrow a^2+b^2=5\Leftrightarrow a^2+\dfrac{4}{a^2}=5\)

\(\Leftrightarrow a^4-5a^2+4=0\)

\(\Leftrightarrow a^4-4a^2-a^2+4=0\)

\(\Leftrightarrow a^2\left(a^2-4\right)\left(a^2-4\right)=0\)

\(\Leftrightarrow\left(a^2-4\right)\left(a^2-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a^2-4=0\\a^2-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a^2=4\\a^2=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=2\\\sqrt{x-1}=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=4\\x-1=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)

Đặt \(t=\sqrt{x^2+1}\Rightarrow x^2+1=t^2\)

         \(\Leftrightarrow\left(4x-1\right)t=2t^2+2x-1\)

         \(\Leftrightarrow2t^2-\left(4x-1\right)t+2x-1=0\)

\(\Delta=\left(4x-1\right)^2-8\left(2x-1\right)=16x^2-24x+9=\left(4x-3\right)^2\)

Do đó ta được:

\(t=\dfrac{4x-1+4x+3}{4}=2x-1\)

hay \(t=\dfrac{4x-1-4x+3}{4}=\dfrac{1}{2}\) ( loại )

Với \(t=2x-1\Leftrightarrow\sqrt{x^2+1}=2x-1\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-1\ge0\\x^2+1=4x^2-4x+1\end{matrix}\right.\) 

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\3x^2-4x=0\end{matrix}\right.\) \(\Leftrightarrow x=\dfrac{4}{3}\)

Vậy phương trình có một nghiệm là \(x=\dfrac{4}{3}\)